Maths Class 12 Integrals, R.D Sharma Question Answer

Page 19.4 Ex.19.1

Q1.

Answer :

(i)
∫x4dx=x4+14+1+C =x55+C

(ii)
∫x54dx=x54+154+1+C=49×94+C

(iii)
∫x-5dx=x-5+1-5+1+C=-14x-4+C=-14×4+C

(iv)
∫dxx3/2=∫x-3/2dx=x-32+1-32+1+C=x-12-12+C=-2x+C

(v)
∫3xdx=3xln 3+C

(vi)
∫dxx23=∫dxx2/3=∫x-2/3 dx=x-23+1-23+1+C=3×13+C

(vii)
∫32 log3xdx=∫3log3 x2dx=∫x2dx=x33+C

(viii)
∫logxx dx=∫1·dx=x+C

Q2.

Answer :

(i)
∫1+cos 2x2dx ∫2 cos2x2dx ∴1+cos2A=2cos2A=∫cosx dx=sin x+C

(ii)
∫1-cos 2x2dx =∫2sin2x2dx ∴1-cos 2x=2sin2x=∫sinx dx=-cos x+C

Q3.

Answer :

∫e6 log x-e5 logxe4 logx -e3 logxdx=∫elogx6-elog x5elogx4-elogx3dx=∫x6-x5x4-x3dx=∫x5x3dx=∫x2dx=x33+C

Q4.

Answer :

∫dxaxbx=∫a-x b-xdx=a-xb-x- loge ab+C

Q5.

Answer :

(i)
∫cos 2x+2 sin2xsin2xdx=∫1-2 sin2x+2 sin2xsin2xdx ∵cos 2x=1-2 sin2x=∫cosec2x dx=-cot x+C

(ii)
∫2 cos2x-cos 2xcos2xdx=∫2 cos2x- 2 cos2x-1cos2xdx ∵cos 2x=2 cos2x-1=∫sec2x dx=tanx+C

Q6.

Answer :

∫elogxxdx=∫xxdx=∫1x dx=∫x-12dx=x-12+1-12+1+C=2x+C

Page 19.14 Ex.19.2

Q1.

Answer :

∫3xx+4x+5dx=∫3×1·x12+4×12+5dx=3∫x32dx+4∫x12dx+5∫dx=3×32+132+1+4×12+112+1+5x+C=3×25×52+4×23×32+5x+C=65×52+83×32+5x+C.

Q2.

Answer :

∫2x+5x-1x13dx=∫2xdx+5 ∫dxx-∫dxx13=∫2xdx+5 ∫dxx-∫x-13dx=2xln 2+5 ln x-x-13+1-13+1+C=2xln 2+5 ln x-32×23+C

Q3.

Answer :

∫x ax2+bx+cdx=∫x12ax2+bx+cdx=∫ax2+12+bx12+1+c x12dx=a∫x52 dx+b∫x32dx+c∫x12dx=ax52+152+1+bx32+132+1+cx12+112+1+C=2a7x72+2b5x32+2c3x32+C

Q4.

Answer :

∫2-3x 3+2×1-2xdx=∫2-3x 3-6x+2x-4x2dx=∫2-3x -4×2-4x+3dx=∫-8×2-8x+6+12×3+12×2-9xdx=∫12×3+4×2-17x+6dx=12×44+4×33-17×22+6x+C=3×4+43×3-172×2+6x+C

Q5.

Answer :

∫mx+xm+mx+xm+mxdx=m∫1xdx+1m∫xdx+∫mxdx+∫xmdx+m∫xdx=mlnx+1mx1+11+1+mxln m+ xm+1m+1+mx1+11+1=m ln x+x22m+mxln m+ xm+1m+1+mx22+C

Q6.

Answer :

∫x-1x2dx=∫x+1x-2dx=∫xdx+∫dxx-2∫dx=x22+ln x-2x+C

Q7.

Answer :

∫1+x3xdx=∫1+x3+3 12x+31x2xdx=∫1+x3+3x+3x2x dx=∫1x+x3x+3xx+3x2xdx=∫x-12+x52+3×12+3x32dx=x-12+1-12+1+x52+152+1+3×12+112+1+3×32+132+1+C=2x+27×72+2×32+65×52+C

Q8.

Answer :

∫x2+elog x+e2xdx=∫x2dx+∫xdx+∫e2xdx=x33+x22+e2xln e2+C

Q9.

Answer :

∫xe+ex+eedx=∫xedx+∫exdx+ee∫1dx=xe+1e+1+ex+x·ee+C

Q10.

Answer :

∫x x3-2xdx=∫x72-2xdx=∫x72-2x-12 dx=x72+172+1-2x-12+1-12+1+C=29×92-4×12+C=29×92-4x+C

Q11.

Answer :

∫1×1+1xdx=∫x-121+1xdx=∫x-12+1x32dx=∫x-12dx+∫x-32dx=x-12+1-12+1+x-32+1-32+1=2x-2x+C

Q12.

Answer :

∫ x6+1×2+1dx=∫ x23+13×2+1dx A3+B3=A+B A2-AB+B2=∫x2+1×4-x2+1×2+1dx=∫x4-x2+1dx=∫x4dx+∫x2dx+∫1dx=x4+14+1-x2+12+1+x+C=x55-x33+x+C

Page 19.15 Ex.19.2

Q13.

Answer :

∫ x-13+x+2x13dx=∫ x-13×13+x12x13+2x13dx=∫x-23+x16+2x-13dx=x-23+1-23+1+x16+116+1+2x-13+1-13+1=x1313+x7676+3×23+C=3×13+67×76+3×23+C

Q14.

Answer :

∫ 1+x2xdx=∫ 1x+xx+2xxdx=∫x-12+x12+2dx=x-12+1-12+1+x12+112+1+2x+C=2x+23×32+2x+C

Q15.

Answer :

∫x3-5xdx=∫ x123-5xdx=∫3×12-5x32dx=3×12+112+1-5×32+132+1+C=2×32-2×52+C

Q16.

Answer :

∫x+1x-2xdx=∫ x2-2x+x-2xdx=∫x2-x-2xdx=∫x32-x12-2x-12dx=x32+132+1-x12+112+1-2x-12+1-12+1+C=25×52-23×32-4×12+C

Q17.

Answer :

∫x5+x-2+2x2dx=∫ x5x2+x-2×2+2x2dx=∫x3+x-4+2x-2dx=x3+13+1+x-4+1-4+1+2x-2+1-2+1+C=x44-13×3-2x+C

Q18.

Answer :

∫3x+42dx=∫ 9×2+2×3x×4+16dx=9∫x2dx+24∫x dx+16∫dx=9×33+24×22+16x+C=3×3+12×2+16x+C

Q19.

Answer :

∫2×4+7×3+6x2x2+2xdx=∫ x22x2+7x+6xx+2dx=∫x2x2+4x+3x+6x+2dx=∫x2xx+2+3x+2x+2dx=∫x2x+3x+2x+2dx=∫2×2+3xdx=2∫x2dx+3∫x dx=2×33+3×22+C=23×3+32×2+C

Q20.

Answer :

∫5×4+12×3+7x2x2+xdx=∫ x25x2+12x+7xx+1dx=∫x5x2+5x+7x+7x+1dx=∫x5xx+1+7x+1x+1dx=∫x5x+7x+1x+1dx=∫5×2+7xdx=5×33+7×22+C

Q21.

Answer :

∫3 sin x-4 cos x+5cos2x-6sin2x+tan2x-cot2xdx=∫ 3 sin x-4 cos x+5 sec2x-6 cosec2x+sec2x-1-cosec2x-1 dx=∫3 sin x-4 cos x+6 sec2x-7 cosec2xdx=-3 cos x-4 sin x+6 tan x-7 -cot x +C=-3 cos x-4 sin x+6 tan x+7 cot x+C

Q22.

Answer :

∫sec2x+cosec2xdx=∫sec2x dx+∫cosec2x dx=tan x-cot x+C

Q23.

Answer :

∫sin3 x-cos3 xsin2 x · cos2 xdx=∫sin3xsin2 x · cos2xdx-∫cos3 xsin2 x · cos2 xdx=∫sin xcos2 xdx-∫cos xsin2 xdx=∫sin xcos x×1cos xdx-∫cos xsin x×1sin xdx=∫sec x tan x dx-∫cosec x cot x dx=sec x–cosec x+C=sec x+cosec x+C

Q24.

Answer :

∫5 cos3x+6 sin3x2 sin2x cos2xdx=∫5 cos3x2 sin2x cos2x+6 sin3x2sin2x cos2xdx=∫52 cos xsin2x+3sin xcos2xdx=52∫cos xsin x×1sin xdx+3∫sin xcos x×1cos xdx=52∫cosec x cot x dx+3∫sec x tan x dx=52-cosec x+3 sec x+C=-52cosec x+3 sec x+C

Q25.

Answer :

∫tan x+cot x2=∫tan2x+cot2x+2 tan x cot xdx=∫tan2x+cot2x+2dx=∫sec2x-1+cosec2x-1+2dx=∫sec2x+cosec2x dx=tan x -cot x+C

Q26.

Answer :

∫1-cos 2×1+cos 2xdx=∫2 sin2x2 cos2xdx ∵1-cos 2θ=2 sin2 θ & 1+cos 2θ=2 cos2θ=∫tan2x dx =∫sec2x-1 dx=∫sec2x dx-∫dx=tan x -x+C

Q27.

Answer :

∫cot xcosec x-cot xdx=∫cos xsin x1sin x-cos xsin xdx=∫cos x1-cos x×1+cos x1+cos xdx=∫cos x+cos2x1-cos2xdx=∫cos x+cos2xsin2x dx=∫cos xsin x×1sin x+cos2xsin2xdx=∫cot x cosec x+cot2xdx=∫cosec x cot x+cosec2x-1dx=-cosec x-cot x-x+C

Q28.

Answer :

∫ cos2x-sin2x1+cos 4xdx=∫cos 2×2 cos2 2xdx ∴1+cos A=2 cos2 A2 & cos2A-sin2A=cos2A=12∫cos 2xcos 2xdx=12x+C=x2+C

Q29.

Answer :

∫dx1-cos x=∫dx1-cos x×1+cosx1+cosx=∫1+cos x1-cos2xdx=∫1+cos xsin2xdx=∫1sin2x+cos xsin x×1sin xdx=∫cosec2x+cosec x cot xdx=-cot x-cosec x+C

Q30.

Answer :

∫dx1-sin x=∫1+sin x1-sin x×1+sin xdx=∫1+sin x1-sin2xdx=∫1+sin xcos2xdx=∫1cos2x+sin xcos x×1cos xdx=∫sec2x+sec x tan xdx=tan x+sec x+C

Q31.

Answer :

∫tan xsec x+tan xdx=∫tan xsec x+tan x×sec x-tan xsec x-tan xdx=∫tan x sec x-tan xsec2x-tan2xdx=∫sec x tan x-tan2x1dx=∫sec x tan x dx-∫sec2x-1dx=sec x-tan x+x+C

Q32.

Answer :

∫cosec xcosec x-cot xdx=∫cosec xcosec x+cot xcosec x-cot x cosec x+cot xdx=∫cosec x cosec x+cot xcosec2x-cot2xdx=∫cosec2x+cosec x cot xdx=-cot x-cosec x+C

Q33.

Answer :

∫dx1+cos 2x ∴1+cosθ=2cos2 θ2=∫dx2 cos2x=12∫sec2x dx=12tan x+C

Q34.

Answer :

∫dx1-cos 2x ∴1-cos A=2sin2 A2=∫dx2 sin2x=12∫cosec2x dx=12-cot x+C=-12cot x+C

Q35.

Answer :

∫tan-1sin 2×1+cos2xdx=∫tan-12 sin x cos x2 cos2xdx ∴sin 2x=2 sinx cosx & 1+cos2x=2 cos2x=∫tan-1 tan x=∫tan-1 tan x=∫x dx=x22+C

Q36.

Answer :

∫cos-1 sin xdx=∫cos-1cosπ2-xdx ∴sin x=cosπ2-x=∫π2-xdx=πx2-x22+C

Q37.

Answer :

∫cot-1 sin 2×1-cos 2xdx=∫cot-12 sin x cos x2 sin2xdx ∴sin 2x=2 sin x cos x & 1-cos 2x=2 sin2x=∫cot-1 cot xdx=∫x dx=x22+C

Q38.

Answer :

∫sin-1 2 tan x1+tan2xdx=∫sin-1 sin 2 xdx ∴sin 2x=2 tan x1+tan2x=2∫x dx=2 x22+C=x2+C

Q39.

Answer :

∫x3+8x-1×2-2x+4dx=∫x3+23 x-1×2-2x+4dx =∫x+2 x2-2x+4 x-1×2-2x+4 dx ∴ a3+b3=a+b a2-ab+b2=∫ x+2 x-1dx=∫x2-x+2x-2dx=∫x2+x-2dx=∫x2 dx+∫x dx-2 ∫1dx=x33+x22-2x+C

Q40.

Answer :

∫a tan x+b cot x2dx=∫a2 tan2x+b2 cot2x+2ab tan x cot xdx=a2∫tan2x dx +b2 ∫cot2x dx+2ab ∫dx=a2∫sec2x-1dx+b2∫cosec2x-1dx+2ab∫dx=a2tan x-x +b2 -cot x-x+2ab x+C=a2 tan x -b2cot x-a2+b2-2abx+C

Q41.

Answer :

∫x3-3×2+5x-7+x2ax2x2dx=∫x32x2-3x22x2+5x2x2-72×2+x2ax2x2dx=∫x2-32+52x-72x-2+ax2dx=12∫x dx-32∫dx+52∫dxx-72∫x-2 dx+12∫axdx=12×22-32x+52ln x-72 x-2+1-2+1+12axln a+C=x24-32x+52ln x+72x+ax2 ln a+C=12×22-3x+5 ln x+7x+ax ln a+C

Q42.

Answer :

∫cos x1+cos xdx=∫cos x1-cos x1+cos x1-cos xdx=∫cos x-cos2x1-cos2 xdx=∫cos x-cos2 xsin2 xdx=∫cos xsin2 x-cos2 xsin2 xdx=∫cot x cosec x – cot2 xdx=∫cot x cosec x-cosec2 x+1dx=∫cot x cosec x dx-∫cosec2 x dx+∫1dx=-cosec x+cot x+x+C

Q43.

Answer :

∫1-cos x1+cos xdx=∫1-cos x21-cos2 xdx=∫1+cos2 x-2cos xsin2 xdx=∫ 1sin2 x+cos2 xsin2 x-2cos xsin2 xdx=∫ cosec2 x+cot2 x-2cot x.cosec xdx=∫ cosec2 x+cosec2 x -1-2cot x.cosec xdx=∫ 2cosec2 x-1-2cot x.cosec xdx=∫2cosec2 x dx-∫1 dx-∫2cot x.cosec x dx=-2cot x-x+2cosec x+C=2cosec x-cot x-x+C

Q44.

Answer :

∫sin2x1+cos xdx=∫1-cos2x1+cos xdx=∫1-cos x 1+cos x1+ cos xdx=∫ 1-cos xdx=x-sin x+C

Q45.

Answer :

Let I=∫cos 2x+2 sin2xcos2xdx=∫1-2 sin2x+2 sin2xcos2x dx=∫sec2x dx=tan x+C

Q46.

Answer :

Let I=∫sin-1 cos x dx=∫sin-1 sin π2-x dx=∫π2-x dx=π2∫dx-∫x dx=π 2x-x22+C

Q47.

Answer :

f’x=x-1×2 f’x=x-x-2∫ f’xdx =∫x-x-2dx fx=x22-x-2+1-2+1+C =x22+1x+Cf1=12 Given⇒122+11+C=12⇒C=-1∴ fx=x22+1x-1

Q48.

Answer :

f’x=x+b, f1=5, f2=13 f’x=x+b∫f’xdx =∫x+bdxfx=x22+bx+C ….(i)f1=5, f2=13 GivenPuting x=1 in (i)f1=122+b1+C5=12+b+C … iiPuting x=2 in (i)f2=222+b2+C13=42+2b+C13=2+2b+C …(iii)Solving (ii) and (iii) we get,b=132 and C=-2Thus, fx=x22+132x-2

Q49.

Answer :

f’x=8×3-2x f2=8f’x=8×3-2x∫f’xdx =∫8×3-2xdx =8∫x3dx-2∫xdxfx=8 x44-2×x22+Cfx=2×4-x2+Cf2=8 Givenf2=2×24-22+C8=32-4+CC=-20∴fx=2×4-x2-20

Q50.

Answer :

f’x=a sinx+b cos xf’0=4, f0=3fπ2=5f’x=a sin x+b cos x∫f’xdx=∫a sin x+b cos xdxfx=-a cos x+b sin x +C …(i)Now puting x=0 in equation (i)f0=-a cos 0+b sin 0 +C3=-a×1+b×0 + C3=-a+C … iiNow puting x=π2 in equation (i)fπ2=-a cos π2+b sin π2 +C5=-a cosπ2+b sin π2+C5=-a×0+b×1+C5=b+C … iiiWe also have f’0=4f’x=a sin x+b cos xf’0=a sin 0+b cos 04=a×0+b×14=b … ivsolving ii, iii and iv we get,b=4C=1a =-2∴ fx=2cos x+4 sin x +1

Q51.

Answer :

fx=x+1x
Integrating both sides
∫fxdx=∫x+1xdx=∫x12+x-12dx=x12+112+1+x-12+1-12+1+C=23×32+2×12+C

Page 19.35 Ex.19.3

Q1.

Answer :

∫2x-35+3x+2dx=∫2x-35dx+∫3x+212dx=2x-35+125+1+3x+212+1312+1+C=2x-3612+293x+232+C

Q2.

Answer :

∫17x-53+15x-4dx=∫7x-5-3+5x-4-12dx=7x-5-3+17-3+1+5x-4-12+15-12+1+C=7x-5-2-14+255x-412+C

Q3.

Answer :

∫12-3x+13x-2dx=∫dx2-3x+∫3x-2-12dx=ln 2-3x-3+3x-2-12+13-12+1+C=ln 2-3x-3+233x-212+C=-13ln 2-3x+233x-2+C

Q4.

Answer :

∫x+3x+14dx=∫x+1+2x+14dx=∫x+1x+14+2x+14dx=∫dxx+13+2∫dxx+14=∫x+1-3 dx+2∫x+1-4dx=x+1-3+1-3+1+2x+1-4+1-4+1+C=-12x+1-2-23x+1-3+C=-12x+12-23x+13+C

Q5.

Answer :

∫x2+5x+2x+2dx=∫x2x+2dx+5∫x dxx+2+2∫dxx+2=∫x2-4+4x+2dx+5∫x+2-2x+2dx+2∫dxx+2=∫x-2x+2x+2dx+∫4x+2dx+5∫1-2x+2dx+2∫dxx+2=∫x-2 dx+4∫dxx+2+5∫dx-10∫dxx+2+2∫dxx+2=∫x-2dx-4∫dxx+2+5∫dx=x22-2x-4 ln x+2+5x+C=x22+3x-4 ln x+2+C

Q6.

Answer :

∫x3x-2dx=∫x3-8+8x-2dx=∫x3-23x-2+8x-2dx=∫x-2×2+2x+4x-2+8x-2dx=∫x2+2x+4dx+8∫dxx-2=x33+2×22+4x+8 ln x-2+C=x33+x2+4x+8 ln x-2+C

Q7.

Answer :

∫x2+x+53x+2dx=19∫9×2+9x+453x+2dx=19∫9×2-43x+2dx+∫9x+63x+2dx+∫433x+2dx=19∫3x-23x+23x+2dx+∫33x+23x+2dx+43∫dx3x+2=19∫3x-2 dx+3∫1dx+43∫dx3x+2=193×22-2x+3x+43 3ln 3x+2+C=1932×2+x-433 ln 3x+2+C=16×2+19x-4327 ln 3x+2+C

Q8.

Answer :

∫dxx+1+x

Rationalise the denominator

=∫x+1-xx+1+xx+1-xdx=∫x+1-xx+1-xdx=∫x+112dx-∫x12dx=x+112+112+1-x12+112+1=23x+132-23×32+C

Q9.

Answer :

∫dx2x+3+2x-3
Rationalise the denominator
=∫2x+3-2x-32x+3+2x-32x+3-2x-3dx=∫2x+3-2x-32x+3-2x-3dx=16∫2x+312dx-16∫2x-312dx=162x+312+1212+1-162x-312+1212+1+C=1182x+332-2x-332+C

Q10.

Answer :
∫x+12x+3dx=12∫2x+22x+3dx=12∫2x+3-12x+3dx=12∫2x+32x+3-12x+3dx=12∫2x+3-12x+3dx=12∫2x+312dx-∫2x+3-12dx=122x+312+1212+1-2x+3-12+12-12+1+C=12132x+332-2x+312+C=162x+332-122x+312+C

Q11.

Answer :

∫2x2x+12dx=∫2x+1-12x+12dx=∫2x+12x+12-12x+12dx=∫dx2x+1-∫2x+1-2dx=log2x+12-2x+1-2+12-2+1+C=log 2x+12+2x+1-12+C=log 2x+12+122x+1+C

Q12.

Answer :

Let I=∫xx+2dx
Putting x + 2 = t
Then, x = t – 2
Difference both sides
dx = dt
Now, integral becomes
I=∫t-2tdt=∫t32-2t12dt=t32+132+1-2t12+112+1+C=25t52-43t32+C=25x+252-43x+223+C

Q13.

Answer :

Let I=∫x-1x+4dx

Putting x + 4 = t
Then, x = t – 4
Difference both sides
dx = dt
Now integral becomes,
I= ∫t-4-1tdt=∫tt-5tdt=∫t12-5t-12dt=t12+112+1-5t-12+1-12+1+C=23t32-10t+C=23x+432-10x+412+C

Q14.

Answer :

∫dxx+a+x+b=∫x+a-x-bx+a+x+bx+a-x+bdx=∫x+a-x+bx+a-x+bdx=1a-b∫x+a12-1a-b∫x+b12dx=1a-bx+a12+112+1-1a-bx+b12+112+1+C=23a-bx+a32-x+b32+C

Q15.

Answer :

∫x4+3×2+1dx=∫x4-1+1+3×2+1dx=∫x4-1×2+1dx+4∫dxx2+1=∫x2-1×2+1×2+1dx+4∫dxx2+1=∫x2-1dx+4∫dxx2+1=x33-x+4 tan-1x+C

Q16.

Answer :

∫2x+3x-12dx=∫2x-2+2+3x-12dx=∫2x-1+5x-12dx=2∫dxx-1+5∫x-1-2 dx=2 ln x-1+5x-1-2+1-2+1+C=2 ln x-1-5x-1+C

Q17.

Answer :

∫sin2 2x+5dx=∫1-cos 4x+102dx ∴sin2A=1-cos2A2=12∫1-cos 4x+10dx=12x-sin 4x+104+C=12x-sin 4x+108+C

Q18.

Answer :

∫sin3 2x+1dx=14∫3 sin 2x+1-sin 32x+1dx ∴sin 3θ=3 sinθ-4sin3θ⇒sin3θ=143sin θ-sin 3θ =34∫sin 2x+1dx-14∫sin 6x+3dx=34-cos 2x+12-14-cos 6x+36+C=-3 8cos 2x+1+124 cos 6x+3+C

Q19.

Answer :

∫cos4 2x dx=∫cos2 2x2dx=∫1+cos 4x22dx ∴cos2 x=1+cos 2×2=14∫1+cos 4×2 dx=14∫1+cos2 4x+2 cos 4xdx=14∫1+1+cos 8×2+2 cos 4xdx=14∫32+cos 8×2+2 cos 4xdx=143×2+sin 8×16+2 sin 4×4+C=3×8+sin 8×64+sin 4×8+C

Q20.

Answer :

∫sin2 bx dx=∫1-cos 2bx2dx ∴sin2 x=1-cos 2×2=12∫1-cos 2bxdx=12x-sin 2bx2b +C=x2-sin 2bx4b+C

Q21.

Answer :

∫sin2 x2 dx=∫1-cos x2dx ∴sin2 x2=1-cos x2=12∫1-cos xdx=12x-sin x +C

Q22.

Answer :

∫cos2 x2 dx=∫1+cos x2dx ∴cos2 x2=1+cos x2=12∫1+cos xdx=12x+sin x +C

Q23.

Answer :

∫sin 4x cos 7x dx=12∫2 cos 7x sin 4x dx=12∫sin 7x+4x-sin 7x-4xdx ∴2 cos A sin B = sin A+B- sin A-B=12∫sin 11x-sin 3x dx=12-cos 11×11+cos 3×3+C=-cos 11×22+cos 3×6

Q24.

Answer :

∫cos 4x cos 3x dx=12∫2 cos 4x cos 3x dx=12∫cos 4x+3x+cos 4x-3xdx ∴2 cos A cos B=cos A+B+cos A-B=12∫cos 7x+cos x dx=12sin 7×7+sin x+C=114sin 7x+12sin x+C

Q25.

Answer :

∫cos2nx dx=∫1+cos 2nx2 dx ∴cos2x=1+cos 2×2=12∫1+cos 2nxdx=12x+sin 2nx2n+C=x2+sin 2nx4n+C

Q26.

Answer :

∫cos mx cos nx dx=12∫2 cos mx cos nxdx=12∫cos mx+nx+cos mx-nxdx ∴2 cos A cos B=cos A+B+cos A-B=12sin m+nxm+n+sin m-nxm-n+C

Q27.

Answer :

∫cos x·cos 2x·cos 3x dx=12∫2 cos 3x·cos 2x cos x dx=12∫cos 3x+2x+cos 3x-2x cos x dx ∴2 cos A cos B=cos A+B+cos A-B=12∫cos 5x+cos x cos x dx=12∫cos 5x·cos x+cos2xdx=14∫2 cos 5x·cos x dx+12∫cos2x dx=14∫cos 6x+cos 4xdx+12∫1+cos 2x2dx=14∫cos 6xdx+14∫cos 4xdx+14∫1+cos 2xdx=14sin 6×6+14sin 4×4+14x+sin 2×2+C=124sin 6x+116sin 4x+14x+18sin 2x+C

Q28.

Answer :

∫sin x·1-cos 2x dx=∫sin x .2 sin2x dx ∴1-cos 2A=2sin2A=2∫sin2x dx=2∫1-cos 2x2dx=12∫1-cos 2xdx=12x-sin 2×2+C

Q29.

Answer :

∫sin x 1+cos 2x dx=∫sin x.2 cos2x dx ∴1+cos 2x=2 cos2x=2∫sin x cos x dx=22∫2 sin x cos x dx=12∫sin 2xdx=12-cos 2×2+C=-122cos 2x+C

Q30.

Answer :

∫1+cos x1-cos x dx=∫2 cos2x22 sin2x2 dx ∴1+cos x=2 cos2x2 & 1-cos x=2 sin2x2 =∫cot2x2 dx=∫cosec2 x2-1 dx=-cot x212-x+C=-2 cot x2-x+C

Q31.

Answer :

∫1-cos x1+cos x dx=∫2 sin2x22 cos2x2 dx 1-cos x=2 sin2x2 & 1+cos x=2 cos2x2=∫tan2x2 dx=∫sec2 x2-1 dx=tan x212-x+C=2 tan x2-x+C

Q32.

Answer :

∫sin mx·cos nx dx=12∫2 sin mx·cos nxdx=12∫sin mx+nx+sin mx-nxdx ∴2 sin A·cos B=sin A+B+sin A-B=12-cos m+nxm+n-cos m-nxm-n+C

Q33.

Answer :

∫ dx1-sinx2=∫1+sin x21-sin x2 1+sin x2 dx=∫1+sin x21-sin2 x2dx=∫1+sinx2cos2x2 dx=∫sec2 x2+sec x2 tan x2dx=tan x212+sec x212+C=2 tan x2+sec x2+C

Q34.

Answer :

∫dx1+cos 3x=∫1-cos 3×1+cos 3x 1-cos 3xdx=∫1-cos 3×1-cos2 3x dx=∫1-cos 3xsin2 3x dx=∫cosec2 3x dx-∫cosec 3x cot 3xdx=-cot 3×3+cosec 3×3+C=13cosec 3x-cot 3x+C=131sin 3x-cos 3xsin 3x+C=13 1-cos 3xsin 3x+C

Q35.

Answer:

∫dx1+sec ax=∫dx1+1cos ax=∫cos ax1+cos ax dx=∫1+cos ax-11+cos ax dx=∫1+cos ax1+cos ax-11+cos axdx=∫dx-∫dx1+cos ax=∫dx-∫1-cos ax1+cos ax1-cos axdx=∫dx-∫1-cos ax1-cos2 axdx=∫dx-∫1-cos axsin2 axdx=∫dx-∫1sin2 ax-cos axsin2 axdx=∫dx-∫1sin2 axdx+∫cos axsin2 axdx=∫dx-∫cosec2 axdx+∫cot ax cosec axdx=x+cot axa -cosec axa+C=x+1acot ax -cosec ax+C=∫dx-∫dx1+cos ax=∫dx-∫dx2 cos2 ax2 ∴1+cos x=2 cos2 x2=∫dx-12∫sec2ax2dx=x-12tan ax2a2=x-tan ax2a+C

Q36.

Answer :

∫ex+12ex dx=∫e2x+2ex+1ex dx=∫e3x+2e2x+ex dx= e3x3+2e2x2+ex+C

Q37.

Answer :

∫ex+1ex2 dx=∫e2x+1e2x+2ex×1ex dx=∫e2x+e-2x+2dx=e2x2+e-2x-2+2x+C=e2x2-e-2×2+2x+C

Q38.

Answer :

∫ex+12 dx=∫e2x+2ex+1 dx=∫e2x dx+2∫exdx+∫dx=e2x2+2ex+x+C

Q39.

Answer :

∫1+cos 4xcot x-tan x dx=∫1+cos 4xcos xsin x-sin xcos x dx=∫2 cos2 2x×sin x cos xcos2x-sin2xdx=∫cos2 2x×2 sin x cos xcos 2xdx=∫cos 2x sin 2xdx=12∫2 sin 2x cos 2xdx=12∫sin 4xdx=12-cos 4×4+C=-18cos 4x+C

Q40.

Answer :

∫dxx+3-x+2
Rationalising the denominator
=∫x+3+x+2x+3-x+2 x+3+x+2 dx=∫x+312+x+212x+3-x+2dx=∫x+312+x+212dx=x+312+112+1+x+212+112+1+C=23x+332+23x+232+C=23x+332+x+232+C

Q41.

Answer :

Let I=∫x+2 3x+5 dxPutting 3x+5=t⇒x=t-53

⇒3dx=dt⇒dx=dt3

∴ I=∫t-53+2tdt3 =13∫t-5+63t dt =19∫t+1 t dt =19∫t32+t12 dt =19t32+132+1+t12+112+1+C =1925t52+23t32+C =19253x+552+233x+532+C ∵t=3x+5 =293x+5323x+55+13+C =293x+5329x+15+515+C =293x+5329x+2015+C =21353x+5329x+20+C

Q42.

Answer :

∫2x+13x+2dx=13∫6x+33x+2dx=13∫6x+4-13x+2dx=13∫23x+23x+2-13x+2dx=13∫23x+2-13x+2dx=13∫23x+2 12dx-∫3x+2 -12dx=1323x+212+13 12+1-3x+2-12+1-12+1×3+C=13493x+232-233x+212+C=4273x+232-293x+212+C=3x+24273x+2-29+C=3x+243x+2-627+C=3x+212x+8-627+C=2276x+13x+2+C

Q43.

Answer :

Let I=∫3x+57x+9dxPutting 7x+9=t⇒x=t-97

& 7dx=dt⇒dx=dt7

∴ I=∫3t-97+5tdt =∫37tt-277t+5tdt7 =37×7∫t12dt-277×7∫t-12dt+57∫t-12dt =37×7t12+112+1-277×7t-12+1-12+1+57t-12+1-12+1+C =27×7t32-277×7×2 t12+10t7+C =27×77x+932-547×77x+912+1077x+9+C ∵ t=7x+9 =27×77x+932+10-547 7x+97+C =277x+932+70-547 7x+97+C =27×77x+932+167×77x+9+C =27×77x+9127x+9+8+C =2497x+9127x+17+C

Page 19.36 Ex.19.3

Q44.

Answer :

Let I=∫ x2+3x-1x+12dx
Putting x + 1 = t
⇒ x = t – 1
& dx = dt

∴ I=∫t-12+3 t-1-1t2dt =∫ t2-2t+1+3t-3-1t2dt =∫t2+t-3t2dt =∫1+1t-3t-2dt =t+log t-3t-2+1-2+1+C =t+log t+3t+C =x+1+log x+1+3x+1+C ∵t=x+1
Let C + 1 = C′
=x+log x+1+3x+1+C′

Q45.

Answer :

∫xx+4dx=∫x+4-4x+4dx=∫x+4-4x+4dx=∫x+412dx-4∫x+4-12dx=x+412+112+1-4x+4-12+1-12+1+C=23x+432-8 x+412+C=x+41223x+4-8+C=x+4122x+8-243+C=x+4122x-163+C=23x-8x+4+C

Q46.

Answer :

Let I=∫2-3×1+3xdx
Putting 1 + 3x = t
⇒ 3x = t – 1

& 3dx=dt⇒dx=dt3

∴ I=∫2-t-1tdt =∫3-ttdt =∫3t-12-t12dt =3∫t-12dt-∫t12dt =3t-12+1-12+1-t12+112+1+C =6t-23t32+C =2t 3-t3+C =2t9-t3+C ∵t=1+3x =231+3x 9-1+3×3+C =23×31+3x 8-3x+C =298-3x 1+3x+C

Q47.

Answer :

Let I=∫5x+3 2x-1dxPutting 2x-1=t⇒2x=t+1⇒x=t+12

& 2dx=dt⇒dx=dt2

∴I=∫5t+12+3·t·dt2 =∫5 t2+52+3×t dt2 =14∫5t+11 t12 dt =14∫5t32+11t12dt =145t32+132+1+11t12+112+1+C =14×25×5 t52+14×11×23 t32+C =12t52+116t32 +C =t322t+113+C =t3223t+113 +C =2x-132232x-1+113+C ∵t=2x-1 =2x-13226x-3+113+C =2x-12322 3x+43+C =2x-1323x+43+C

Q48.

Answer :

∫2x-1x-12dx=∫2x-2+2-1x-12dx=∫2 x-1x-12+1x-12dx=2∫dxx-1+∫x-1-2 dx=2 ln x-1+x-1-2+1-2+1+C=2 ln x-1-1x-1+C

Q49.

Answer :

∫tan2 2x-3dx=∫sec2 2x-3-1dx=∫sec2 2x-3dx-∫1dx=tan 2x-32-x+C

Q50.

Answer :

Let I= ∫1cos2x 1-tan x2dx=∫sec2x1-tan x2dx=∫sec2x dx1-tan x2

Let 1 – tan x = t
-sec2x dx=dt⇒sec2x dx=-dt

∴ I=∫-dtt2 =-∫t-2 dt =-t-2+1-2+1+C =1t+C =11-tan x+C

Page 19.46 Ex.19.4

Q1.

Answer :

∫11-cos 2xdx= ∫12sin2xdx ∵1-cos 2x=2sin2x= 12∫cosec x dx= 12ln cosec x-cotx+C= 12 ln 1sinx-cosxsinx+C= 12 ln 2sin2x2sinx+C ∵1-cos x=2sin2x2= 12 ln 2sin2x22sinx2 cosx2+ C ∵ sin x=2sinx2 cosx2= 12 ln tanx2+C

Q2.

Answer :

∫11+cosxdx= ∫12cos2x2dx= 12∫secx2 dx= 12×2 ln tanx2+secx2+C= 2 ln 1+sinx2cosx2+C= 2 ln sinx4+cosx42cos2x4-sin2x4+C ∵ 1+sinθ=sin2θ2+cos2θ2+2sinθ2cosθ2=sinθ2+cosθ22& cosθ=cos2θ2-sin2θ2= 2 ln sinx4+cosx42cosx4-sinx4cosx4+sinx4+C= 2 ln sinx4+cosx4cosx4-sinx4+C= 2 ln 1+tanx41-tanx4+C= 2 ln tanπ4+x4+C

Q3.

Answer :

∫1+cos2x1-cos2xdx= ∫2cos2x2sin2xdx= ∫cotx dx= ln sinx+C

Q4.

Answer :

∫1-cosx1+cosxdx= ∫2sin2x22cos2x2dx ∵1-cos x=2sin2x2 & 1+cosx=2cos2x2= ∫tanx2 dx= -2 ln cosx2+C

Q5.

Answer :

We know that, tan 5x=tan 2x+3x⇒ tan 5x=tan 2x+tan 3×1-tan 2x tan 3x⇒ tan 5x-tan 2x tan 3x tan 5x=tan 2x+tan 3x⇒tan 2x tan 3x tan 5x = tan 5x -tan 2x-tan 3x ∴∫tan 2x tan 3x tan 5x=∫tan 5x-tan 2x-tan 3xdx = 15 ln sec 5x-12 ln sec 2x-13 ln sec 3x+C

Q6.

Answer :

Let I=∫1+tan x tan x+θdx = ∫1+tanxtan x+tan θ1-tan x tan θdx = ∫1+tan2 x1-tan x tan θdx = ∫sec2x dx1-tan x tan θPutting tanx=t⇒ sec2x=dtdx ⇒dx=dtsec2x∴ I= ∫11-t tanθdt = -1tan θ ln 1-t tan θ+C ∵ ∫1ax+bdx=1aln ax+b+C = -cot θ ln 1-tanx tan θ+C = cot θ ln 11-tan x tan θ+C = cot θ ln cosx cosθcos x cos θ-sin x sin θ+C = cot θ ln cos xcos x+θ+C’ Let C’=C+cot θ ln cosθ

Q7.

Answer :

Let I=∫sinx-asinx-bdxPutting x-b=t ⇒ x=b+t& dx=dt∴ I= ∫sinb+t-asintdt =∫sinb-a+tsintdt = ∫sinb-acos tsint+∫cosb-a sin tsin tdt = ∫sinb-acot t dt+∫cosb-adt = sinb-a ln sin t+t cosb-a+C = sinb-a ln sinx-b+x-bcosb-a+C ∵t=x-b

Q8.

Answer :

Let I=∫sinx-αsinx+αdxPutting x+α=t ⇒x=t-α& dx=dt∴I=∫sin t-2αsin tdt = ∫sin t cos 2αsin t-cos t sin 2αsin tdt = cos 2α∫dt-sin 2α∫cot t dt = tcos 2α-sin 2α ln sin t+C = x+αcos 2α-sin 2α ln sin x+α+C ∵ t=x+α = xcos 2α-sin 2α ln sin x+α+C

Q9.

Answer :

Let I=∫sin 2xsinx-π6 sinx+π6dx = ∫sin 2xsin2x-sin2π6 dx ∵sin A+B sinA-B=sin2A-sin2B = ∫sin 2xsin2x-14dxPutting sin2x-14=t⇒ 2sin x cos x dx=dt⇒ sin 2x dx=dt∴ I= ∫1tdt = ln t +C = ln sin2x-14+C ∵t=sin2x-14

Q10.

Answer :

Let I=∫cosxcosx-adxPutting x-a=t ⇒ x=a+t ⇒ dx=dt∴ I= ∫cosa+tdtcost = ∫cos a cos tcos t-sin a sin tcos tdt = ∫cos a-sin a tan tdt = tcos a-sin a ln sec t+C = x-acos a-sin a ln secx-a+C ∵t=x-a

Q11.

Answer :

∫1-sin 2×1+sin 2xdx=∫cos2x+sin2x-2 sin x cos xcos2x+sin2x+2 sin x cos x dx= cos x-sin x2cos x+sin x2dx= ∫cos x-sin xcos x+sin xdx= ∫1-tan x1+tan xdx= ∫tan π4-xdx=1-1ln sec π4-x ∵∫tan ax+bdx=1aln sec ax+b+C= -ln cos π4-x-1+C= ln cos π4-x+C

Q12.

Answer :

Let I=∫e3xe3x+1dxPutting e3x+1=t ⇒3e3x=dtdx⇒dx=dt3e3x∴ I= ∫e3x3te3xdt = 13∫1tdt = ln t3+C = ln e3x+13+C

Q13.

Answer :

Let I=∫secx tanx3 sec x+5dxPutting sec x=t ⇒dtdx=sec x tan x⇒dt=sec x tan x dx∴I= ∫dt3t+5 = 13 ln 3t+5+C = 13 ln 3 sec x+5+C ∵t=sec x

Q14.

Answer :

Let I=∫1-cotx1+cotxdx = ∫1-cosxsinx1+cosxsinxdx = ∫sinx-cosxsinx+cosxdxPutting sinx+cosx=t⇒ cosx-sinxdx=dt⇒ sinx-cosxdx=-dt∴ I= ∫-dtt = -ln t +C = -ln sinx+cosx+C

Q15.

Answer :

Note: Here , we are considering log x as loge x.Let I=∫secx cosecxlog tanxdxPutting log tan x=t⇒ sec2xtanx=dtdx⇒secx cosecx dx=dt∴ I= ∫1tdt =log t+C = log log tanx+C

Q16.

Answer :

Here, we are considering log x as logex .Let I=∫1×3+logxdxPutting log x=t⇒ 1x=dtdx⇒dxx=dt∴ I= ∫dt3+t = log 3+t+C = log 3+log x+C ∵t= log x

Q17.

Answer :

Let I=∫ex+1ex+xdxPutting ex+x=t⇒ex+1=dtdx⇒ex+1dx=dt∴ I=∫1tdt = ln t +C = ln ex+x+C ∵ t=ex+x

Q18.

Answer :

Here, we are considering log x as loge x .Let I=∫1x log xdxPutting logx=t⇒1x=dtdx⇒1xdx=dt∴ I= ∫1tdt = log logx+C

Q19.

Answer :

Let I=∫sin 2xacos2x+bsin2xdx = ∫sin 2xa1-sin2x+bsin2x dx = ∫sin 2xb-asin2x+a dxPutting sin2x=t ⇒2sin x.cos x= dtdx⇒ sin 2x=dtdx⇒sin 2x dx=dt∴I= ∫1b-at+adt = 1b-a ln b-at+a+C ∵∫1ax+bdx=1alnax+b+C = 1b-a ln b-asin2x+a+C ∵t= sin2x = 1b-a ln bsin2x+a1-sin2x+C = 1b-a ln bsin2x+acos2x+C

Q20.

Answer :

Let I=∫ cos x2+3sin xdxPutting sinx=t ⇒cosx=dtdx⇒cosx dx=dt∴ I= ∫dt2+3t = 13ln 2+3t+C ∵∫1ax+bdx=1aln ax+b+C = 13 ln 2+3 sinx+C ∵ t=sin x

Q21.

Answer :

Let I=∫1-sinxx+cosxdxPutting x+cosx=t⇒1-sinx=dtdx⇒1-sinxdx=dt∴ I= ∫1tdt =ln t+C = ln x+cosx+C ∵ t= x+ cos x

Q22.

Answer :

Let I=∫ab+cexdxDividing numerator and denominator by ex⇒ I=∫ae-xbe-x+cdxPutting e-x=t⇒-e-x=dtdx⇒e-xdx=-dt∴ I=∫-a bt+cdt = -ab ln bt+c+C ∵ ∫1ax+bdx=1aln ax+b+C = -ab ln be-x+c+C ∵ t=e-x+C

Q23.

Answer :

Let I=∫1ex+1dx =∫e-x1+e-xdxPutting e-x=t⇒-e-x=dtdx⇒e-xdx=-dt∴ I= ∫-11+tdt = -ln 1+t+C = – ln 1+e-x+C

Q24.

Answer :

Note:Here we are considering log x as loge xLet I=∫cot xlog sin xdxPutting log sin x=t⇒ cot x=dtdx⇒cot x dx=dt∴I= ∫1tdt =log t+C = log log sinx+C ∵t=log sin x

Page 19.47 Ex.19.4

Q25.

Answer :

Let I=∫e2xe2x-2dxPutting e2x=t⇒2e2x=dtdx⇒e2xdx=dt2∴ I= 12∫1t-2dt = 12 ln t-2+C = 12 ln e2x-2+C ∵ t= e2x

Q26.

Answer :

∫2cosx-3sinx6cosx+4sinxdx⇒∫2cosx-3sinx23cosx+2sinxdtLet, 3cosx+2sinx=t ⇒2cosx-3sinx=dtdx ⇒2cosx-3sinxdx=dtNow, ∫2cosx-3sinx23cosx+2sinxdt= ∫dt2t=12logt+C= 12 log 3cosx+2sinx+C

Q27.

Answer :

Let I=∫cos2x+x+1×2+sin2x+2xdxPutting x2+sin2x+2x=t⇒ 2x+2cos 2x+2=dtdx⇒x+cos 2x+1dx=dt2∴ I= 12∫1tdt =12lnt+C = 12 ln x2+sin2x+2x+C ∵t=x2+sin 2x+2x

Q28.

Answer :

∫1cosx+acosx+bdx= 1sina-b∫sina-bcosx+acosx+bdx= 1sina-b∫sinx+a-x+bcosx+a cosx+bdx= 1sina-b∫sinx+acosx+bcosx+acosx+b-cosx+asinx+bcosx+acosx+bdx= 1sina-b∫tanx+a-tan x+bdx= 1sina-b∫tanx+adx-∫tan x+bdx= 1sina-b-ln cosx+a+ln cosx+b+C= 1sina-b ln cosx+bcosx+a+C

Q29.

Answer :

Let I=∫-sinx+2cosx2sinx+cosxdxPutting 2sinx+cosx=t⇒2cosx-sinx=dtdx⇒-sinx+2cosxdx=dt∴ I=∫1tdt =lnt+C = ln 2sinx+cosx+C ∵t=2sin x+cos x

Q30.

Answer :

∫cos4x-cos2xsin4x-sin2xdx= ∫-2sin4x+2x2sin4x-2x22cos4x+2x2sin4x-2x2dx ∵cos A-cos B=-2sin A+B2sin A-B2 & sin A-sin B=2cos A+B2sin A-B2= -∫sin 3xcos 3xdx= -∫tan 3x dx= -ln sec 3×3+C= 13 ln sec 3x-1+C= 13 ln cos 3x+C

Q31.

Answer :

Note: Here, we are considering log x as logexLet I=∫secxlog secx+tanxdxPutting log secx+tanx=t⇒ secx tanx+sec2xsecx+tanx=dtdx⇒ secxsecx+tanxsecx+tanx=dtdx⇒secx dx=dt∴ I= ∫dtt =logt+C =log log secx+tanx+C

Q32.

Answer :

Note : Here, we are considering log x as logexLet I=∫cosec xlog tanx2dxPutting log tan x2=t⇒12sec2x2tanx2=dtdx⇒12 sinx2.cosx2=dtdx⇒1sinx=dtdx⇒cosec x dx=dt∴ I= ∫dtt =logt+C = log log tanx2+C

Q33.

Answer :

Note: Here, we are considering log x as loge x .Let I=∫1x logx loglogxdxPutting loglogx=t⇒1xlogx=dtdx⇒1xlogxdx=dt∴I=∫dtt =logt+C = logloglogx+C ∵t=loglogx

Q34.

Answer :

Let I=∫cosec2x1+cot xdxPutting cotx=t⇒-cosec2x=dtdx⇒cosec2x dx=-dt∴ I= ∫-dt1+t =- ln 1+t+C =- ln 1+cot x+C ∵ t=cot x

Q35.

Answer :

Let I=∫10×9+10x loge1010x+x10dxPutting 10x+x10=t⇒10x loge10+10×9=dtdx⇒10x loge10+10x9dx=dt∴ I= ∫1tdt =ln t+C = ln 10x+x10+C ∵t=10x+x10

Q36.

Answer :

Let I=∫ex-1+xe-1ex+xedxPutting ex+xe=t⇒ ex+exe-1=dtdx⇒ eex-1+xe-1=dtdx⇒ex-1+xe-1dx=dte∴ I= 1e∫1tdt =1e ln t+C = 1e ln ex+xe+C ∵ t=ex+xe

Q37.

Answer :

Note: Here, we are considering log x as loge xLet I=∫1+tanxx+log secxdxPutting x+log secx=t⇒1+secx tanxsecx=dtdx⇒1+tanxdx=dt∴ I= ∫1tdt = log t+C = log x+log secx+C ∵t=x+log sec x

Q38.

Answer :

Let I=∫sin 2xa2+b2sin2xdxPutting sin2x=t⇒2sinx.cosx=dtdx⇒sin 2x=dtdx⇒sin 2x dx=dt∴ I= ∫1a2+b2tdt = 1b2 ln a2+b2t+C = 1b2 ln a2+b2 sin2x+C ∵ t= sin2x

Q39.

Answer :

Note: Here, we are considering log x as loge xLet I=∫x+1xx+logxdxPutting x+logx=t⇒1+1x=dtdx⇒x+1xdx=dt∴ I= ∫1tdt = log t+C = log x+logx+C

Q40.

Answer :

Let I=∫11-x22+3 sin-1xdxPutting sin-1x=t⇒11-x2=dtdx⇒11-x2dx=dt∴ I= ∫12+3tdt = 13 ln 2+3t+C = 13 ln 2+3sin-1x+C ∵ t=sin-1x

Q41.

Answer :

Let I=∫sec2xtanx+2dxPutting tan x=t⇒sec2x=dtdx⇒sec2x dx=dt∴ I= ∫1t+2dt = ln t+2+C = ln tanx+2+C ∵ t= tan x

Q42.

Answer :

Let I=∫2 cos 2x+sec2xsin 2x+tan x-5dxPutting sin 2x+tan x-5=t⇒2cos 2x+sec2x=dtdx⇒2cos 2x+sec2xdx=dt∴ I= ∫1tdt = ln t+C = ln sin 2x+tan x-5+C ∵ t=sin 2x+tan x-

Q43.

Answer :

∫sin 2xsin 5x sin 3xdx= ∫sin 5x-3xsin 5x sin 3xdx=∫sin 5x cos 3x-cos 5x sin 3xsin 5x sin 3xdx=∫ sin 5x cos 3xsin 5x sin 3x-cos 5x sin 3xsin 5x sin 3xdx= ∫cot 3x-cot 5x dx=∫cot 3x dx-∫cot 5x dx= 13 ln sin 3x-15 ln sin 5x+C

Q44.

Answer :

Note: Here, we are considering log x as loge xLet I=∫1+cot xx+log sin xdxPutting x+log sinx=t⇒1+cotx=dtdx⇒1+cot xdx=dt∴ I=∫1tdt = log t+C = log x+log sinx+C ∵t= x+log sin x

Q45.

Answer :

Let I=∫1-sin2xx+cos2xdxPutting x+cos2x=t⇒ 1-2cosx.sinx=dtdx⇒1-sin 2xdx=dt∴ I=∫1tdt = ln t+C = ln x+cos2x+C ∵t= x+cos2x

Q46.

Answer :

Let I=∫1+tanx1-tanxdx = ∫1+sinxcosx1-sinxcosxdx = ∫cosx+sinxcosx-sinxdxPutting cosx-sinx=t⇒ -sinx-cosxdx=dt⇒ sinx+cosxdx=-dt∴ I= -∫1tdt = – ln t +C = – ln cosx-sinx+C ∵ t=cos x-sin x

Q47.

Answer :

Let I=∫cos 2xcosx+sinx2dx = ∫cos2x-sin2xcosx+sinx2dx = ∫cos x-sin xcos x+sin xdxPutting cos x+sin x=t ⇒-sinx+cosx=dtdx⇒cosx-sinxdx=dt∴ I=∫1tdt = ln t+C = ln cos x+sin x+C ∵ t=cos x+sin x

Q48.

Answer :

We have,I=∫cos x-sin x1+sin 2x dx=∫cos x-sin xsin2x+cos2x+2sin x cos x dx=∫cos x-sin xsin x+cos x2 dxPutting sin x+cos x=t⇒cos x-sin xdx=dt∴I=∫1t2 dt=-1t+C=-1sin x+cos x+C

Q49.

Answer :

Let I=∫1xx+1dxPutting x+1=t⇒12x=dtdx⇒1xdx=2dt∴ I=2∫1tdt = 2 ln t+C = 2 ln x+1+C ∵ t=x+1

Q50.

Answer :

∫sec xsec 2xdx= ∫cos 2xcos xdx= ∫2cos2x-1cos xdx= ∫2cosx-sec x dx= 2 sin x-ln sec x+tan x+C

Q51.

Answer :

∫1cos 3x-cos xdx= ∫14cos3x-4cosxdx ∵cos 3x=4 cos 3x-3 cos x= ∫14cos xcos2x-1dx= -14∫1cos x sin2xdx= -14∫sin2x+cos2xcos x sin2x dx= -14∫sec x dx+∫cot x cosec x dx= -14ln secx+tanx-cosec x+C= 14cosec x-lnsec x+tan x+C

Q52.

Answer :

∫1sin x cos2xdx=∫sin2x+cos2xsin x cos2xdx=∫tan x sec x+cosec x dx= sec x+ln cosec x-cot x+C= sec x+ln tanx2+C ∵cosecx-cotx=1-cosxsin x=tanx2

Q53.

Answer :

∫1cosx+acosx+bdx=1sina-b∫sinx+acosx+b-cosx+asinx+bcosx+a cos x+bdx=1sina-b∫tanx+adx-∫tanx+bdx=1sina-bln secx+a-lnsecx+b =1sina-blncos x+bcos x+a+C

Page 19.56 Ex.19.5

Q1.

Answer :

∫log xxdxLet, log x=t⇒1x=dtdxNow, ∫log xxdx=∫t·dt=t22+C=log x22+C

Q2.

Answer :

∫log 1+1xx1+xdxLet, log 1+1x=t⇒11+1x×-1×2=dtdx⇒xx+1×-1×2=dtdx⇒-dxxx+1=dt⇒dxxx+1=-dtNow, ∫log 1+1xx1+xdx=∫t ·-dt=-t22+C=-12log1+1×2+C

Q3.

Answer :

∫1+x2xdxLet, 1+x=t⇒12x=dtdx⇒dxx=2dtNow, ∫1+x2xdx =2∫t2dt=23t3+C=231+x3+C

Q4.

Answer :

∫1+ex·exdxLet 1+ex=t⇒ex=dtdx⇒exdx=dtNow, ∫1+ex·exdx=∫t·dt=t12+112+1+C=23t32+C=231+ex32+C

Q5.

Answer :

∫cos2 x13 sin x dxLet, cos x=t⇒-sin x=dtdx⇒sinx dx=-dtNow, ∫cos2 x13 sin x dx=-∫t23dt=-t23+123+1+C=-35 t53+C=-35cos53x+C

Q6.

Answer :

∫ex dx1+ex2Let 1+ex=t⇒ex =dtdx⇒ex dx=dtNow, ∫ex dx1+ex2=∫dtt2=∫t-2 dt=t-2+1-2+1+C=-1t+C=-11+ex+C

Q7.

Answer :

∫cot3x cosec2x dxLet, cot x=t⇒-cosec2x =dtdx⇒cosec2x dx=-dtNow, ∫cot3x cosec2x dx=∫t3-dt=-t44+C=-cot4x4+C

Q8.

Answer :

∫esin-1 x21-x2 dxLet esin-1 x=t Differentiating both sides w.r.t. x,esin-1 x ×11-x2 dx=dtNow, ∫esin-1 x21-x2 dx=∫esin-1 x·esin-1 x1-x2dx =∫t ·dt=t22+C=esin-1 x22+C

Q9.

Answer :

∫1+sin xx-cos xdxLet, x-cos x=t⇒1+sin x=dtdx⇒1+sin x dx=dtNow, ∫1+sin xx-cos xdx=∫dtt=∫t-12dt=t-12+1-12+1+C=2t+C=2x-cos x+C

Q10.

Answer :

∫dx1-x2sin-1 x2Let, sin-1 x=t⇒11-x2=dtdx⇒11-x2 dx=dtNow,∫dx1-x2sin-1 x2=∫dtt2=∫t-2dt=t-2+1-2+1+C=-1t+C=-1sin-1 x+C

Q11.

Page 19.57 Ex.19.5

Q17.

Answer :

∫1x log x2 dxLet log x=t⇒1x dx=dtNow, ∫1x log x2 dx=∫t2dt=t33+C=log x33+C

Q18.

Answer :

∫sin5 x cos x dxLet sin x=t⇒cos x=dtdx⇒cos x dx=dtNow, ∫sin5 x cos x dx=∫t5dt=t66+C=16 sin6 x+C

Q19.

Answer :

∫tan32x·sec2x dxLet tan x=t⇒sec2x=dtdx⇒sec2x dx=dtNow, ∫tan32x·sec2x dx =∫ t32 dt=t32+132+1+C=25 t52+C=25 tan52 x+C

Q20.

Answer :

∫x3x2+13 dx=∫x2.xx2+13dxLet x2+1=t⇒x2=t-1⇒2x dx=dt⇒x dx=dt2Now, ∫x2.xx2+13dx=12∫t-1t3dt=12∫1t2-1t3 dt=12∫t-2-t-3dt=12t-2+1-2+1-t-3+1-3+1+C=12-1t+12t2+C=12-1×2+1+12×2+12+C=12-2 x2+1+12 x2+12=14-2×2-2+1×2+12=-141+2x2x2+12+C

Q21.

Answer :

∫4x+2 x2+x+1 dx=2∫2x+1 x2+x+1 dxLet x2+x+1=t⇒2x+1=dtdx⇒2x+1 dx=dtNow, 2∫2x+1 x2+x+1 dx=2∫t dt=2∫t12dt=2 t12+112+1+C=2×23t32+C=43 t32+C=43×2+x+132+C

Q22.

Answer :

∫4x+32×2+3x+1dxLet 2×2+3x+1=t⇒4x+3=dtdx⇒4x+3 dx=dtNow, ∫4x+32×2+3x+1dx=∫dtt=∫t-12dt=t-12+1-12+1+C=2 t+C=2 2×2+3x+1+C

Q23.

Answer :

∫ dx1+x=∫x dxx 1+xLet 1+x=t⇒x=t-1⇒12x=dtdx⇒dxx =2dt

Now, ∫xx1+xdx=2∫t-1tdt=2∫1-1tdt=2 t-log t+C=2 1+x-2 log 1+x+CLet C + 2 =C’=2x-2 log 1+x+C’

Q24.

Answer :

We have,I=∫x+x+1x+2 dxLet, x+1=t2Differentiating both sides we getdx=2tdtNow, integration becomesI=∫t2-1+tt2+12t dt=2∫t3+t2-tt2+1 dt=2∫t3+t-t+t2+1-1-tt2+1 dt=2∫t3+t+t2+1-t-t-1t2+1 dt=2∫t3+tt2+1 dt++2∫t2+1t2+1 dt+2∫-2t-1t2+1 dt=2∫tdt+2∫dt-2∫2tt2+1 dt-2∫1t2+1 dt=t2+2t-2log t2+1-2tan-1t+C=x+1+2x+1-2log x+2-2tan-1x+1+C

Q25.

Answer :

∫1+cos xx+sin x3dxLet x+sin x=t⇒1+cos x=dtdx⇒1+cos x dx=dtNow, ∫1+cos xx+sin x3dx=∫dtt3=∫t-3dt=t-3+1-3+1+C=-12 t2+C=-12 x+sin x2+C

Q26.

Answer :

∫cos x-sin x1+sin 2xdx⇒∫cos x-sin xcos2 x+sin2 x+2 sin x.cos x dx⇒∫cos x-sin xcos x+sin x2dxLet cos x+sin x=t⇒-sin x+cos x=dtdx⇒-sin x+cos x dx=dtNow, ∫cos x-sin xcos x+sin x2dx=∫dtt2=∫t-2dt=t-2+1-2+1+C=-1t+C=-1sin x +cos x+C

Q27.

Answer :

∫sin 2xa+b cos 2x2dxLet a+b cos2x=t⇒-b sin 2x dx × 2=dt⇒sin 2x dx=-dt2bNow, ∫sin 2xa+b cos 2x2dx=-12b∫dtt2=-12b∫t-2dt=-12bt-2+1-2+1+C=12b×1t+C=12b a+b cos 2x+C

Q28.

Answer :

∫log x2 dxx=∫2 log xx dx=2∫log xxdxLet log x=t⇒1x=dtdx⇒1x dx=dtNow, 2∫log xxdx=2∫t dt=2t22+C=t2+C=log x2+C

Q29.

Q56.

∫x tan-1 x21+x4 dxLet tan-1 x2=t⇒11+x22×2x=dtdx⇒x dx1+x4=dt2Now, ∫x tan-1 x21+x4 dx=12∫t. dt=12×t22+C=tan-1 x224+C

Q57.

Answer :

∫sin-1 x31-x2 dxLet sin-1 x=t⇒11-x2dx=dtNow, ∫sin-1 x31-x2 dx =∫ t3 dt=t44+C=sin-1 x44+C

Q58.

Answer :

∫ sin 2+3 log xxdxLet 2+3 log x=t⇒3x=dtdx⇒dxx=dt3Now, ∫ sin 2+3 log xxdx=13∫ sin t dt=13 -cos t+C=-13cos 2+3 log x+C

Q59.

Answer :

∫ x.ex2 dxLet x2=t⇒2x dx=dt⇒x dx=dt2Now, ∫ x.ex2 dx=12∫et dt=12et+C=12ex2+C

Q60.

Answer :

Let I=∫dxx+1 x2+2x+2=∫dxx+1 x2+2x+1+1=∫dxx+1 x+12+1Putting x+1=t⇒dx=dtNow, integral becomesI=∫dtt t2+1=∫t·dtt2 t2+1Again putting t2=p⇒2t dt=dp⇒t dt=dp2Now, integral becomesI=12 ∫dpp p+1=12∫dpp2+p=12∫dpp2+p+14-14=12∫dpp+122-122=12 12×12 log p+12-12p+12+12 +C=12 log pp+1+C=12 log t2t2+1+C=12 log x+12x+12+1+C= log x+12x+12+1+C= log x+1×2+2x+2+C

Q61.

Answer :

∫x5 dx1+x3=∫x3.x2 dx1+x3Let 1+x3=t ⇒x3=t-1⇒3×2=dtdx⇒x2 dx=dt3Now, ∫x3.x2 dx1+x3=13∫t-1t dt=13∫t-1tdt=13 ∫t12-t-12dt=13t12+112+1-t-12+1-12+1+C=1323t32-2t+C=29 1+x332-231+x312+C

Q62.

Answer :

∫ 4×3 5-x2 dx=4∫x2×x 5-x2 dxLet 5-x2=t ⇒x2=5-t⇒2x=-dtdx⇒x dx=-dt2Now, 4∫x2×x 5-x2 dx=4-2 ∫5-t.t dt=-2∫5t12+2 ∫t32 dt=-10 t12+112+1+2 t32+132+1+C=-203t32+45t52+C=-2035-x232+455-x252+C

Q63.

Answer :

∫ ecos2. sin 2x dxLet cos2 x=t⇒2 cos x.-sin x=dtdx⇒sin 2x dx=-dtNow, ∫ ecos2. sin 2x dx =-∫et dt=-et+C=-ecos2 x +C

Q64.

Answer :

∫ x sin-1 x21-x4dxLet sin-1 x2=t⇒1×2×1-x4=dtdx⇒x dx1-x4=dt2Now, ∫ x sin-1 x21-x4dx=12∫tdt=t24+C=sin-1 x224+C

Q65.

Answer :

∫dxx+x=∫dxx 1+xLet 1+x=t⇒12x=dtdx⇒dxx=2dtNow, ∫dxx 1+x=∫2dtt=2∫dtt=2 log t+C=2 log 1+x+C

Q66.

Answer :

∫ sec4 x.tan x dx=∫ sec2 x.sec2 x tan x dx=∫ 1+tan2 x sec2 x. tan x dx=∫ tan x+tan3 x sec2 x dxLet tan x=t⇒sec2 x dx=dtNow, ∫ tan x+tan3 x sec2 x dx=∫t+t3 dt=t22+t44+C=tan2 x2+tan4 x4+C

Q67.

Answer :

∫dxx2x4+134=∫dxx2 x41+1×434=∫ dxx2.x3 1+1×434=∫1+1×4-34x5dxLet 1+1×4=t⇒-4x5dx=dt⇒dxx5=-dt4Now, ∫1+1×4-34x5dx=-14 ∫t-34 dt=-14 t-34+1-34+1+C=-t14+C=-1+1×414+C

Q68.

Answer :

∫cos5 xsin xdx=∫ cos4 x .cos xsin xdx=∫cos2 x2.cos xsin xdx=∫1-sin2 x2×cos xsin xdx=∫ 1-sin4 x-2 sin2 xsin xcos x dxLet sin x=t⇒cos x dx=dtNow, ∫ 1-sin4 x-2 sin2 xsin xcos x dx=∫ 1+t4-2t2tdt=∫1t+t3-2tdt=log t +t44-2t22+C=log t+t44-t2+C=log sin x+sin4 x4-sin2 x+C

Q69.

Answer :

∫sin5 xcos4 xdx=∫sin4 x. sin xcos4 xdx=∫sin2 x2. sin xcos4 xdx=∫ 1-cos2 x2 sin xcos4 xdx=∫ 1+cos4 x-2 cos2 xcos4 xsin x dx=∫ 1cos4 x+1-2cos2 xsin x dxLet cos x=t⇒-sin x=dtdx⇒sin x dx=-dtNow, ∫ 1cos4 x+1-2cos2 xsin x dx=-∫ t-4+1-2t-2dt=-t-4+1-4+1+t-2t-2+1-2+1+C=–13t3+t+2t+C=13t3-t-2t+C=13 cos3 x-cos x-2cos x+C

Q70.

Answer :

∫e2x dx1+ex⇒∫ex.ex1+exdxLet 1+ex=t ⇒ex=t-1⇒ex dx=dtNow, ∫ex.ex1+exdx=∫ t-1.dtt=1-1tdt=t-log t+C=1+ex-log 1+ex+CLet C+1=C’=ex-log 1+ex+C’

Q71.

Answer :

∫ sec2 xxdxLet x=t⇒12x=dtdx⇒dxx=2dtNow, ∫ sec2 xxdx=2∫ sec2 t dt=2 tan t+C=2 tan x+C

Q72.

Answer :

∫ tan3 x 2x.sec 2xdx=∫ tan2 2x. sec 2x tan 2x dx=∫ sec2 2x-1 sec 2x tan 2x dxLet sec 2x=t⇒sec 2x tan 2x×2=dtdx⇒sec 2x tan 2x dx=dt2Now, ∫ tan3 x 2x.sec 2xdx=12∫t2-1 dt=12t33-t+C=12 sec3 2×3-sec2x+C=16 sec3 2x-sec 2×2+C

Page 19.63 Ex.19.6

Q1.

Answer :

∫x2 x+2 dxLet x+2=t⇒x=t-2⇒dx=dtNow, ∫x2 x+2 dx=∫t-22 t dt=∫42-4t+4t12 dt=∫t2+12-4t1+12+4t12dt=∫t52-4t32+4t12dt=t52+152+1-4t32+132+1+4t12+112+1+C=27t72-85t52+83t32+C=27x+272-85x+252+83x+232+C

Q2.

Answer :

∫x2x-1dxLet x-1=t2⇒x=t2+1⇒1=2t dtdx⇒dx=2t dtNow, ∫x2x-1dx=∫t2+12t2t dt=2∫t4+2t2+1dt=2t4+14+1+2t2+12+1+t+C=2t55+2t33+t+C=23t5+10t3+15t15+C=215t3t4+10t2+15+C=215x-1 3x-12+10x-1+15+C=215x-1 3×2-2x+1+10x-10+15+C=215x-1 3×2-6x+3+10x-10+15+C=215x-13×2+4x+8 +C

Q3.

Answer :

∫x2 dx3x+4Let 3x+4=t ⇒x=t-43⇒1=13.dtdx⇒dx=dt3Now, ∫x2 dx3x+4=13∫t-432tdt=127∫t2t-8tt+16tdt=127∫t32-8t12+16t-12dt=127 t32+132+1+8t12+112+1+16t-12+1-12+1+C=127 25t52-8×23t32+32t12+C=2135t52-1681t32+3227t12+C=21353x+452-16813x+432+32273x+412+C

Q4.

Answer :

∫2x-1x-12dxLet x-1=t⇒x=t+1⇒1=dtdxNow, ∫2x-1x-12dx=∫2t+1-tt2dt=∫2t+1t2dt=2∫dtt+∫t-2 dt=2 log t+t-2+1-2+1+C=2 log x-1-1x-1+C

Q5.

Answer :

∫2×2+3 x+2 dxLet x+2=t⇒x=t-2⇒dx=dt∫2t-22+3t dt=∫2t t2-4t+4+3tdt=2∫t52-4t32+4t12 dt+3∫t12 dt=2t52+152+1-4t32+132+1+4t12+112+1+3t12+112+1+C=227t72-85t52+83t32+2t32+C=47t72-165t52+163t32+2t32+C=47t72-165t52+223t32+C=47x+272-165x+252+223x+232+C

Q6.

Answer :

∫x2+3x+1x+12 dxLet x+1=t⇒x=t-1⇒1=dtdx⇒dx=dtNow, ∫x2+3x+1x+12 dx=∫t-12+3t-1+1t2dt=∫t2-2t+1+3t-3+1t2dt=∫t2+t-1t2dt=∫1+1t-t-2 dt=t+log t-t-2+1-2+1+C=t+log t+1t+C=x+1+log x+1+1x+1+CLet 1+C=C’=x+log x+1+1x+1+C’

Q7.

Answer :

∫x21-xdxLet 1-x=t ⇒x=1-t⇒1=-dtdx⇒dx=-dtNow, ∫x21-xdx=∫1-t2tdt=∫1-t2-2ttdt=∫1t+t2t-2ttdt=∫t-12+t32-2t12dt=t-12+1-12+1+t32+132+1-2t12+112+1+C=2t12+25t52-43t32+C=2t121+t25-23t+C=2t1215+3t2-10t15+C=21-x 15+31-x2-101-x15+C=2151-x 15+312+x2-2x-10+10x+C=2151-x 15+3+3×2-6x-10+10x+C=2151-x 3×2+4x+8+C

Q8.

Answer :

∫x 1-x23 dxLet 1-x=t ⇒x=1-t⇒1=-dtdx⇒dx=-dtNow, ∫x 1-x23 dx=-∫1-t·t23 dt=-∫t23-t24dt=∫t24-t23 dt=t2525-t2424+C=24t25-25t24600+C=t2460024t-25+C=1-x24600 241-x-25+C=-1600 1-x24 1+24x+C

Page 19.67 Ex.19.7

Q1.

Answer :

∫ tan³ x sec² x dx
Let tan x = t
⇒ sec² x dx = dt
Now, ∫ tan³ x sec² x dx
= ∫ t³.dt
=t44+C=tan4 x4+C

Q2.

Answer :

∫ tan x. sec⁴ x dx
= ∫ tan x. sec² x . sec² x dx
= ∫ tan x (1 + tan² x) sec² x dx
Let tan x = t
⇒ sec² x dx = dt
Now, ∫ tan x (1 + tan² x) sec² x dx
= ∫ t (1 + t²) dt
= ∫ (t + t³) dt
=t22+t44+C=12tan2 x+14 tan4 x+C

Q3.

Answer :

∫ tan⁵ x sec⁴ x dx
= ∫ tan⁵ x. sec² x . sec2 x dx
= ∫ tan⁵ x (1 + tan² x) sec² x dx
Let tan x = t
⇒ sec²x dx = dt
Now, ∫tan⁵x (1+tan² x) sec² x dx
= ∫ t⁵ (1 + t²) dt
= ∫ (t⁵ + t⁷) dt
=t66+t88+C=tan6 x6+tan8 x8+C

Q4.

Answer :

∫ sec⁶x tan x dx
=∫ sec⁶x.sec x tan x dx
Let sec x = t
⇒ sec x tan x dx = dt
Now, ∫ sec⁶x.sec x tan x dx
= ∫ t⁶. dt
=t66+C=sec6 x6+C

Q5.
Answer :

∫ tan⁵ x dx
= ∫ tan⁴ x. tan x dx
= ∫(sec² x – 1)² . tan x dx
= ∫ (sec⁴ x – 2 sec² x + 1) tan x dx
= ∫ tan x . sec⁴x dx – 2 ∫ sec² x . tan x dx+ ∫ tan x dx
= ∫ sec²x. sec²x . tan x dx – 2 ∫ tan x sec²x dx + ∫ tan x dx
= ∫ (1 + tan²x) . tan x . sec²x dx – 2 ∫ tan x . sec²x dx + ∫ tan x dx
Let I1=∫ (1 + tan²x) . tan x . sec²x dx – 2 ∫ tan x . sec²x dx
And I2=∫ tan x dx
∫ tan⁵x dx=I1 + I2
Now, I1=∫ (1 + tan²x) . tan x . sec²x dx – 2 ∫ tan x . sec²x dx
Let tan x = t
⇒ sec²x dx = dt
I1=∫ (1 + tan²x) . tan x . sec²x dx – 2 ∫ tan x . sec²x dx
∫ (1 + t²) . t. dt – 2 ∫ t. dt
∫ (t + t³) dt – 2 ∫ t dt
t22+t44-2t22+C1=t44-t22+C1=tan4 x4-tan2 x2+C1
And I₂=∫ tan x dx
=logsec x+C2
∫tan5 x dx=tan4 x4-tan2x2+C1+logsec x+C2 =tan4 x4-tan2x2+logsec x+C1+C2 =tan4 x4-tan2x2+logsec x+C ∴C= C1+C2

Q6.

Answer :

∫ tan x ·sec4 x dx=∫tan x · sec2 x ·sec2 x dx=∫tan x·1+tan2 x sec2 x dxLet tan x=t⇒sec2 x dx=dtNow, ∫tan x·1+tan2 x sec2 x dx=∫t 1+t2 dt=∫t+t52dt=∫t12+t52dt=23t32+27t72+C=23tan32 x+27tan72 x+C

Page 19.68 Ex.19.7

Q7.

Answer :

∫ sec⁴ 2x dx
= ∫ sec² 2x . sec² 2x dx
= ∫ (1 + tan² 2x) . sec² 2x dx
Let tan 2x = t
⇒sec² 2x . 2 dx = dt
⇒sec2 2x . dx=dt2Now, ∫1+tan22x.sec22x dx=12∫1+t2 dt=12t+t33+C=t2+t36+C=tan 2×2+tan3 2×6+C

Q8.

Answer :

∫ cosec⁴ 3x dx
= ∫ cosec² 3x . cosec² 3x dx
= ∫ (1 + cot² 3x) cosec2 3x dx
Let cot (3x) = t
⇒–cosec² (3x) × 3 dx = dt
⇒cosec2 3xdx=-dt3Now, ∫1+cot2 3x=-13∫1+t2 dt=-13 t+t33+C=-t3-t39+C=-cot 3×3-cot3 3×9+C

Q9.
Answer :

∫ cotⁿ x cosec² x dx
Let cot x = t
⇒ –cosec² x dx = dt
⇒ cosec² x dx = –dt
Now, ∫cotnx cosec2x dx=-∫tn dt =-tn+1n+1+C=-cotn+1 xn+1+C

Q10.
Answer :

∫ cot⁵ x . cosec⁴ x dx
= ∫ cot⁵ x . cosec² x . cosec² x dx
= ∫ cot⁵x . (1 + cot²x) . cosec²x dx
Let cot x = t
⇒ – cosec²x dx = dt
⇒ cosec²x dx = –dt
Now, ∫ cot⁵x . cosec4 x dx
= ∫ t⁵ (1 + t²) dt
= ∫(t⁵ + t⁷) dt
=-t66+t88+C=-cot6 x6+cot8 x8+C

Q11.
Answer :

∫ cot⁵ x dx
= ∫ cot⁴x . cot x dx
= ∫ (cosec²x – 1)² cot x dx
= ∫ (cosec⁴x – 2 cosec²x + 1) cot x dx
= ∫ cosec⁴x . cot x dx – 2 ∫ cot x . cosec²x dx + ∫ cot x dx
= ∫ cosec²x . cosec²x . cot x . dx – 2 ∫ cot x cosec²x dx + ∫ cot x dx
=∫ (1 + cot ² x) . cot x . cosec²x dx – 2 ∫ cot x cosec²x dx + ∫ cot x dx
= ∫ (cot x + cot³x) cosec²x dx – 2 ∫ cot x cosec²x dx + ∫ cot x dx
Now, let I₁= ∫ (cot x + cot³x) cosec²x dx – 2 ∫ cot x cosec²x dx
And I₂= ∫ cot x dx
First we integrate I₁
I₁= ∫ (cot x + cot³x) cosec²x dx – 2 ∫ cot x cosec²x dx
Let cot x = t
⇒ – cosec²x dx = dt
⇒ cosec²x dx = – dt

I₁= ∫ (t + t³) (– dt) – 2∫ t (–dt)
= –∫(t + t³) + 2∫t dt
=-t22-t44+2.t22+ C1=t22-t44+C1=cot2 x2-cot4 x4+C1
Now we integrate I₂
I₂= ∫ cot x dx
= logsin x+C2
Now, ∫ cot⁵x dx=I₁ + I₂
= -14cot4x+12cot2x+logsin x+C1+C2
= -14cot4x+12cot2x+logsin x+C ∴C=C1+C2

Q12.

Answer :

∫ cot⁶x dx
= ∫ cot⁴x . (cosec² x – 1) dx
= ∫ cot⁴ x × cosec²x dx – ∫ cot⁴x dx
= ∫ cot⁴ x . cosec²x dx – ∫ cot²x . cot²x dx
= ∫ cot⁴ x – cosec²x dx – ∫ (cosec² x – 1) cot²x dx
= ∫ cot⁴ x . cosec²x dx – ∫ cot²x . cosec²x dx + ∫ cot²x dx
= ∫ cot⁴x . cosec²x dx – ∫ cot² x . cosec²x dx + ∫ (cosec²x – 1) dx
Now, let I1= ∫ cot⁴ x . cosec² x dx – ∫ cot²x . cosec²x dx
And I2= ∫ (cosec²x – 1) dx
First we integrate I1
I₁= ∫ cot⁴x . cosec²x dx – ∫ cot² x . cosec²x dx
Let cot x = t
⇒ –cosec² x dx = dt
⇒ cosec² dx = – dt
I₁=– ∫ t⁴ dt + ∫ t² dt
=-t55+t33+C1=-cot5 x5+cot3 x3+C1
Now we integrate I2
I₂= ∫ (cosec²x – 1) dx
= – cot x – x + C₁
Now, ∫ cot⁶ x dx=I₁ + I₂
= -15cot5x+13cot3x-cot x-x+C1+C2
= -15cot5x+13cot3x-cot x-x+C ∴C=C1+C2−14cot4x+12cot2x+log|sin x|+C [∴C=C1+C2]

Page 19.71 Ex.19.8

Q1.

Answer :

∫ sin⁴ x cos³ x dx
= ∫ sin⁴x . cos²x cos x dx
= ∫ sin⁴x . (1 – sin²x ) cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ sin⁴x . (1 – sin² x ) cos x dx
= ∫ t⁴ (1 – t²) dt
= ∫ (t⁴ – t⁶) dt
=t55-t77+C=sin5 x5-sin7 x7+C

Q2.

Answer :

∫ sin⁵x dx
= ∫ sin⁴x . sin x dx
= ∫ (1 – cos²x)² sin x dx
= ∫ (1 – cos⁴x – 2 cos²x) sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos⁴x – 2 cos²x) sin x dx
=–∫ (1 + t⁴– 2t²) dt
=-t+t55-2t33+C=-t-t55+2t33+C=-cos x+23cos3 x-cos5 x5+C

Q3.

Answer :

∫ cos⁵x dx
= ∫ cos⁴x . cos x dx
= ∫ (1 – sin²x)² cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin²x)²cos x dx
= ∫ (1 – t²)² . dt
= ∫ (1 + t⁴ – 2t²) dt
= ∫ dt + ∫ t⁴ dt – 2 ∫t² dt
=t+t55-2t33+C=sin x+sin5 x5-23sin3 x+C

Q4.

Answer :

∫ sin⁵x cos x dx
Let sin x = t
cos x dx = dt
Now, ∫ sin⁵x cos x dx
= ∫ t⁵ . dt
=t66+C=sin6 x6+C

Q5.

Answer :

∫ sin³x . cos⁶x dx
= ∫ sin²x . cos⁶x . sin x dx
= ∫ (1 – cos²x) . cos²x . sin x dx
Let cos x = t
⇒ –sin x dx = dt
Now, ∫ (1 – cos²x) . cos²x . sin x dx
= –∫ (1 – t²) . t⁶ dt
= ∫ (t² – 1) t⁶ dt
= ∫ (t⁸ – t⁶) dt
=t99-t77+C=cos9 x9-cos7 x7+C

Q6.

Answer :

∫ cos⁷x dx
= ∫ cos⁶x . cos x dx
= ∫ (cos²x)³ cos x dx
= ∫ (1 – sin²x)³ . cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin²x)³.cos x dx
= ∫ (1 – t²)³ dt
= ∫ (1 – t⁶ – 3t² + 3t⁴) dt
=t-t77-3t33+3t55+C=sin x-17sin7 x-sin3 x+35sin5x+C

Q7.

Answer :

∫ x . cos³ x² sin x² dx
Let x² = t
⇒ 2x dx = dt
⇒x dx=dt2Now, ∫x. cos3x2 sin x2dx=12∫ cos3 t. sin t . dtAgain let cos t = p⇒-sin t dt = dp⇒sin t dt = -dpSo, 12∫ cos3 t. sin t . dt =-12p3 dp=-12 p44+C=-p48+C=-cos4 t8+C=-cos4 x28+C

Q8.

Answer :

∫ sin⁷x dx
= ∫ sin⁶x . sin x dx
= ∫ (sin² x)³ sin x dx
= ∫ (1 – cos²x)³ sin x dx
Let cos x = t
⇒ –sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos²x)³ sin x dx
= ∫ (1 – t²)³ . (–dt)
= –∫ (1 – t⁶ – 3t² + 3t⁴) dt
=-t-t77-t3+3t55+C=-cos x-cos7 x7-cos3 x+35cos5 x+C=-cos x+17cos7 x+cos3 x-35cos5 x+C

Q9.

Answer :

∫ sin³x . cos⁵x dx
= ∫ sin²x . cos⁵x . sin x dx
= ∫ (1 – cos²x) . cos⁵x sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos²x) . cos⁵x sin x dx
= –∫ (1 – t²) t⁵ dt
= –∫ (t⁵ – t⁷) dt
= ∫(t⁷ – t⁵) dt
=t88-t66+C=cos8 x8-cos6 x6+C

Q10.

Answer :

∫dxsin4 x.cos2 xDividing numerator & denominator by sin2 x=∫1sin2 xsin4 x.cot2 xdx=∫cosec6 xcot2dx=∫cosec4 x.cosec2 x dxcot2 x=∫1+cot2 x2.cosec2 x dxcot2 xLet cot x=t⇒-cosec2 x=dtdx⇒-cosec2 x dx=dtNow, ∫1+cot2 x2.cosec2 x dxcot2 x=∫1+t2t2 -dt=-∫1+t4+2t2t2dt=-∫t-2+t2+2dt=-t-2+1-2+1+t33+2t+C=–1t+t33+2t+C=-13t3-2t+1t+C=-13cot3 x-2 cot x+1cot x+C=-13cot3 x-2 cot x+tan x+C

Q11.

Answer :

∫dxsin3 x.cos5 xdxDividing numerator & denominator by cos8 x=∫1cos8 xdxsin3 xcos3 x=∫sec8 xtan3 xdx=∫sec6 x.sec2 x dxtan3 x=∫1+tan2 x3. sec2 x dxtan3 xLet tan x=t⇒ sec2 x dx=dtNow, ∫1+tan2 x3. sec2 x dxtan3 x=∫1+t23t3.dt=∫1+t6+3t2+3t4t3dt=∫1t3+t3+3t+3tdt=∫t-3 dt+∫t3 dt+3∫dtt+3∫t dt=t-3+1-3+1+t3+13+1+3 log t+3t22+C=-12 tan x-2 +14tan4 x+3 log tan x+32 tan2 x+C

Q12.

Answer :

∫dxsin3 x . cos xDividing numerator & denominator by sin4 x=∫1sin4 xdxsin3 x.cos xsin4 x=∫cosec4 x dxcot x=∫cosec2 x.cosec2 x dxcot x=∫1+cot2 x.cosec2 x dxcot xLet cot x=t⇒-cosec2 x=dtdx⇒cosec2 x dx=-dtNow, ∫1+cot2 x.cosec2 x cot xdx=∫1+t2.-dtt=-∫1t+tdt=-log t-t22+C=-log cot x-cot2 x2+C=log cot x-1-cosec2 x-12+C=log 1cot x-cosec2 x2+12+C=log tan x-12sin2 x+C’ ∴C’=C+12

Q13.

Answer :

∫dxsin x. cos3 xDividing numerator & denominaor by cos4 x=∫1cos4 x dxsin x.cos3 xcos4 x=∫sec4 x dxtan x=∫sec2 x . sec2 x dxtan x=∫1+tan2 x . sec2 x tan xdxLet tan x = t⇒sec2 x = dxdt⇒sec2 x dx =dtNow, ∫1+tan2 x . sec2 x tan xdx =∫1+t2 tdt=∫1t+tdt=log t+t22+C=log tan x+tan2 x2+C

Page 19.76 Ex.19.9

Q1.

Answer :

∫dxa2-b2x2= 1b2∫dxa2b2-x2 = 1b2×12ab log ab+xab-x+C ∴∫dxa2-x2= 12a log a+xa-x+C= 12ab log a+bxa-bx+c

Q2.

Answer :

∫dxa2x2-b2 = 1a2∫dxx2-ba2= 1a2×12ba log x-bax+ba+C ∴∫dxx2-a2=12a log x-ax+a+C= 12ab log ax-bax+b+C

Q3.

Answer :

∫dxa2x2+b2= 1a2∫dxx2+ba2 = 1a2×abtan-1xba+C ∴∫dxa2+x2=1atan-1xa+C= 1abtan-1axb+C

Q4.

Answer :

∫x2-1×2+4dx = ∫x2+4-4-1×2+4dx = ∫x2+4×2+4dx-5∫dxx2+22= ∫dx-5∫dxx2+22= x-52tan-1×2+C ∴∫dxx2+a2= 1atan-1xa+C

Q5.

Answer :

∫dx1+4×2= ∫dx1+2x2let 2x=t⇒2dx=dt⇒dx=dt2Now, ∫dx1+2×2= 12∫dt1+t2 = 12 log t+1+t2+C ∵ ∫dxx2+a2=log x+x2+a2+C= 12 log 2x+1+4×2+C

Q6.

Answer :

∫dxa2+b2x2= ∫dxb2a2b2+x2= 1b∫dxx2+ab2=1b log x+x2+a2b2+C= 1blog x+b2x2+a2b+C= 1blog bx+b2x2+a2b+C= 1blog bx+b2x2+a2-log b+C= 1b log bx+b2x2+a2-log bb+Clet C-log bb=C’= 1blog bx+b2x2+a2+C’

Q7.

Answer :

∫dxa2-b2x2= ∫dxb2a2b2-x2= 1b∫dxab2-x2= 1bsin-1xba+C

Q8.

Answer :

∫dx2-x2+1let 2-x=t⇒ -dx=dt⇒ dx=-dtNow,∫dx2-x2+1 =-∫dtt2+1= -log t+t2+1+C= -log 2-x+2-x2+1+C

Q9.

Answer :

∫dx2-x2-1let 2-x=t⇒ -dx=dt⇒ dx=-dtNow, ∫dx2-x2-1= ∫-dtt2-1= -log t+t2-1+C= -log 2-x+2-x2-1+C

Q10.

Answer :

∫x4+1×2+1dx= ∫x4-1+1+1×2+1dx= ∫x4-1×2+1+2×2+1dx= ∫x2-1×2+1×2+1+2×2+1dx= ∫x2-1+2×2+1dx= x33-x+2tan-1x+C

Page 19.79 Ex.19.10

Q1.

Answer :

∫dx4x2+12x+5= 14∫dxx2+3x+54= 14∫dxx2+3x+322-322+54= 14∫dxx+322-94+54= 14∫dxx+322-12let x+32=t⇒dx=dtNow, 14∫dxx+322-12= 14∫dxt2-12= 14×12×1 log t-1t+1+C= 18 log x+32-1x+32+1+C= 18 log x+12x+52+C= 18 log 2x+12x+5+C

Q2.

Answer :

∫dxx2-10x+34= ∫dxx2-10x+25-25+34= ∫dxx-52+9= ∫dxx-52+32let x-5=t⇒ dx=dtNow, ∫dxx-52+32= ∫dtt2+32= 13tan-1t3+C= 13tan-1x-53+C

Q3.

Answer :

∫dx1+x-x2= ∫-dxx2-x-1= ∫-dxx2-x+14-14-1= ∫-dxx-122-54= ∫dx54-x-122= ∫dx522-x-122let x-12=t⇒ dx=dtNow, ∫dx522-x-122= ∫dt522-t2= 12×52 log 52+t52-t+C
= 15 log 5+2t5-2t+C= 15 log 5+2x-125-2x-12+C= 15 log 5-1+2×5+1-2x+C

Q4.

Answer :

∫dx2x2-x-1= 12∫dxx2-x2-12= 12∫dxx2-x2+142-142-12= 12∫dxx-142-116-12= 12∫dxx-142-1+816= 12∫dxx-142-342let x-14=t⇒ dx=dt
Now, 12∫dxx-122-342= 12∫dtt2-342= 12∫dtt2-342= 12×34×12 log t-34t+34+C= 23×12 log x-14-34x-14+34+C= 23×12 log x-1x+12+C= 13 log 2x-12x+1+C= 13 log x-12x+1+log2+C= 13 log x-12x+1+13 log 2+C= 13 log x-12x+1+C’ ∵C’=13 log 2+C

Q5.

Answer :

∫dx4x2-4x+3= ∫dx4x2-4x+1-1+3= ∫dx2x2-2×2x×1+1+2=∫dx2x-12-22

= 12 tan-12x-12×12+C= 122 tan-12x-12+C

Q6.

Answer :

∫dxx2+6x+13= ∫dxx2+2×x×3+9-9+13= ∫dxx+32+22= 12 tan-1x+32+C

Page 19.82 Ex.19.11

Q1.

Answer :

∫sec2 x dx1-tan2 xlet tan x=t⇒ sec2 x dx=dtNow, ∫sec2 x dx1-tan2 x= ∫dt1-t2= 12 log 1+t1-t+C= 12 log 1+tanx1-tanx+C

Q2.

Answer :

∫exdx1+e2xlet ex=t⇒exdx=dtNow, ∫exdx1+e2x=∫dt1+t2= tan-1t+C= tan-1ex+C

Q3.

Answer :

∫cos x dxsin2 x+4sin x+5let sin x =t⇒ cos x dx=dtNow,∫cos x dxsin2 x+4sin x+5 = ∫dtt2+4t+5= ∫dtt2+2×t×2+4+1= ∫dtt+22+12= 11 tan-1t+21+C= tan-1sin x+2+C

Q4.

Answer :

∫ex dxe2x+5ex+6let ex=t⇒ ex dx=dtNow, ∫ex dxe2x+5ex+6=∫dtt2+5t+6= ∫dtt2+5t+522-522+6= ∫dtt+522-254+6= ∫dtt+522-25+244= ∫dtt+522-122= 12×12 log t+52-12t+52+12+C= log t+2t+3+C= log ex+2ex+3+C

Q5.

Answer :

∫e3x dx4e6x-9let e3x=t⇒ e3x×3dx=dt⇒ e3x dx=dt3Now, ∫e3x dx4e6x-9= 13∫dt4t2-9= 13∫dt2t2-32= 13×12×3 log 2t-32t+3×12+C= 136 log 2t-32t+3+C= 136 log 2e3x-32e3x+3+C

Q6.

Answer :

∫dxex+e-x= ∫dxex+1ex= ∫ex dxe2x +1let ex=t⇒ ex dx=dtNow, ∫ex dxe2x+1= ∫dt1+t2= tan-1t+c= tan-1ex+c

Q7.

Answer :

∫x dxx4+2×2+3let x2=t⇒ 2x dx=dt⇒ x dx =dt2Now, ∫x dxx4+2×2+3= 12∫dtt2+2t+3= 12∫dtt2+2t+1+2= 12∫dtt+12+22 = 12×12 tan-1t+12+C ∵∫dxx2+a2=1atan-1xa+C= 122 tan-1×2+12+C

Q8.

Answer :

∫3×51+x12dxlet x6=t⇒ 6×5 dx=dt⇒ x5 dx=dt6Now, ∫3×51+x12dx= 36∫dt1+t2= 12 tan-1t+C

= 12 tan-1×6+C

Q9.

Answer :

∫x2dxx6-a6let x3=t⇒ 3×2 dx=dt⇒ x2 dx=dt3Now, ∫x2dxx6-a6= 13∫dtt2-a32= 13×12a3 log t-a3t+a3+C= 16a3 log x3-a3x3+a3+C

Q10.

Answer :

∫x2x6+a6dx⇒ ∫x2dxx32+a32let x3=t⇒ 3x2dx=dt⇒ x2dx=dt3Now, ∫x2x6+a6dx= 13∫dtt2+a32= 13a3tan-1ta3+C= 13a3tan-1x3a3+C

Q11.

Answer :

∫dxxx6+1= ∫x5dxx6x6+1let x6=t⇒ 6x5dx=dt⇒ x5dx=dt6Now, ∫dxx6x6+1= 16∫dttt+1= 16∫dtt2+t= 16∫dtt2+t+14-14= 16∫dtt+122-122= 16×12×12 log t+12-12t+12+12+C= 16 log tt+1+C= 16 log x6x6+1+C

Q12.

Answer :

∫x dxx4-x2+1Let x2=t⇒ 2x dx=dt⇒ x dx=dt2Now, ∫x dxx4-x2+1=12∫dtt2-t+1= 12∫dtt2-t+122-122+1= 12∫dtt-122+34= 12∫dtt-122+322= 12×23tan-1t-1232+C= 13tan-12t-13+C= 13tan-12×2-13+C

Q13.

Answer :

∫x dx3x4-18×2+11let x2=t⇒ 2x dx=dt⇒ x dx=dt2Now, ∫x dx3x4-18×2+11= 12∫dt3t2-18t+11= 13×2∫dtt2-6t+113= 16∫dtt2-6t+9-9+113= 16∫dtt-32-163= 16∫dtt-32-432= 16×12×43 log t-3-43t-3+43+C= 348 log x2-3-43×2-3+43+C

Q14.

Answer :

∫exdx1+ex2+exlet ex=t= exdx=dt= ∫dt1+t2+t= ∫dt2+t+2t+t2= ∫dtt2+3t+2= ∫dtt2+3t+322-322+2= ∫dtt+322-94+2= ∫dtt+322-14= ∫dtt+322-122

= 12×12 log t+32-12t+32+12+C= log t+1t+2+C= log ex+1ex+2+C

Page 19.85 Ex.19.12

Q1.

Answer :

∫dx2x-x2= ∫dx2x-x2-1+1= ∫dx1-x2-2x+1= ∫dx1-x-12 = sin-1x-1+C ∵∫dxa2-x2=sin-1xa+C

Q2.

Answer :

∫dx8+3x-x2⇒ ∫dx8-x2-3x⇒ ∫dx8-x2-3x+322-322⇒ ∫dx8-x-322+94⇒ ∫dx4122-x-322⇒ sin-1x-32412+C⇒ sin-12x-341+C

Q3.

Answer :

∫dx5-4x-2×2= ∫dx252-2x-x2= 12∫dx52-2x-x2= 12∫dx52-x2+2x= 12∫dx52-x2+2x+1-1= 12∫dx52-x+12+1= 12∫dx72-x+12= 12∫dx722-x+12= 12sin-1x+127+C= 12sin-127x+1+C

Q4.

Answer :

∫dx3x2+5x+7=∫dx3x2+53x+73=13∫dxx2+53x+562-562+73=13∫dxx+562-2536+73=13∫dxx+562+-25+8436=13∫dxx+562+5936=13∫dxx+562+59362=13 log x+56+x+562+5936+C=13 log x+56+x2+53x+73+C

Q5.

Answer :

Let I=∫dxx-α β-x=∫dxβx-x2-αβ+αx=∫dx-x2+α+β x-αβ=∫dx-x2-α+β x+αβ=∫dx-x2-α+β x+α+β22-α+β22+αβ=∫dx-x-α+β22+α+β22-αβ=∫dx-x-α+β22+α+β2-4αβ4=∫dx-x-α+β22+α-β22=∫dxα-β22-x-α+β22=sin-1 x-α+β2α-β2+C=sin-1 2x-α-βα-β+C

Q6.

Answer :

∫dx7-3x-2×2=12∫dx72-32x-x2=12∫dx72-x2-32x=12∫dx722-x2+32x+342-342=12∫dx722-x+342+916=12∫dx72+916-x+342=12∫dx56+916-x+342=12∫dx6542-x+342=12 sin-1 x+34654+C=12sin-1 4x+365+C

Q7.

Answer :

∫dx16-6x-x2=∫dx16-x2+6x=∫dx16-x2+6x+32-32=∫dx16+9-x+32=∫dx52-x+32=sin-1 x+35+C

Q8.

Answer :

∫dx7-6x-x2=∫dx7-x2+6x=∫dx7-x2+6x+32-32=∫dx7+9-x+32=∫dx42-x+32=sin-1x+34+C

Q9.

Answer :

∫dx5x2-2x=∫dx5x2-25x=15∫dxx2-25x+152-152=15∫dxx-152-152=15 log x-15+x-152+152+C=15 log 5x-15+5×2-2×5+C

Page 19.89 Ex.19.13

Q1.

Answer :

∫x dxx4+a4=∫x dxx22+a22let x2=t⇒2x dx=dt⇒x dx=dt2Now, ∫x dxx22+a22=12∫dtt2+a22=12 log t+t2+a4+C=12 log x2+x4+a4+C

Q2.

Answer :

∫sec2x dx4+tan2 xlet tan x=t⇒sec2x dx=dtNow, ∫sec2x dx4+tan2 x=∫dt22+t2=log t+4+t2+C=log tan x+4+tan2x+C

Q3.

Answer :

∫ex dx16-ex2let ex=t⇒ex dx=dtNow, ∫ex dx16-ex2= ∫dt16-t2=∫dt42-t2=sin-1 t4+C=sin-1 ex4+C

Q4.

Answer :

∫cos x dx4+sin2xlet sin x=t⇒cos x dx=dtNow,∫cos x dx4+sin2x =∫dt22+t2=log t+4+t2+C=log sin x+4+sin2x+C

Q5.

Answer :

∫sin x dx4 cos2x-1let cos x=t⇒-sin x dx=dt⇒sin x dx=-dtNow, ∫sin x dx4 cos2x-1=∫-dt4t2-1=∫-dt4t2-14=-1 2∫dtt2-122=-12 log t+t2-14+C=-1 2 log t+4t2-12+C=-12 log 2t+4t2-12+C=-12log 2t+4t2-1-log 2+C=-12 log 2t+4t2-1+log 22+Clet C´=log 22+C=-12 log 2cost+4cos2t-1+C´

Q6.

Answer :

∫x dx4-x4⇒∫x dx22-x22let x2=t⇒2x dx=dt⇒x dx=dt2Now, ∫x dx22-x22=12∫dt22-t2=12×sin-1 12+C=12sin-1 x22+C

Q7.

Answer :

∫dxx4-9 log x2let log x=t⇒1x dx=dtNow, ∫dxx4-9 log x2=∫dt4-9t2=∫dt22-3t2=13 sin-1 3t2+C=13 sin-1 3 log x2+C

Q8.

Answer :

∫sin 8x dx9+sin4 4x⇒∫2 sin 4x·cos 4 x9+sin24x2dxlet sin2 4x=t⇒2 sin 4x·cos 4x × 4 dx=dt⇒2 sin 4x cos 4x dx=dt4Now, ∫2 sin 4x·cos 4 x9+sin24x2dx=14∫dt9+t2=14∫dt32+t2=14 log t+32+t2+C=14 log sin2 4x+9+sin4 4x+C

Q9.

Answer :

∫cos 2 x·dxsin2 2x+8let sin 2x=t⇒cos 2x×2·dx=dt⇒cos 2x·dx=dt2Now, ∫cos 2 x·dxsin2 2x+8 =12∫dtt2+222=12log t+t2+8+C=12 log sin 2x+sin2 2x+8+C

Q10.

Answer :

∫sin 2 x dxsin4x+4 sin2x-2let sin2x=t⇒2 sin x cos x dx=dt⇒sin 2 x dx=dtNow, ∫sin 2 x dxsin4x+4 sin2x-2=∫dtt2+4t-2=∫dtt2+4t+4-4-2=∫dtt+22-62=log t+2+t+22-6+C=log t+2+t2+4t-2+C=log sin2x+2+sin4x+4 sin2x-2+C

Page 19.90 Ex.19.13

Q11.

Answer :

∫sin 2 x dxcos4x-sin2 x+2⇒∫2 sin x cos x dxcos4x-1-cos2x+2⇒∫2 sin x cos xcos4x+cos2x+1Let cos2x =t⇒2 cos x ×-sin x dx=dtsin 2x dx=-dtNow, ∫sin 2 x dxcos4x-sin2 x+2=∫-dtt2+t+1=∫-dtt2+t+122-122+1=-∫dtt+122+34=-∫dtt+122+322=-log t+12+t+122+322+C=-log t+12+t2+t+1+C=-log cos2x+12+cos4x+cos2x+1+C

Q12.

Answer :

∫cos x dx4-sin2xlet sin x=t⇒cos x dx=dtNow, ∫cos x dx4-sin2x=∫dt4-t2=∫dt22-t2=sin-1 t2+C=sin-1 sin x2+C

Q13.

Answer :

∫dxx23x23-22=∫dxx23x132-22Let x13=t⇒13 x-23 dx=dt⇒13×23 dx=dt⇒dxx23=3 dtNow, ∫dxx23x23-22=3∫dtt2-22=3 log t+t2-22+C=3 log x13+x23-4+C

Q14.

Answer :

∫dx1-x2 9+sin-1 x 2let sin-1x=t⇒11-x2 dx=dtNow, ∫dx1-x2 9+sin-1 x 2 =∫dt9+t2=∫dt32+t2=log t+32+t2+C=log sin-1x+9+sin-1×2+C

Q15.

Answer :

∫cos x dxsin2x-2 sin x-3let sin x=t⇒cos x dx=dtNow, ∫cos x dxsin2x-2 sin x-3=∫dtt2-2t-3=∫dtt2-2t+1-1-3=∫dtt-12-22=log t-1+t-12-22+C=log t-1+t2-2t-3+C=log sin x-1+sin2x-2 sin x-3+C

Q16.

Answer :

∫cosec x-1 dx=∫1sin x-1dx=∫1-sin xsin xdx=∫1-sin x 1+sin xsin x 1+sin xdx=∫1-sin2xsin2x+sinxdx=∫cos x dxsin2x+sin xLet sin x=t⇒cos x dx=dtNow, ∫cos x dxsin2x+sin x=∫dtt2+t ∫dtt2+t=∫dtt2+t+122- 122=∫dtt+122-122=log t+12+t+122-122+C=log t+12+t2+t+C=log sin x+12+sin2x+sinx+C

Q17.

Answer :

∫sin x-cosxsin 2x dx=∫sin x-cos x1+sin 2x-1dx=∫sin x-cos xsin2x+cos2x+2 sin x cos x-1dx=∫sin x-cos xsin x+cos x2-1dxlet sin x+cos x=t⇒cos x-sin x dx=dt⇒sin x-cos xdx=-dtNow, ∫sin x-cos xsin x+cos x2-1dx=-∫dtt2-12=-log t+t2-1+C=-log sin x+cos x+sin x+cos x2-1+C=-log sin x+cos x+sin2 x+cos2 x +2sinx.cosx-1+C=-log sin x+cos x+sin 2x+C

Page 19.94 Ex.19.14

Q1.

Answer :

∫xx2+3x+2dxx=A ddxx2+3x+2+Bx=A 2x+3+Bx=2 Ax+3A+B

Comparing the Coefficients of like powers of x we get
2A=1⇒A=123A+B=032+B=0B=-32x=12 2x+3-32

Now, ∫xx2+3x+2dx=∫122x+3-32×2+3x+2dx=12∫2x+3dxx2+3x+2-32∫dxx2+3x+2=12∫2x+3dxx2+3x+2-32∫dxx2+3x+322- 322+2=12∫2x+3dxx2+3x+2 -32∫dxx+322-94+2=12∫2x+3 dxx2+3x+2-32∫dxx+322-122=12 log x2+3x+2-32×12×12 log x+32-12x+32+12+C=12 log x2+3x+2-32 log x+1x+2+C

Q2.

Answer :

∫x+1 dxx2+x+3x+1=Addxx2+x+3+Bx+1=A 2x+1+Bx+1 =2 Ax+A+B

Comparing Coefficients of like powers of x
2A=1A=12A+B=112+B=1B=12x+1=12 2x+1+12

Now, ∫x+1 dxx2+x+3=∫12 2x+1dxx2+x+3+12∫dxx2+x+3=12∫2x+1dxx2+x+3+12∫dxx2+x+122- 122+3=12∫2x+1dxx2+x+3 +12∫dxx+122+3 -14=12∫2x+1 dxx2+x+3+12∫dxx+122+1122=12 log x2+x+3+12×211 tan-1 x+12112+C=12 log x2+x+3+111 tan-1 2x+111+C

Q3.

Answer :

∫x-3×2+2x-4dxx-3=Addxx2+2x-4+Bx-3=A 2x+2+Bx-3 =2 A x+2A+B

Comparing Coefficients of like powers of x

2A=1A=122A+B=-32×12+B=-3B=-4

Now, ∫x-3×2+2x-4dx=∫122x+2 -4×2+2x-4dx=12∫2x+2 dx x2+2x-4-4∫dxx2+2x+1-1-4=12∫2x+2 dxx2+2x-4-4∫dxx+12- 52=12 log x2+2x-4-425 log x+1-5x+1+5+C=12 log x2+2x-4-25 log x+1-5x+1+5+C

Q4.

Answer :

∫2x-3 dxx2+6x+132x-3=Addxx2+6x+13+B2x-3=A 2x+6+B2x-3 =2 A x+6A+B

Comparing Coefficients of like powers of x
2A=2A=16 A+B=-36+B=-3B=-9∴ 2x-3=1 2x+6-9

∴ ∫2x-3×2+6x+13dx=∫2x+6-9 x2+6x+13dx=∫2x+6×2+6x+13dx -∫9 dxx2+6x+13=∫2x+6 dxx2+6x+13-9∫dxx2+6x+32-32+13=∫2x+6 dxx2+6x+13-9∫dxx+32+22=log x2+6x+13-9×12 tan-1 x+32+C=log x2+6x+13-92 tan-1 x+32+C

Q5.

Answer :

∫x-13×2-4x+3dxx-1=Addx3x2-4x+3+Bx-1=A 6x-4+Bx-1 =6 A x+B-4 A

Comparing the Coefficients of like powers of x
6 A=1A=16B-4 A=-1B-4×16=-1B=-1+23B=13

Now, ∫x-1 dx3x2-4x+3=∫166x-4+133×2-4x+3dx=16∫6x-4 dx3x2-4x+3+13∫dx3x2-4x+3=16∫6x-4 dx3x2-4x+3+19∫dxx2-43x+1=16∫6x-4 dx3x2-4x+3+19∫dxx2-43x+232 232+1=16∫6x-4 dx3x2-4x+3+19∫dxx-232-49+1=16∫6x-4 dx3x2-4x+13+19∫dxx-232+532=16 log 3×2-4x+3+19×35 tan-1 x-2353+C=16 log 3×2-4x+3+135 tan-1 3 x-25+C=16 log 3×2-4x+3+515 tan-1 3x-25+C

Q6.

Answer :

∫2x dx2+x-x22x=Addx2 +x-x2+B2x=A 0+1-2x+B2x=-2 A x+A+B

Comparing the Coefficients of like powers of x
-2 A=2A=-1A+B=0-1+B=0B=1

Now, ∫2x dx2+x-x2=∫-11-2x+1-x2+x+2dx=-∫1-2x-x2+x+2dx+∫dx-x2+x+2=-I1+I2 … 1 say whereI1=∫1-2x-x2+x+2dxI2=∫dx-x2+x+2I1=∫1-2x-x2+x+2dxlet -x2+x+2=t⇒1-2x dx=dtI1=∫dttI1=log t+C1=log 2+x-x2+C1 … 2I2=∫dx-x2+x+2I2=∫-dxx2-x-2I2=∫-dxx2-x+122-122-2I2=∫-dxx-122-322I2=-12×32log x-12-32x-12+32+C2I2=-13 log x-2x+1+C2 … 3from 1 2 and 3∫2×2+x-x2dx=-log 2+x-x2-13log x-2x+1+C1+C2=-log 2+x-x2+13 log 1+xx-2+Cwhere C =C1+C2

Q7.

Answer :

∫1-3x dx3x2+4x+21-3x=Addx3x2+4x+2+B1-3x=A 6x+4+B1-3x=6 A x+4 A+B

Comparing the Coefficients of like powers of x

6 A=-3A=-124 A+B=14×-12+B=1B=3

1-3x=-126x+4+3Now, ∫1-3x dx3x2+4x+2=∫-126x+4+33×2+4x+2dx=-12∫6x+4 dx3x2+4x+2+3∫dx3x2+4x+2=-12 I1+3I2 say … 1whereI1=∫6x+43×2+4x+2 and I2=∫dx3x2+4x+2I1=∫6x+43×2+4x+2dxlet 3×2+4x+2=t⇒6x+4 dx=dtI1=∫dtt=log t+C1=log 3×2+4x+2+C1 … 2I2=∫dx3x2+4x+2I2=13∫dxx2+43x+23I2=13∫dxx2+4xx+232-232+23I2=13∫dxx-232-49+23I2=13∫dxx+232+29I2=13∫dxx+232+232I2=13×32 tan-1 x+2323+C2I2=12 tan-1 3x+22+C2 … 3from 1, 2 and 3∫1-3x dx3x2+4x+2=-12 log 3×2+4x+2+3×12 tan-1 3x+22+C1+C2=-12 log 3×2+4x+2+32 tan-1 3x+22+C Where C=C1+C2

Q8.

Answer :

∫2x+5 dxx2-x-22x+5=Addxx2-x-2+B2x+5=A 2x-1+B2x+5=2 A x+B-A

Comparing the Coefficients of like powers of x

2 A=2A=1B-A=5B-1=5B=6

∴ 2x+5=1·2x-1+6∴ ∫2x+5×2-x-2dx⇒∫2x-1+6×2-x-2dx⇒∫2x-1×2-x-2dx+6∫dxx2-x-2=I1+6 I2 say … 1whereI1=∫2x-1×2-x-2dx I2=∫dxx2-x-2I1=∫2x-1×2-x-2dxlet x2-x-2=t⇒2x-1 dx=dtI1=∫dttI1=log tI1=log x2-x-2+C1 … 2I2=∫dxx2-x-2I2=∫dxx2-x+122- 122-2I2=∫dxx-122-14-2I2=∫dxx-122-322I2=12×32 log x-12-32x-12+32I2=13 log x-2x+1+C2 … 3∫2x+5 dxx2-x-2=log x2-x-2+63 log x-2x+1+C1+C2=log x2-x-2+2 log x-2x+1+C Where C=C1+C2

Q9.

Answer :

∫3 sin x-2 cos x dx5-cos2x-4 sin x=∫3 sin x-2 cos x dx5-1-sin2x-4 sin x=∫3 sin x-2 cos x dx sin2x-4 sin x+4Let sin x=t⇒cos x dx=dt∫3t-2 dtt2-4t+43t-2=Addxt2-4t+4+B3t-2=A 2t-4+B3t-2=2 A t+B-4 A

Comparing the Coefficients of like powers of t

2 A=3A=32B-4 A=-2B-4×32=-2B=-2+6B=4

3t-2=32 2t-4+4∴ ∫3t-2 dtt2-4t+4=∫322t-4+4t2-4t+4dt=32∫2t-4t2-4t+4dt+4∫dtt2-4t+4=32 I1+4 I2 … 1whereI1=∫2t-4 dtt2-4t+4, I2=∫dtt2-4t+4I1=∫2t-4 dtt2-4t+4Let t2-4t+4=p⇒2t-4 dt=dpI1=∫2t-4 dtt2-4t+4=∫dpp=log p+C1=log t2-4t+4+C1 … 2I2=∫dtt2-4t+4I2=∫dtt-22I2=∫t-2-2 dtI2=t-2-2+1-2+1+C2I2=-1t-2+C2 … 3from 1, 2 and 3∫3 sin x-2 cos x dx5-cos2x-4 sinx=32 log t2-4t+4+4×-1t-2+C1+C2=32 log sin2x-4 sin x+4+42-t+C Where C=C1+C2=32log sin x-22+42-sin x+C=32×2 log sin x-2+42-sin x+C=3 log 2-sin x+42-sin x+C

Q10.

Answer :

∫ax3+bx x4+c2dx=∫ax3 x4+c2dx+∫bx x22+c2dx=I1+I2 sayWhereI1=∫ax3 x4+c2dx & I2=∫bx x22+c2dxNow, I1=∫ax3 x4+c2dxlet x4+c2=t⇒4×3 dx=dt⇒x3 dx=dt4I1=a4∫dtt=a4 log t+C1=a4 log x4+c2+C1Now, I2=∫bx x22+c2dxlet x2=p⇒2x dx=dp⇒x dx=dp2

I2=b2∫dpp2+c2=b2×1c tan-1 pc+C2=b2c tan-1 x2c+C2∫ax3+bx x4+c2dx=a4 log x4+c2+b2c tan-1 x2c+C1+C2 =a4 log x4+c2+b2c tan-1 x2c+C Where C=C1+C2

Q11.

Answer :

∫x+22×2+6x+5dxx+2=Addx2x2+6x+5+Bx+2=A 4x+6+Bx+2=4 A x+6 A+B

Comparing the Coefficients of like powers of x

4 A=1A=146 A+B=26×14+B=2B=12

∴ ∫x+22×2+6x+5dx=∫144x+6+122×2+6x+5dx=14∫4x+62×2+6x+5dx+12∫12×2+6x+5dx=14∫4x+62×2+6x+5dx+14∫dxx2+3x+52=14∫4x+62×2+6x+5dx+14∫dxx2+3x+322-322+52=14∫4x+62×2+6x+5dx+14∫dxx+322-94+52=14∫4x+62×2+6x+5dx+14∫dxx+322+14=14∫4x+62×2+6x+5dx+14∫dxx+322+122=14 log 2×2+6x+5+14×2 tan-1 x+3212+C=14 log 2×2+6x+5+12 tan-1 2x+3+C

Page 19.96 Ex.19.15

Q1.

Answer :

∫x2+x+1×2-xdxx2+x+1×2-x=1+2x+1×2-x∴∫x2+x+1×2-xdx=∫1+2x+1×2-xdx=∫1+2x-1+2×2-xdx=∫dx+∫2x-1 dxx2-x+∫2 dxx2-x+122-122=∫dx+∫2x-1 dxx2-x+2∫dxx-122-122=x+log x2-x+2×12×12logx-12-12x-12+12=x+log x2-x+2 log x-1x+C

Q2.

Answer :

∫x2+x-1×2+x-6dxx2+x-1×2+x-6=1+5×2+x-6 ∫x2+x-1×2+x-6dx=∫dx+5∫dxx2+x-6=∫dx+5∫dxx2+x+122-122-6=∫dx+5∫dxx+122-14-6=∫dx+5∫dxx+122-522=x+5×12×52 log x+12-52x+12+52+C=x+log x-2x+3+C

Q3.

Answer :

We have,I=∫1-x2 x 1-2xdx=∫-x2+1-2×2+xdx=∫12dx+∫1-x2-2×2+xdx=12∫dx+12∫2-x-2×2+xdx=12∫dx+∫2-x-2×2+xdx=12I1+I2 saywhere I1=∫dx & I2=∫2-x-2×2+xdxNow, I1=∫dx =x+C1I2=∫2-x-2×2+xdxLet 2-x=A ddx -2×2+x+B⇒2-x=A -4x+1+B⇒2-x=-4Ax+A+B

Comparing coefficients of like terms

-1=-4 A ⇒A=14& A+B=2⇒14+B=2⇒B=2-14 =8-14 =74

∴∫2-x-2×2+xdx=∫14-4x+1+74-2×2+xdx =∫14-4x+1-2×2+xdx+∫74-2×2+xdx =14log -2×2+x+74∫1-2×2-x2+116-116dx =14log -2×2+x-78∫1x-142-142dx =14log x-2x+1-78×12×14logx-14-14x-14+14+C2 =14log x+14log -2x+1-74logx-12x+C2 =14log x+14log -2x+1-74log x-12+74log x+C2 =2 log x+14log -2x+1-74log 2x-1+C3 , where C3=C2+74log 2 =2 log x+14log -2x+1-74log 1-2x+C3 =2 log x-32log 1-2x+C3Thus, I= 12x+C1+2 log x-32log 1-2x+C3 =12x+log x-34log 1-2x+C, where C=12C1+C3

Q4.

Answer :

Let I∫x2+1×2-5x+6dxDividing Numerator by Denominatorx2-5x+6×2+1 1 x2-5x+6 – + – 5x-5×2+1×2-5x+6=1+5x-5×2-5x+6 ….. 1Also 5x-5×2-5x+6=5x-5x-2 x-3Let 5x-5x-2 x-3=Ax-2+Bx-3⇒5x-5x-2 x-3=A x-3 +B x-2x-2 x-3⇒5x-5=A x-3+B x-2let x=35×3-5=A×0+B 3-210=Blet x=25×2-5=A 2-3+B×0A=-55x-5x-2 x-3=-5x-2+10x-3 …..2from 1 and 2I=∫dx-5∫dxx-2+10∫dxx-3=x-5 log x-2+10 log x-3+C

Q5.

Answer :

Let I=∫x2x2+7x+10dxNow,x2+7x+10×2 1 x2+7x+10 – – – -7x-10 ∴x2x2+7x+10=1-7x+10×2+7x+10⇒x2x2+7x+10=1-7x+10×2+2x+5x+10x2x2+7x+10=1-7x+10x x+2 +5 x+2x2x2+7x+10=1- 7x+10x+2 x+5 ….. 1Consider,7x+10x+2 x+5=Ax+2+Bx+57x+10=A x+5+B x+2let x+5=0x=-5⇒7 -5+10=A×0+B -5+2-25=B -3⇒B=253let x+2=0x=-27 -2+10=A -2+5⇒-4=A 3⇒A=-437x+10x+2 x+5=-43 x+2+253 x+5 …..2from 1 and 2x2x2+7x+10=1+43 x+2-253 x+5⇒∫x2 dxx2+7x+10=∫dx+43∫dxx+2-253∫dxx+5=x+43 log x+2-253 log x+5+C

Q6.

Answer :

Let I=∫x2+x+1×2-x+1dxNow,x2-x+1×2+x+1 1 x2-x+1 – + – 2x Therefore,x2+x+1×2-x+1=1+2xx2-x+1⇒∫x2+x+1×2-x+1 dx=∫dx+∫2x-1+1×2-x+1 dx=∫dx+∫2x-1×2-x+1 dx+∫dxx2-x+1=∫dx+∫2x-1 dxx2-x+1+∫dxx2-x+122-122+1=∫dx+∫2x-1 dxx2-x+1+∫dxx-122+322=x+ log x2-x+1+23 tan-1 2x-13+C

Q7.

Answer :

Let I=∫x-12×2+2x+2 dx=∫x2-2x+1×2+2x+2 dxHere,x2+2x+2×2-2x+1 1 x2+2x+2 – – – -4x-1 Therefore,x2-2x+1×2+2x+2=1-4x+1×2+2x+2 ….. 1Let 4x+1=Addx x2+2x+2+B4x+1=A 2x+2+B4x+1=2A x+2A+BEquating Coefficients of like terms2A=4A=22A+B=12×2+B=1B=-3∫x2-2x+1×2+2x+2 dx=∫dx-2∫2x+2×2+2x+2 dx+3∫dxx2+2x+2=∫dx-2∫2x+2×2+2x+2 dx+3∫dxx+12+12=x-2 log x2+2x+2+31 tan-1 x+11+C=x-2 log x2+2x+2+3 tan-1 x+1+C

Q8.

Answer :

Let I=∫x3+x2+2x+1×2-x+1 dxx2-x+1×3+x2+2x+1 x+2 x3-x2+x – + – 2×2+ x+1 2×2-2x+2 – + – 3x-1Therefore,x3+x2+2x+1×2-x+1=x+2+3x-1×2-x+1 ….. 1Let3x-1=Addx x2-x+1+B3x-1=A 2x-1+B3x-1=2A x+B-AEquating Coefficients of like terms2A=3A=32B-A=-1B-32=-1B=12∫x3+x2+2x+1×2-x+1 dx =∫x+2 dx+∫32 2x-1+12×2-x+1 dx=∫x+2 dx+32 ∫2x-1×2-x+1 dx+12∫dxx2-x+1=∫x+2 dx+32∫2x-1 dxx2-x+1+12∫dxx2-x+14-14+1=∫x+2 dx+32∫2x-1 dxx2-x+1 +12∫dxx-122+322=x22+2x+32 logx2-x+1+12×23 tan-1 x-1232+C=x22+2x+32 log x2-x+1+13 tan-1 2x-13+C

Q9.

Answer :

Let I=∫x2 x4+4×2+4 dx=∫x6+4x2x2+4 dxNow,x2+4×6+4×2 x4-4×2+20 x6+4×4 – – -4×4+4×2 -4×4-16×2 + + 20×2 20×2 + 80 – – -80 Therefore, x2 x4+4×2+4=x4-4×2+20-80×2+4I=∫x2 x4+4×2+4 dx=∫x4-4×2+20 dx-80∫dxx2+ 22=∫x4 dx-4∫x2 dx+20∫dx-80∫dxx2+22=x4+14+1-4 x33+20 x-80×12 tan-1 x2+C=x55-43 x3+20x-40 tan-1 x2+C

Q10.

Answer :

Let I=∫x2 dxx2+6x+12Now,x2+6x+12×2 1 x2+6x+12 – – – -6x-12Therefore,x2x2+6x+12=1-6x+12×2+6x+12 ….. 1Let 6x+12=Addx x2+6x+12+B⇒6x+12=A 2x+6+B⇒6x+12=2A x+6A+BEquating Coefficients of like terms2A=6A=36A+B=1218+B=12B=-6∴x2x2+6x+12=1-3 2x+6-6×2+6x+12I=∫x2 dxx2+6x+12=∫dx-3∫2x+6 dxx2+6x+12+6∫dxx2+6x+12=∫dx-3 ∫2x+6 dxx2+6x+12+6∫dxx2+6x+9+3=∫dx-3∫2x+6 dxx2+6x+12+6∫dxx+32+32=x-3 log x2+6x+12+63 tan-1 x+33+C=x-3 log x2+6x+12+23 tan-1 x+33+C

Page 19.99 Ex.19.16

Q1.

Answer :

Let I=∫x dxx2+6x+10x=A ddx x2+6x+10+Bx=A 2x+6+Bx=2A x+6A+BEquating Coefficients of like terms2A=1A=126A+B=06×12+B=0B=-3I=∫x dxx2+6x+10=∫12 2x+6-3×2+6x+10dx=12∫2x+6 dxx2+6x+10-3∫dxx2+6x+32-32+10=12∫2x+6 dxx2+6x+10-3∫dxx+32+12let x2+6x+10=t⇒2x+6 dx=dtI=12∫dtt-3∫dxx+32+1=12×2t-3 log x+3+x+32+1+C=t-3 log x+3+x2+6x+10+C=x2+6x+10-3 log x+3+x2+6x+10+C

Q2.

Answer :

Let I=∫2x+1 dxx2+2x-1=∫2x+2-1 dxx2+2x-1=∫2x+2 dxx2+2x-1-∫dxx2+2x-1=∫2x+2 dxx2+2x-1-∫dxx2+2x+1-1-1=∫2x+2 dxx2+2x-1-∫dxx+12-22let x2+2x-1=t⇒2x+2 dx=dtI=∫dtt-∫dxx+12-22=2t-log x+1+x+12-22+C=2×2+2x-1-log x+1+x2+2x-1+C

Q3.

Answer :

Let I=∫x+1 dx4+5x-x2Also, x+1=A ddx 4+5x-x2+Bx+1=A 5-2x+Bx+1=-2A x+5A+BEquating Coefficients of like terms-2A=1⇒A=-12And5A+B=1⇒-52+B=1B=72I=∫x+1 dx4+5x-x2=∫-12 5-2x+724+5x-x2dx=-12∫5-2x dx4+5x-x2+72∫dx4-x2-5x=-12∫5-2x dx4+5x-x2+72∫dx4-x2-5x+522-522=-12∫5-2x dx4+5x-x2+72∫dx4-x-522+254=-12∫5-2×4+5x-x2dx+72∫dx414-x-522=-12∫5-2×4+5x-x2dx+72∫dx4122-x-522let 4+5x-x2=t⇒5-2x dx=dtThen,I=-12∫dtt+72∫dx4122-x-522=-12×2t+72×sin-1 x-52412+C=-t+72 sin-1 2x-541+C=-4+5x-x2+72 sin-1 2x-541+C

Q4.

Answer :

Let I=∫6x-53×2-5x+1dxPutting 3×2-5x+1=t⇒6x-5 dx=dtThen,I=∫dtt=2t+C=23×2-5x+1+C

Q5.

Answer :

Let I=∫3x+1 dx5-2x-x2Consider, 3x+1=A ddx 5-2x-x2+B⇒3x+1=A -2-2x+B⇒3x+1=-2A x-2A+BEquating Coefficients of like terms-2A=3⇒A=-32And-2A+B=1⇒-2×-32+B=1⇒B=1-3⇒B=-2∴I=∫-32 -2-2x-25-2x-x2dx=-32∫-2-2x dx5-2x-x2-2∫dx5-2x-x2=-32∫-2-2x dx5-2x-x2-2∫dx5-x2+2x=-32∫-2-2x dx5-2x-x2-2∫dx5-x2+2x+1-1=-32∫-2-2x dx5-2x-x2-2∫dx6-x+12=-32∫-2-2x dx5-2x-x2-2∫dx62-x+12let 5-2x-x2=t⇒-2-2x dx=dt∴I=-32∫dtt-2∫dx62-x+12=-32×2t-2 sin-1 x+16+C=-35-2x-x2-2 sin-1 x+16+C

Q6.

Answer :

Let I=∫x dx8+x-x2Consider, x=Addx 8+x-x2+Bx=A 1-2x+Bx=-2A x+A+BEquating Coefficients of like terms-2A=1⇒A=-12AndA+B=0⇒-12+B=0⇒B=12∴x=-12 1-2x+12Then,I=-12∫1-2x dx8+x-x2+12∫dx8+x-x2=-12∫1-2x dx8+x-x2+12∫dx8-x2-x=-12∫1-2x dx8+x-x2+12∫dx8-x2-x+14-14=-12∫1-2x dx8+x-x2+12∫dx8+14-x-122=-12∫1-2x dx8+x-x2+12∫dx3322-x-122let 8+x-x2=t⇒1-2x dx=dt∴I=-12∫dtt+12∫dx3322-x-122=-12×2t+12 sin-1 x-12332+C=-t+12 sin-1 2x-133+C=-8+x-x2+12 sin-1 2x-133+C

Q7.

Answer :

Let I=∫x+2 dxx2+2x-1Consider,x+2=A ddx x2+2x-1+B⇒x+2=A 2x+2+B⇒x+2=2A x+2A+BEquating Coefficients of like terms2A=1⇒A=12And2A+B=2⇒2×12+B=2⇒B=1Then,I=∫12 2x+2+1×2+2x-1dx=12∫2x+2 dxx2+2x-1+∫dxx2+2x-1let x2+2x-1=t⇒2x+2 dx=dt∴I=12∫dtt+∫dxx2+2x-1=12∫t-12dt+∫dxx2+2x+1-2=12 t-12+1-12+1+∫dxx+12-22=t+ log x+1+x+12-22+C=x2+2x-1+ log x+1+x2+2x-1+C

Q8.

Answer :

Let I=∫x+2×2-1dx=∫xx2-1dx+2∫dxx2-1let x2-1=t⇒2x dx=dt⇒x dx=dt2Then,I=12∫dtt+2∫dxx2-12=12∫t-12 dt+2∫dxx2-12=12 t-12+1-12+1+2 log x+x2-1+C=t+2 log x+x2-1+C=x2-1+2 log x+x2-1+C

Q9.

Answer :

Let I=∫x-1×2+1 dx=∫x dxx2+1-∫dxx2+1Putting x2+1=t⇒2x dx=dt⇒x dx=dt2Then,I=12∫dtt-∫dxx2+12=12∫t-12dt-∫dxx2+12=12 t-12+1-12+1-∫dxx2+12=t- log x+x2+1+C=x2+1- log x+x2+1+C

Q10.

Answer :

Let I=∫x dxx2+x+1Consider,x=A ddx x2+x+1+B⇒x=A 2x+1+B⇒x=2A x+A+BEquating Coefficient of like terms2A=1⇒A=12AndA+B=0⇒12+B=0⇒B=-12∴I=∫12 2x+1-12×2+x+1 dx=12∫2x+1×2+x+1dx-12∫dxx2+x+14-14+1Putting x2+x+1=t⇒2x+1 dx=dtThen,I=12∫dtt-12∫dxx+122+322=12∫t-12 dt-12 log x+12+x+122+322+C=12t-12+1-12+1-12 log x+12+x2+x+1+C=t-12 log x+12+x2+x+1+C=x2+x+1-12 log x+12+x2+x+1+C

Q11.

Answer :

Let I=∫x+1×2+1 dx=∫x dxx2+1+∫dxx2+1Putting, x2+1=t⇒2x dx=dt⇒x dx=dt2Then,I=12∫dtt+∫dxx2+1=12∫t-12dt+∫dxx2+1=12 t-12+1-12+1+log x+x2+1+C=t+ log x+x2+1+C=x2+1+ log x+x2+1+C

Q12.

Answer :

Let I=∫2x+5 dxx2+2x+5Consider,2x+5=A ddx x2+2x+5+B⇒2x+5=A 2x+2+B⇒2x+5=2A x+2A+BEquating Coefficients of like terms2A=2⇒A=1And 2A+B=5⇒B=3∴I=∫2x+2+3×2+2x+5 dx=∫2x+2 dxx2+2x+5+3∫dxx2+2x+5let x2+2x+5=t⇒2x+2 dx=dtThen,I=∫dtt+3∫dxx2+2x+1+4=∫t-12 dt+3 ∫dxx+12+22=t-12+1-12+1+3 log x+1+x+12+4+C=2t+3 log x+1+x2+2x+5+C=2×2+2x+5+3 log x+1+x2+2x+5+C

Q13.

Answer :

Let I=∫3x+1 dx5-2x-x2Consider,3x+1=A ddx 5-2x-x2+B⇒3x+1=A -2-2x+B⇒3x+1=-2A x+-2A+BEquating Coefficients of like terms-2A=3⇒A=-32And-2A+B=1⇒-2×-32+B=1⇒B=-2∴I=∫-32 -2-2x-25-2x-x2 dx=-32∫-2-2x dx5-2x-x2-2∫dx5-2x-x2=-32∫-2-2x dx5-2x-x2-2∫dx5-x2+2x=-32∫-2-2x dx5-2x-x2-2∫dx5-x2+2x+1-1=-32∫-2-2x dx5-2x-x2-2 ∫dx6-x+12=-32∫-2-2x dx5-2x-x2-2∫dx62-x+12Putting, 5-2x-x2=t⇒-2-2x dx=dtThen,I=-32∫dtt-2 sin-1 x+16+C1=-32×2t-2 sin-1 x+16+C=-35-2x-x2-2 sin-1 x+16+C

Q14.

Answer :

Let I=∫1-x1+x dx=∫1-x1-x1+x1-x dx=∫1-x1-x2 dx=∫dx1-x2-∫x dx1-x2Putting 1-x2=t⇒-2x dx=dt⇒x dx =-dt2Then,I=∫dx1-x2+12∫dtt=sin-1 x+12×2t+C=sin-1 x+1-x2+C

Q15.

Answer :

Let I=∫2x+1 dxx2+4x+3Consider,2x+1=A ddx x2+4x+3+B⇒2x+1=A 2x+4+B⇒2x+1=2A x+4A+BEquating Coefficients of like terms2A=2⇒A=1And4A+B=1⇒4+B=1⇒B=-3∴I=∫2x+4-3×2+4x+3dx=∫2x+4 dxx2+4x+3 -3∫dxx2+4x+4-4+3=∫2x+4 dxx2+4x+3-3∫dxx+22-12Let x2+4x+3=t⇒2x+4 dx=dtThen,I=∫dtt-3∫dxx+22-12=∫t-12 dt-3 ∫dxx+22-12=t-12+1-12+1-3 log x+2+x+22-1+C=2t-3 log x+2+x2+4x+3+C=2×2+4x+3-3 log x+2+x2+4x+3+C

Q16.

Answer :

Let I=∫2x+3 dxx2+4x+5=∫2x+4-1×2+4x+5dx=∫2x+4 dxx2+4x+5-∫dxx2+4x+5=∫2x+4 dxx2+4x+5-∫dxx+22+1Consider, x2+4x+5=t⇒2x+4 dx=dt∴I=∫dtt-∫dxx+22+12=∫t-12 dt-∫dxx+22+12=t-12+1-12+1- log x+2+x+22+1+C=2×2+4x+5- log x+2+x2+4x+5+C

Q17.

Answer :

Let I=∫5x+3 dxx2+4x+10Consider,5x+3=A ddx x2+4x+10+B⇒5x+3=A 2x+4+B⇒5x+3=2A x+4A+BEquating Coefficients of like terms2A=5⇒A=52And4A+B=3⇒4×52+B=3⇒B=-7∴I=52∫2x+4 dxx2+4x+10-7∫dxx2+4x+10=52∫2x+4 dxx2+4x+10 -7∫dxx2+4x+4-4+10=52∫2x+4 dxx2+4x+10-7∫dxx+22+62Putting, x2+4x+10=t⇒2x+4 dx=dtThen,I=52∫dtt-7 log x+2+x+22+6+C=52∫t-12 dt-7 log x+2+x2+4x+10+C=52×2t-7 log x+2+x2+4x+10+C=5×2+4x+10-7 log x+2+x2+4x+10+C

Page 19.103 Ex.19.17

Q1.

Answer :

Let I =∫ 14 cos2 x+9 sin2 xdxDividing numerator and denominator by cos2 x⇒I=∫ 1cos2 x4+9 tan2 xdx =∫ sec2 x 4+9 tan2 xdxLet tan x=t⇒sec2 x dx=dt∴I =∫ dt4+9t2 =19∫ dt49+t2 =19∫ dt232+t2 =19×32tan-1 t23+C =16tan-1 3t2+C =16tan-1 3 tan x2+C

Q2.

Answer :

Let I=∫14 sin2 x+5 cos2 xdxDividing numerator & denominator by cos2 x⇒I=∫sec2 x 4 tan2 x+5dxLet tan x=t⇒sec2 x dx=dt∴I=∫ dt4t2+5 =14∫ dtt2+54 =14∫dtt2+522 =14×25 tan-1 t5×2+C =125tan-1 2 tan x5+C

Q3.

Answer :

Let I=∫ 2 2+sin 2xdx =∫ 2 2+2 sin x cos xdx =∫ 11+sin x cos xdxDividing numerator and denominator by cos2 x⇒I=∫ sec2 x dxsec2 x+tan x =∫ sec2 x dx1+tan2 x+tan xLet tan x=t⇒sec2 x dx=dt∴I=∫ dtt2+t+1 =∫dtt2+t+14-14+1 =∫ dtt+122+322 =23tan-1 t+1232+C =23tan-1 2t+13+C =23tan-1 2 tan x+13+C

Q4.

Answer :

Let I=∫ cos x cos 3xdx =∫cos x4 cos3x-3 cos xdx cos 3A=4 cos3 A-3 cos A =∫ 14 cos2 x-3dxDividing numerator and denominator by cos2 x⇒I=∫ sec2 x4-3 sec2 x dx =∫ sec2 x4-31+tan2 x dx =∫ sec2 x1-3 tan2 x dx =∫ sec2 x 1-3 tan x2 dxLet 3 tan x=t⇒3 sec2 x dx=dt⇒sec2 x dx=dt3∴I=13 ∫ dt12-t2 =13×12ln 1+t1-t+C =123ln 1+3 tan x1-3 tan x+C

Q5.

Answer :

Let I=∫ 11+ 3 sin2 xdxDividing numerator and denominator by cos2 x⇒I=∫ sec2 xsec2 x+3 tan2 xdx =∫ sec2 x 1+tan2 x+3 tan2 xdx =∫ sec2 x 1+4 tan2 xdx =∫ sec2 x 1+2 tan x2dxLet 2 tan x=t⇒2 sec2 x dx=dt⇒sec2 x dx=dt2∴I=12∫ dt1+t2 =12 tan-1 t+C =12 tan-1 2 tan x+C

Q6.

Answer :

Let I=∫ 13+2 cos2 xdxDividing numerator and denominator by cos2 x⇒I=∫ sec2 x3 sec2 x+2 dx =∫ sec2 x 3 1+tan2 x+2dx =∫ sec2 x 3 tan2 x+5dx =∫ sec2 x 52+3 tan x2dxLet 3 tan x=t⇒3 sec2 x dx=dt⇒sec2 x dx=dt3∴I=13∫ dt52+t2 =13×15 tan-1 t5+C =115 tan-1 3 tan x5+C

Q7.

Answer :

Let I=∫ 1sin x-2 cos x 2 sin x+cos xdxDividing numerator and denominator by cos2 x⇒I=∫ sec2 x sin x-2 cos xcos x×2 sin x+cos xcos xdx =∫ sec2 x tan x-2 2 tan x+1dxLet tan x=t⇒sec2 x dx=dt∴I=∫ dtt-2 2t+1 =∫ dt2t2+t-4t-2 =∫ dt2t2-3t-2 =12∫ dtt2-32t-1 =12∫ dtt2-32t+342-342-1 =12∫ dtt-342-916-1 =12∫ dtt-342-542 =12×12×54 log t-34-54t-34+54+C =15 ln t-2t+12+C =15ln t-222t+1+C =15ln t-22t+1+15 ln 2+C =15 ln t-22t+1+C’ where C’=C+15ln 2 =15 ln tan x-22 tan x+1+C

Q8.

Answer :

Let I=∫ sin 2xsin4 x+cos4 x dx =∫ 2 sin x cos xsin4 x+cos4 x dxDividing numerator & denominator by cos4 x⇒I=∫ 2 sin x cos xcos4 xtan4 x+1dx =∫ 2 tan x. sec2 xtan2 x2+1 dxLet tan2 x=t⇒2 tan x sec2 x dx=dt=∫ dtt2+1∴I=tan-1 t+C =tan-1 tan2 x+C

Q9.

Answer :

Let I=∫ 1cos xsin x+2 cos xdxDividing numerator and denominator by cos2 x⇒I=∫ sec2 x cos xcos x×sin x+2 cos xcos xdx =∫ sec2 x tan x+2dxLet tan x+2=t⇒sec2 x dx=dt∴I=∫ dtt =ln t+C =ln tan x+2+C

Q10.

Answer :

Let I=∫ 1sin2 x+sin 2xdx =∫ 1sin2 x+2 sin x cos xdxDividing numerator and denominator by cos2 x⇒I=∫ sec2 x tan2 x+2 tan xdxLet tan x=t⇒sec2 x dx=dt∴ I=∫ dtt2+2t =∫ dtt2+2t+1-1 =∫ dtt+12–12 =12ln t+1-1t+1+1+C =12ln tt+2+C =12ln tan xtan x+2+C

Q11.

Answer :

Let I=∫ 1cos 2x+3 sin2 xdx =∫ 11-2 sin2 x+3 sin2 xdx =∫ 11+sin2 xdxDividing numerator and denominator by cos2 x⇒I=∫sec2 x sec2 x+tan2 xdx =∫sec2 x 1+tan2 x+tan2 xdx =∫ sec2 x 1+2 tan2dx =∫ sec2 x 1+2 tan x2dxLet 2 tan x=t⇒2 sec2 x dx=dt⇒sec2 x dx=dt2∴I=12 ∫ dt1+t2 =12 tan-1 t+C =12 tan-1 2 tan x+C

Page 19.105 Ex.19.18

Q1.

Answer :

Let I=∫ 15+4 cos xdxPutting cos x= 1-tan2 x21+tan2 x2 I =∫ 15+41-tan2 x21+tan2 x2dx =∫ 1+tan2 x25 1+tan2 x2+41-tan2 x2dx =∫ sec2 x2 dx5+5 tan2 x2+4-4 tan2 x2 =∫ sec2 x2 dxtan2 x2+9Let tan x2=t⇒12 sec2 x2dx=dt⇒sec2 x2dx=2dt∴I=2 ∫ dtt2+32 =23tan-1 t3+C =23tan-1 tan x23+C

Q2.

Answer :

Let I=∫ 15-4 sin xdxPutting sin x= 2 tan x21+tan2 x2⇒I=∫ 15-4×2 tan x21+tan2 x2dx =∫ 1+tan2 x251+tan2 x2-8 tan x2dx =∫ sec2 x2 5 tan2 x2-8 tan x2+5dxLet tan x2=t⇒12 sec2x2dx=dt⇒sec2 x2dx=2dt∴I=2 ∫ dt5t2-8t+5 =25∫ dtt2-85t+1 =25∫ dtt2-85t+452-452+1 =25 ∫ dtt-452-1625+1 =25 ∫ dtt-452+352 =25×53 tan-1 t-4535+C =23 tan-1 5t-43+C =23tan-1 5 tan x2-43+C

Q3.

Answer :

Let I = ∫ 11-2 sin xdxPutting sin x=2 tan x21+tan2 x2⇒I =∫11-2 ×2 tan x21+tan2 x2dx =∫ 1+tan2 x2 1+tan2 x2-4 tan x2dx =∫ sec2 x2tan2 x2-4 tan x2+1 dxLet tan x2=t⇒sec2 x2×12dx=dt⇒sec2 x2dx=2dt∴I=2∫ dtt2-4t+1 =2∫ dtt2-4t+4-4+1 =2 ∫ dtt-22-3 =2 ∫ dtt-22-32 =2×123ln t-2-3t-2+3+C =13ln tan x2-2-3tan x2-2+3+C

Q4.

Answer :

Let I =∫ 14 cos x-1dxPutting cos x= 1-tan2 x21+tan2 x2⇒I=∫ 141-tan2 x21+tan2 x2-1dx =∫ 141-tan2 x2-1+tan2 x21+tan2 x2 =∫ 1+tan2 x2dx4-4 tan2 x2-1-tan2 x2 =∫ sec2 x2 dx3-5 tan2 x2Let tan x2=t⇒12 sec2 x2dx=dt⇒ sec2 x2dx=2dt∴I=2 ∫ dt3-5 t2 =25 ∫ dt35-t2 =25 ∫ dt352-t2 =25×523ln 35+t35-t+C =115ln 3+5 t3-5 t+C =115ln 3+5 tan x23-5 tan x2+C

Page 19.106 Ex.19.18

Q5.

Answer :

Let I=∫ 11-sin x+cos xdxPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2 =∫ 11-2 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx =∫ 1+tan2 x21+tan2 x2-2 tan x2+1-tan2 x2dx =∫ sec2 x22-2 tan x2dx =12∫ sec2 x21-tan x2dxLet 1-tan x2=t⇒-sec2 x2×12dx=dt⇒sec2 x2dx=-2dt∴I=12 ∫ -2 dtt =-∫ dtt =- ln t+C =-ln 1-tan x2+C

Q6.

Answer :

Let I=∫ 13+2 sin x+cos xdxPutting sin x=2 tan x21+tan2 x2and cos x=1-tan2 x21+tan2 x2⇒I =∫ 13+2×2 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx =∫ 1+tan2 x231+tan2 x2+4 tan x2+1-tan2 x2dx =∫ sec2 x23+3 tan2 x2+4 tan x2+1-tan2 x2 dx =∫ sec2 x22 tan2 x2+4 tan x2+4dx =12∫ sec2 x2tan2 x2+2 tan x2+2dxLet tan x2=t⇒ sec2 x2×12 dx=dtsec2 x2dx=2dt∴I=12 ∫ 2 dtt2+2 t+2 =∫ dtt2+2t+1+1 =∫ dtt+12+12 =tan-1 t+11+C =tan-1 1+tan x2+C

Q7.

Answer :

Let I=∫ 113+3 cos x+4 sin xdxPutting cos x =1-tan2 x21+tan2 x2 and sin x=2tan x21+tan2 x2∴I =∫ 113+3 1-tan2 x21+tan2 x2+4×2tan x21+tan2 x2dx =∫ 1+tan2 x2131+tan2 x2+3-3 tan2 x2+8 tan x2 dx =∫ sec2 x2 13 tan2 x2-3 tan2 x2+16+8 tan x2dx =∫ sec2 x2 10 tan2 x2+8 tan x2+16dxLet tan x2=t⇒12 sec2 x2dx=dt⇒sec2 x2dx=2dt∴I=∫ 2 dt10t2+8t+16 =∫ dt5t2+4t+8 =15 ∫ dtt2+45t+85 =15∫ dtt2+45t+252-252+85 =15∫ dtt+252-425+85 =15∫ dtt+252+-4+4025 =15∫ dtt+252+652 =15×56tan-1 t+2565+C =16tan-1 5t+26+C =16 tan-1 5 tan x2+26+C

Q8.

Let I=∫ dxcos x-sin xPutting cos x=1-tan2 x21+tan2 x2 and sin x=2 tan x21+tan2 x2⇒I=∫ dx1-tan2 x21+tan2 x2-2 tan x21+tan2 x2     =∫ sec2 x2dx1-tan2 x2-2 tan x2Let tan x2=t⇒12 sec2 x2dx=dtsec2 x2dx=2 dt∴I=∫ 2 dt1-t2-2t     =∫ -2 dtt2+2t-1     =∫ -2dtt2+2t+1-2     =-∫ 2dtt+12-22     =∫ 2dt22-t-12     =222ln 2+t+12-t-1+C     =12 ln 2+tan x2+12-tan x2-1+C

Q9.

Answer :

Let I=∫1sin x+cos xdxPutting sin x =2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2 =∫ 12 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx =∫ sec2 x21-tan2 x2+2 tan x2 dxLet tan x2=t⇒ 12sec2 x2dx=dt⇒sec2 x2dx=2dt∴I=2∫ dt1-t2+2t =-2 ∫ dtt2-2t-1 =-2 ∫ dtt2-2t+1-2 =2∫ dt22-t-12 =2×122ln 2+t-12-t+1+C =12ln 2+tan x2-12-tan x2+1+C

Q10.

Answer :

Let I=∫ 15-4 cos xdxPutting cos x=1-tan2 x21+tan2 x2 ⇒I=∫ 15-4 1-tan2 x21+tan2 x2dx =∫ 1+tan2 x25 1+tan2 x2-4+4 tan2 x2dx =∫ sec2 x2 9 tan2 x2+1dxLet tan x2=t⇒12 sec2 x2dx=dt⇒sec2 x2dx=2dt∴I=2∫dt9t2+1 =29∫dtt2+19 =29∫ dtt2+132 =29×3 tan-1 t13+C =23 tan-1 3t+C =23 tan-1 3 tan x2+C

Q11.

Answer :

Let I=∫ 12+sin x+cos xdxPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2⇒I =∫ 12+2 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx =∫ 1+tan2 x2 21+tan2 x2+2 tan x2+1-tan2 x2dx =∫ sec2 x2 2+2tan2 x2+2 tan x2+1-tan2 x2dx =∫ sec2 x2 tan2 x2+2 tan x2+3dxLet tan x2=t⇒12 sec2 x2dx=dt⇒sec2 x2dx=2dt∴ I=2∫ dtt2+2t+3 =2∫ dtt2+2t+1+2 =2∫ dtt+12+22 =2×12 tan-1 t+12+C =2 tan-1 tan x2+12+C

Q12.

Answer :

Let I=∫ 1sin x+3 cos xdxPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2⇒I =∫ 12 tan x21+tan2 x2+31-tan2 x21+tan2 x2dx =∫ 1+tan2 x2 2 tan x2+3-3tan2 x2dx =∫sec2 x2-3tan2 x2+2 tan x2+3dx

Let tan x2=t⇒12 sec2 x2dx=dt⇒sec2 x2dx=2dt∴ I=2∫ dt-3t2+2t+3=-23∫ dtt2-23t-1=-23∫dtt2-23t+132-132-1=-23∫ dtt-132-232=-23 ×1223log t-13-23t-13+23+C

=-12log t-33t+13+C=-12log 3t-33t+1+C=12log 3t+13t-3+C=12log 3tanx2+13tanx2-3+Cor, 12log 1+3tanx23-3tanx2+C

Q13.

Answer :

Let I=∫ dx3 sin x+cos xPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2⇒I =∫ 132 tan x21+tan2 x2+1-tan2 x21+tan2 x2dx =∫ 1+tan2 x2 23 tan x2+1-tan2 x2dx =∫sec2 x2-tan2 x2+23 tan x2+1dx

Let tan x2=t⇒12 sec2 x2dx=dt⇒sec2 x2dx=2dt∴ I=2∫ dt-t2+23t+1=-2∫ dtt2-23t-1=-2∫dtt2-23t+32-32-1=-2∫ dtt-32-22=-22×2log t-3-2t-3+2+C

=-12logtanx2-2-3tanx2+2-3+C=12logtanx2+2-3tanx2+2-3+C

Q14.

Answer :

Let I=∫ dxsin x-3 cos xPutting sin x=2 tan x21+tan2 x2 and cos x=1-tan2 x21+tan2 x2⇒I =∫ 12 tan x21+tan2 x2-31-tan2 x21+tan2 x2dx =∫ 1+tan2 x2 2 tan x2-3+3tan2 x2dx =∫sec2 x23tan2 x2+2 tan x2-3dx

Let tan x2=t⇒12 sec2 x2dx=dt⇒sec2 x2dx=2dt∴ I=2∫ dt3t2+2t-3=23∫ dtt2+23t-1=2∫dtt2+23t+132-1-132=2∫ dtt+132-232=22×23log t+13-23t+13+23+C
=32log tanx2-13tanx2+33+C=32log 3 tanx2-13 tanx2+3+C

Q15.

Answer :

Let I=∫ 15+7 cos x+sin xdxPutting cos x=1-tan2 x21+tan2 x2 and sin x=2 tan x21+tan2 x2 ⇒I = ∫ 15+7 1-tan2 x21+tan2 x2+2 tan x21+tan2 x2 dx =∫ sec2 x251+tan2 x2+7-7 tan2 x2+2 tan x2dx =∫ sec2 x2-2 tan2 x2+2 tan x2+12dxLet tan x2=t⇒12 sec2 x2dx=dt⇒sec2 x2dx=2dt∴I=∫ 2 dt-2t2+2t+12 =∫ dt-t2+t+6 =∫ -dtt2-t-6 =∫ -dtt2-t+122-122-6 =∫ -dtt-122-14-6 =∫ -dtt-122-522 =∫ dt522-t-122 =12×52log 52+t-1252-t+12+C =15log 2+t3-t+C =15log 2+tan x23-tan x2+C

Page 19.110 Ex.19.19

Q1.

Answer :

Let I= ∫11-cot xdx =∫11-cos xsin xdx =∫sin xsin x-cos xdx =12∫2 sin xsin x-cos x dx =12∫sin x+cos x+sin x-cos xsin x-cos xdx =12∫sin x+cos xsin x-cos xdx+12∫dxPutting sin x -cos x =t⇒cos x+sin x dx=dt∴ I=12∫1tdt+12∫dx =12 ln t+x2+C =x2+12 ln sin x-cos x+C

Q2.

Answer :

Let I= ∫11-tan xdx =∫11-sin xcos xdx =∫cos x cos x-sin xdx =12∫2 cos x cos x-sin xdx =12∫cos x+sin x+cos x-sin xcos x-sin xdx =12∫cos x+sin xcos x-sin xdx+12∫dxPutting cos x-sin x=t⇒-sin x-cos xdx=dt⇒sin x+cos xdx=-dt∴ I=-12∫dtt+x2+C =-12 ln cos x-sin x+x2+C =x2-12 ln cos x-sin x+C

Q3.

Answer :

Let I=∫3+2 cos x +4 sin x 2 sin x+cos x+3dxLet 3+2 cos x+4 sin x=A 2 sin x+cos x+3 +B 2 cos x-sin x +C⇒3+2 cos x+4 sin x=2A-B sin x+A+2B cos x+3A+C

Comparing the coefficients of like terms
2A-B=4 … 1A+2B=2 … (2)3A+C=3 … (3)

Multiplying eq (1) by 2 and adding it to eq (2) we get ,
⇒4A-2B+A+2B=8+2⇒5A=10⇒A=2

Putting value of A = 2 in eq (1)

⇒2×2-B=4⇒B=0Putting value of A in eq (3) ⇒3×2+C=3⇒ C=-3

∴ I=∫2 2 sin x+cos x+3-32 sin x+cos x+3dx =2∫dx-3∫12 sin x+cos x+3dxSubstituting sin x=2 tan x21+tan2 x2 and cos x =1-tan2 x21+tan2 x2∴ I=2∫dx-3∫12×2 tan x21+tan2 x2+1-tan2 x21+tan2 x2+3dx =2∫dx-3∫1+tan2 x2 4 tan x2+1-tan2 x2+3 1+tan2 x2dx =2∫dx-3∫sec2 x22 tan2 x2+4 tan x2+4 dx =2∫dx-32∫sec2 x2 tan2 x2+2 tan x2+2dxPutting tan x2=t⇒12 sec2 x2 dx=dt⇒sec2 x2 dx=2dt∴ I=2∫dx-32∫2t2+2t+2 dt =2∫dx-3∫1t2+2t+1+1dt =2∫dx-3∫1t+12+12dt =2x-31 tan-1 t+11+C =2x-3 tan-1 tan x2+1+C ∵ t= tan x2

Q4.

Answer :

Let I=∫dxp+q tan x =∫1p+q sin xcos xdx =∫cos x q sin x+p cos xdxLet cos x=A q sin x+p cos x+B q cos x-p sin x⇒cos x=Ap+Bq cos x+Aq-Bp sin x

Comparing coefficients of like terms

Ap+Bq=1 … 1Aq-Bp=0 … 2

Multiplying eq (1) by p and eq (2) by q and then adding

⇒Ap2+Bpq=p⇒Aq2-Bpq=0⇒A=pp2q2

Putting value of A in eq (1)
p2p2+q2+Bq=1⇒Bq=1-p2p2+q2⇒Bq=p2+q2-p2p2+q2⇒B=qp2+q2∴ I=∫pp2+q2×q sin x+p cos xq sin x +p cos x+qp2+q2×q cos x-p sin xq sin x+p cos xdx =pp2+q2∫dx+qp2+q2∫q cos x-p sin xq sin x +p cos xdxPutting q sin x+p cos x=t⇒q cos x-p sin x dx=dt∴ I=pp2+q2∫dx+qp2+q2∫1tdt =pp2+q2 x+qp2+q2 ln q sin x+p cos x+C

Q5.

Answer :

Let I=∫5 cos x+62 cos x+sin x+3dx& let 5 cos x+6=A 2 cosx+sin x+3+B-2 sin x+cos x+C ….(1) ⇒5 cos x+6=A-2B sin x+2A+B cos x +3A+C

Comparing coefficients of like terms

A-2B=0 … 22A+B=5 … (3)3A+C=6 … (4)

Multiplying eq (3) by 2 and then adding to eq (2)

4A + 2B + A – 2B = 10
⇒A = 2

Putting value of A in eq (2) and eq (4) we get,
B = 1& C = 0

By putting the values of A,B and C in eq (1) we get ,∴ I=∫2 2 cos x+sin x+3+-2 sin x+cos x2 cos x+sin x+3dx =2∫dx+∫ -2 sin x+cos x2 cos x+sin x+3dxPutting 2 cos x+sin x+3=t⇒-2 sin x+cos xdx=dt∴ I=2∫dx+∫1tdt =2x+ln 2 cos x+sin x+3+C

Q6.

Answer :

Let I=∫2 sin x+3 cos x3 sin x+4 cos xdx& let 2 sin x+3 cos x=A 3 sin x+4 cos x+B 3 cos x-4 sin x …(1)⇒2 sin x+3 cos x=3A-4B sin x+4A+3B cos x

By comparing the coefficients of like terms we get,

3A-4B=2 … 24A-3B=3 … 3

Multiplying eq (2) by 3 and eq (3) by 4 and then adding,

9A-12B+16A+12B=6+12⇒25A=18⇒A=1825Putting value of A=1825 in eq 2 we get,3×1825-4B=2⇒5425-2=4B⇒425×4=B⇒B=125
Thus, substituting the values of A,B and C in eq (1) we get ,

I =∫18253 sin x+4 cos x+125 3 cos x-4 sin x3 sin x+4 cos xdx =1825∫dx+125∫3 cos x-4 sin x3 sin x+4 cos xdxPutting 3 sin x+4 cos x=t⇒3 cos x-4 sin x dx=dt∴ I=1825∫dx+125∫1tdt =18×25+125 ln t+C =18×25+125 ln 3 sin x+4 cos x+C

Page 19.111 Ex.19.19

Q7.

Answer :

Let I=∫13+4 cot xdx =∫13+4 cos xsin xdx =∫sin x 3 sin x+4 cos xdxLet sin x=A3 sin x+4 cos x+B 3 cos x-4 sin x …(1)⇒sin x=3A-4B sin x+4A+3B cos xBy comparing the coefficients of both sides we get ,3A-4B=1 … 24A+3B=0 … 3

Multiplying eq (2) by 3 and equation (3) by 4 , then by adding them we get

9A-12B+16A+12B=3+0⇒25A=3⇒A=325Putting value of A in eq 3 we get,4×325+3B=0⇒3B=-1225⇒B=-425
Thus, by substituting the value of A and B in eq (1) we getI=∫3253 sin x+4 cos x-4253 cos x-4 sin x3 sin x+4 cos xdx =325∫dx-425∫3 cos x-4 sin x3 sin x+4 cos xdxPutting 3 sin x+4 cos x=t⇒3 cos x-4 sin xdx=dt∴I=325∫dx-425∫dtt =325x-425 ln t+C =3×25-425 ln 3 sin x+4 cos x+C

Q8.

Answer :

Let I=∫2 tan x+33 tan x+4dx =∫2 sin xcos x+33 sin xcos x+4dx =∫2 sin x+3 cos x3 sin x+4 cos xdxLet 2 sin x+3 cos x=A 3 sin x+4 cos x+B 3 cos x-4 sin x ….(1) ⇒2 sin x+3 cos x=3A-4B sin x+4A+3B cos xEquating the coefficients of like terms3A-4B=2 … 24A+3B=3 … 3

Multiplying equation (2) by 3 and equation (3) by 4 ,then by adding them we get

9A-12B=616A+12B=12 25A=18⇒A=1825Putting value of A in eq 2 we get,⇒B=125
Thus, by substituting the values of A and B in eq (1) we get,I= ∫1825 3 sin x+4 cos x+1253 cos x-4 sin x3 sin x+4 cos xdx =1825∫dx+125∫3 cos x-4 sin x3 sin x+4 cos xdxPutting 3 sin x+4 cos x=t⇒3 cos x-4 sin xdx=dt∴ I=1825x+125∫1tdt =18×25+125 ln t+C =18×25+125 ln 3 sin x+4 cosx+C

Q9.

Answer :

Let I=∫dx4 +3 tan x=∫dx4+3 sin xcos x=∫cos x dx4 cos x+3 sin xConsider,cos x=A 4 cos x+3 sin x+Bddx4 cos x+3 sin x⇒cos x=A 4 cos x+3 sin x+B -4 sin x+3 cos x⇒cos x=4A+3B cos x+3A-4B sin xEquating the coefficients of like terms4A+3B=1 …..13A-4B=0 …..2
Solving (1) and (2), we get
A=425 and B=325
∫4254 cos x+3 sin x+-4 sin x+3 cos x3254 cos x+3 sin xdx=425∫dx+325∫-4 sin x+3 cos x4 cos x+3 sin xdxlet 4 cos x+3 sin x=t⇒-4 sin x+3 cos xdx=dtThen,I=425∫dx+325∫dtt=4×25+325 log t+C=4×25+325 log 4 cos x+3 sin x+C

Q10.

Answer :

Let I=∫8 cot x+13 cot x+2dx =∫8 cos xsin x+13 cos xsin x+2dx =∫8 cos x+sin x3 cos x+2 sin xdxNow, let 8 cos x+sin x=A 3 cos x+2 sin x+B -3 sin x+2 cos x …(1) ⇒8 cos x+sin x=3A cos x+2A sin x-3B sin x+2B cos x ⇒8 cos x+sin x=3A+2B cos x+2A-3B sin x Equating the coefficients of like terms we get, 2A-3B=1 … 23A+2B=8 … 3

Solving eq (2) and eq (3) we get,
A = 2, B = 1
Thus, by substituting the values of A and B in eq (1) we get ,
I=∫2 3 cos x+2 sin x+1-3 sin x+2 cos x3 cos x+2 sin xdx =2∫3 cos x+2 sin x3 cos x+2 sin xdx+∫-3 sin x+2 cos x3 cos x+2 sin xdx =2∫dx+∫-3 sin x+2 cos x3 cos x+2 sin xdxPutting 3 cos x+2 sin x=t⇒-3 sin x+2 cos xdx=dt∴ I=2∫dx+∫1tdt =2x+ln t+C =2x+ln 3 cos x+2 sin x+C

Q11.

Answer :

Let I=∫4 sin x+5 cos x5 sin x+4 cos xdx& let 4 sin x+5 cos x=A 5 sin x+4 cos x+B 5 cos x-4 sin x …(1) ⇒4 sin x+5 cos x=5A-4B sin x+4A+5B cos xBy equating the coefficients of like terms we get, 5A-4B=4 … 24A+5B=5 … 3

By solving eq (2) and eq (3) we get,
A=4041, B=941Thus, by substituting the values of A and B in eq (1) , we getI=∫40415 sinx +4 cos x+9415 cos x-4 sin x5 sin x+4 cos xdx =4041∫dx+941∫5 cos x-4 sin x5 sin x+4 cos xdxPutting 5 sin x+4 cos x=t⇒5 cos x-4 sin xdx=dt∴I=4041x+941∫1tdt =40 41x+941 ln t+C =4041x+941 ln 5 sin x+4 cos x+C

Page 19.121 Ex.19.20

Q1.

Answer :

∫ x cosx dxTaking x as the first function and cos x as the second function.= x∫cosx dx-∫ddxx∫cosx dxdx= x sinx-∫sinx dx= x sinx+cosx+C

Q2.

Answer :

∫ log x+1dx= ∫1. log x+1dxTaking log x+1 as the first function and 1 as the second function.= log x+1∫1 dx-∫ddxlogx+1∫1 dxdx= x log x+1-∫xx+1dx= x log x+1-∫x+1x+1-1x+1dx= x log x+1-x+log x+1+C

Q3.

Answer :

∫x3 logx dxTaking log x as the first function and x3 as the second function.= logx∫x3dx-∫ddxlogx∫x3dxdx= log xx44-∫1xx44dx= x4x logx-x416+C

Q4.

Answer :

∫xex dxTaking x as the first function and ex as the second function.=x∫exdx-∫ddxx∫ex dxdx= xex-∫1exdx= xex-ex+C= x-1ex+C

Q5.

Answer :

∫xe2x dxTaking x as the first function and e2x as the second function .= x∫e2xdx-∫ddxx∫e2x dxdx= x e2x2-∫e2x2dx= x2e2x-e2x4+C= e2xx2-14+C

Q6.

Answer :

∫x2 e-x dxTaking x2 as the first function and e-x as the second function.= x2∫e-xdx-∫ddxx2∫e-xdxdx= -x2 e-x-∫2xe-x-1dx= -x2 e-x+2∫x e-x dx= – x2 e-x+2-x e-x+∫e-x dx= -x2 e-x+2-x e-x -e-x+C= -e-xx2+2x+2+C

Q7.

Answer :

∫x2 cosx dxTaking x2 as the first function and cos x as the second function.= x2∫cosx dx-∫ddxx2∫cosx dxdx= x2sinx-∫2x sinx dx= x2sinx-2x∫sinx-∫ddxx∫sinx dxdx= x2sinx-2-xcosx+∫cosx dx= x2sinx+2x cosx-2 sinx+C

Q8.

Answer :

∫x2 cos 2x dxTaking x2 as the first function and cos 2x as the second function.= x2∫cos 2x dx-∫2x∫cos 2x dxdx= x2 sin 2×2-∫2x sin 2x2dx= x22sin 2x-∫x sin 2x dx= x22sin 2x-x∫sin2x-∫∫sin 2x dxdx= x22sin 2x–x cos 2×2+∫cos 2x2dx= x22sin 2x+x cos 2×2-sin 2×4+C

Q9.

Answer :

∫x sin 2x dx Taking x as the first function and sin 2x as the second function.= x∫sin2x dx-∫ddxx∫sin 2x dxdx= -x cos 2×2+∫cos 2x2dx= -x cos 2×2+sin 2×4+C

Q10.

Answer :

∫log log xxdxTaking log log x as the first function and 1x as the second function.= log log x∫1xdx-∫ddx log log x∫1xdxdx= log x.log log x-∫1x log xlog xdx= log x.log log x-∫1xdx= log x.log log x-log x+ C

= log xlog log x-1+C

Q11.

Answer :

∫x2 cosx dxTaking x2 as the first function and cos x as the second function.= x2∫cosx dx-∫ddxx2∫cosx dxdx= x2sinx-∫2x sinx dx= x2sinx-2x∫sinx-∫ddxx∫sin x dxdx= x2sinx+2xcosx-2∫cos x dx= x2sinx+2x cosx-2 sinx+C

Q12.

Answer :

∫x cosec2x dx Taking x as the first function and cosec2 x as the second function.= x∫cosec2x dx-∫ddxx∫cosec2x dxdx= -x cotx+∫cotx dx= -x cotx+log sinx+c

Q13.

Answer :

∫x cos2x dxTaking x as the first function and cos2x as the second function.= x∫1+cos 2x 2dx-∫ddxx∫1+cos 2x 2dxdx= x2x+sin2x2-∫12x+sin2x2dx= x2x+sin2x2-x24-cos2x8+C= x22+x sin2x2-x24+cos2x8+C= x24+x sin2x2+cos2x8+C

Q14.

Answer :

∫xn logx dxTaking log x as the first function and xn as the second function.= logx∫xndx -∫ddx logx∫xn dxdx= logxxn+1n+1-∫1xxn+1n+1dx= logxxn+1n+1-∫xnn+1dx= logxxn+1n+1-xn+1n+12+C

Q15.

Answer :

∫1xn logx dxTaking log x as the first function and 1xnas the second function.= logx∫1xndx-∫ddxlogx∫1xndxdx= logxx-n+1-n+1-∫1xx-n+1-n+1dx= logxx-n+1-n+1-∫x-n-n+1dx= logxx-n+1-n+1-x-n+1-n+12+C=logxx1-n1-n-x1-n1-n2+C

Q16.

Answer :

∫x2 sin2x dxTaking x2 as the first function and sin2 x as the second function.= x2∫1-cos2x2-∫ddxx2∫1-cos2x2dxdx= x22x-sin2x2-∫2xx-sin2x22dx= x22x-sin2x2-∫x2dx+∫xsin2x2dx Here, taking x as the first function and sin 2x as the second function.=x32-x2sin2x4-x33+12x∫sin 2x-∫ddxx∫sin 2x dxdx= x32-x2sin2x4-x33+12-xcos2x2+∫cos2x dx4= x36-x2sin2x4-x cos2x4+sin2x8+C

Q17.

Answer :

∫2×3·ex2dx=∫x2·ex2·2x dxLet x2=t⇒2x dx=dt=∫tI·eIIt dt=t·et-∫1·et dt=t et-et+C=x2 ex2-ex2+C=ex2x2-1+C

Q18.

Answer :

∫x3 cosx2 dx Let x2=t ⇒2x=dtdx⇒dx=dt2x= 12∫t cost dtTaking t as the first function and cos t as the second function .= 12tsint-∫sint dt= 12tsint+cost …(1) Substituting the value of t in eq (1) = x2sinx22+cosx22+c

Q19.

Answer :

∫xsin x·cos x dx=12∫x2 sin x cos x dx=12∫x·sin 2x dxTaking x as the first function and sin 2x as the second function . =12x∫sin 2x dx -∫ddxx∫sin 2x dxdx=12x×-cos 2×2-∫1·-cos 2x2dx=12-x cos 2×2+sin 2×4+C=-x cos 2×4+sin 2×8+C

Q20.

Answer :

Let I=∫sin x·log cos x dxLet cos x =t⇒-sin x dx=dt⇒sin x dx=-dt ∴ I=-∫log t dt =-∫1·log t dtTaking log t as the first function and 1 as the second function . =log t∫1 dt-∫ddt log t∫1dtdt =-log t ·t-∫1t×t dt =-log t·t-t+C =-tlog t-1+C ….(1) Substituting the value of t in eq (1) =-cos xlog cos x-1+C =cos x 1-log cos x+C

Q21.

Answer :

∫log x2 x ·dxTaking log x2 as the first function and x as the second function . =log x2∫x dx-∫ddxlog x2∫xdxdx=log x2·x22-∫2 log xx×x22 dx=log x2×x22-∫xII log xI dx=log x2×x22-logx ∫x dx -∫ddxlog x∫x dxdx=log x2×x22-log x·x22-∫1x×x22dx=log x2×x22-log x·x22+x24+C=x22log x2-log x+12 +C

Q22.

Answer :

Let I =∫ex dx =∫x·exxdxLet x=t⇒12xdx=dt⇒dxx=2 dt ∴I=2∫t·et dtTaking t as the first function and et as the second function . =2t∫etdt-∫ddtt∫etdtdt =2t·et-∫1·et dt+C …(1)Substituting the value of t in eq(1) =2x ex-ex+C =2exx-1+C

Q23.

Answer :

Let I=∫log x+2 dxx+22Let log x+2=t⇒x+2=et⇒1x+2dx=dt∴I=∫tetdt =∫t e-t dtTaking t as the first function and e-t as the second function. =t∫e-t-∫ddtt∫e-2tdtdt =t×e-t-1-∫1·e-t dt =-t e-t+e-t-1+C =-e-tt+1+C =-t+1et+C …(1)Substituting the value of t in eq (1) =-log x+2+1x+2+C =-log x+2 x+2-1x+2+C

Q24.

Answer :

∫x+sin x1+cos xdx=∫x1+cos x+sin x1+cos xdx=∫x2 cos2 x2+2 sin x2 cos x22 cos2 x2dx=12∫xI·sec2II x2dx+∫tan x2dx=12x·tan x212-∫1×2 tan x2dx+log sec x212+C=x tan x2-log sec x212+log sec x212+C=x tan x2+C

Q25.

Answer :

∫log10 x dx=∫log xlog 10dx=1log 10∫1·log x dxTaking log x as the first function and 1 as the second function=1log 10logx ∫1 dx-∫ddxlog x∫1 dxdx=1log 10log x·x-∫1x·x dx=1log 10x log x-x+C=1log 10xlog x-1+C

Q26.

Answer :

Let I=∫cos x dx =∫x· cos xxdxLet x=t⇒12xdx=dt⇒dxx=2dt∴I=2∫t·cos t·dtTaking t as the first function and cos t as the second function . =2 t·sin t-∫1·sin t dt =2 t·sin t+cos t+C ….(1) Substituting the value of t in eq (1) =2 x·sin x+cos x+C

Q27.

Answer :

Let, I=∫sin3x dx …..1Consider, x=t …..2Differentiating both sides we get,12xdx=dt⇒dx=2x dt⇒dx=2t dtTherefore, 1 becomes,I=∫sin3t 2t dt =2∫tsin3t dt =2∫t 3sint-sin3t4 dt Since, sin3A=3sinA-4sin3A =32∫t sint dt-12∫t sin3t dt =32t∫ sint dt-∫ dtdt∫sint dtdt-12t∫ sin3t dt-∫ dtdt∫sin3t dtdt =32-t cost +∫cost dt-12-t cos3t 3+13∫cos3t dt =32-t cost +sint -12-t cos3t 3+19sin3t +C =-32t cost+32sint+16t cos3t -118sin3t+C =-32xcosx+32sinx+16xcos3x-118sin3x+C

Note: The answer in indefinite integration may vary depending on the integral constant.

Q28.

Answer :

Let, I=∫cos3x dx …..1Consider, x=t …..2Differentiating both sides we get,12xdx=dt⇒dx=2x dt⇒dx=2t dtTherefore, 1 becomes,I=∫cos3t 2t dt =2∫t cos3t dt =2∫t 3cost+cos3t4 dt Since, cos3A=4cos3A-3cosA =32∫t cost dt+12∫t cos3t dt =32t∫ cost dt-∫ dtdt∫cost dtdt+12t∫ cos3t dt-∫ dtdt∫cos3t dtdt =32t sint -∫sint dt+12t sin3t 3-13∫sin3t dt =32t sint+cost +12t sin3t 3+19cos3t +C =32t sint+32cost+16t sin3t +118cos3t+C=32xsinx+32cosx+16xsin3x+118cos3x+C

Note: The final answer in indefinite integration may vary based on the integration constant.

Q29.

Answer :

Let I =∫cosec3x dx =∫cosec2x·cosec x dx =∫cosec2x·1+cot2x dxLet cot x=t⇒-cosec2x dx=dt∴ I=-∫1+t2dt =-t21+t2-122 log t+1+t2+C …(1)Substituting the value of t in eq (1) =-cot x2·cosec x-12 log cot x+cosec x+C =-12cosec x cot x-12 log cos xsin x+1sin x+C =-12 cosec x cot x-12 log 2 cos2 x22 sin x2 cos x2+C =-12 cosec x cot x-12 log cot x2+C =-12 cosec x cot x+12 log tan x2+C ∵ log cot x2=log 1tan x2 ⇒-log tan x2

Q30.

Answer :

Let I =∫x sin3x dx
sin (3A) = 3 sin A – 4 sin3 A
sin3 A=143 sin A-sin 3A∴I=14∫ x.3 sin x-sin 3xdx =34∫ xI.sin xII dx-14∫xI.sin 3xIIdx =34x-cos x-∫1.-cos xdx-14x-cos 3×3-∫1.-cos 3x3dx =-3x cos x4+34sin x+x cos 3×12-136sin 3x+C

Q31.

Answer :

Let I=∫x cos3x dx

As we know ,cos 3x=4 cos3 x-3cosx⇒cos3 x=14cos 3x+3cos x

∴I=14∫x.cos 3x+3 cos xdx =14∫ xI.cosII 3x dx+34 ∫ xI.cos xII dx =14x.∫cos 3x dx-∫ddxx.∫cos 3x dxdx+34x∫cos x-∫ddxx.∫cos x dxdx =14x.sin 3×3-∫1.sin 3x3dx+34xsin x-∫1.sin x dx =x sin 3×12+cos 3×36+34x sin x+34cos x+C

Q32.

Answer :

∫ x tan² x dx
= ∫ x (sec² x – 1) dx
=∫ xI. sec2 xII dx-∫ x dx=x∫sec2x-∫ddxx∫sec2x dxdx-x22+C1=x.tan x-∫1.tan x dx-x22+C1=x tan x-log sec x-x22+C1+C2=x tan x-log sec x-x22+C where C=C1+C2

Q33.

Answer :

∫ xsec 2x-1sec 2x+1dx=∫ x 1cos 2x-11cos 2x+1dx=∫ x 1-cos 2×1+cos 2xdx=∫ x 2 sin2 x2 cos2 xdx ∵1-cos 2x=2sin2 x and 1+cos 2x=2 cos2x=∫ x. tan2 x dx=∫ x.sec2 x-1 dx=∫ xI.sec2 xII dx-∫ x dx=x∫sec2x dx-∫ddxx∫sec 2x dxdx-x22+C1=x tan x-∫1. tan x dx-x22+C1=x tan x-log sec x-x22+C2+C1=x tan x-log sec x-x22+C where C=C1+C2

Q34.

Answer :

∫ x+1ex .log x ex dxLet x ex=t⇒x.ex+1.exdx=dt∴∫ x+1ex .log x ex dx=∫1II.log tIdt =log t∫1 dt-∫ddtlog t-∫1dtdt =log t×t-∫1t×t dt =t log t-t+C …(1)Substituting the value of t in eq (1)⇒∫ x+1ex .log x ex dx=x ex.log x ex-x ex+C =x exlogx ex-1+C

Q35.

Answer :

∫ sin^-1 (3x – 4x³)dx
Let x = sin θ
⇒ dx = cos θ.dθ
& θ = sin^-1 x
∫ sin^-1 (3x – 4x³)dx =∫ sin–^-1 (3 sin θ – 4 sin³ θ) . cos θ dθ
= ∫ sin^-1 (sin 3θ) . cos θ dθ
=3∫ θI.cos θII dθ=3θ∫cos θdθ-∫ddθθ-∫cos θ dθdθ =3θ.sin θ-∫1.sin θ dθ=3θ.sin θ+cos θ+C=3θ.sin θ+1-sin2 θ+C=3sin-1 x.x+1-x2+C ∵θ= sin-1x

Q36.

Answer :

∫ sin-1 2×1+x2dxLet x=tan θdx=sec2 θ dθ∴∫ sin-1 2×1+x2dx=∫ sin-1 2 tan θ1+tan2 θ. sec2 θ dθ =∫ sin-1 sin 2θ.sec2 θ dθ =∫ 2θ sec2 θ dθ =2∫ θI sec2 θII dθ =2θ∫sec2θ dθ-∫ddθθ∫sec2θ dθdθ =2θ.tan θ-∫1.tan θ dθ =2θ tan θ-log sec θ+C =2θ tan θ-log 1+tan2 θ12+C =2tan-1 x×x-log 1+x212+C =2 x tan-1 x-2×12log 1+x2+C =2 x tan-1 x-log 1+x2+C

Q37.

Answer :

Let I= ∫tan-1 3x-x31-3×2 dx=∫3 tan-1 x dx=3∫tan-1 x×1 dx=3 tan-1 x×x-∫11+x2×x dx=3x tan-1 x-3∫x1+x2 dxlet 1+x2=t⇒2x dx=dtThen,I=3x tan-1x-32∫dtt=3x tan-1x-32 log t+C=3x tan-1x-32 log 1+x2+C

Q38.

Answer :

∫ x2II.sin-1Ix dx=sin-1x∫ x2 dx-∫ddxsin-1x∫x2 dxdx=sin-1x.x33-∫11-x2 x33dxLet 1-x2=t⇒x2=1-t⇒-2x dx=dt⇒x dx=-dt2∴∫ x2.sin-1x dx=sin-1 x.x33-13∫ x2.x1-x2dx =sin-1 x.x33-16∫ 1-ttdt =sin-1 x.x33+16∫t-12 dt-16∫t12 dt =sin-1 x.x33+16×2t-16×23t32+C =sin-1 x.x33+1-x23-191-x232+C ∵ 1-x2=t

Q39.

Answer :

∫ x.sin-1 x1-x2dxLet sin-1 x=θx=sin θdx=cos θ dθ∴∫ x.sin-1 x1-x2dx=∫ sin θ.θ1-sin2 θ.cos θ dθ =∫ sin θ.θcos θ.cos θ dθ =∫ θI.sin θII dθ =θ∫sin θ dθ-∫ddθθ∫sin θ dθdθ =θ-cos θ-∫ 1.-cos θ dθ =-θ cos θ+sin θ+C =-θ 1-sin2 θ+sin θ+C =-sin-1 x 1-x2+x+C ∵sin-1 x=θ

Q40.

Answer :

Let I=∫ x2 tan-1 x1+x2dx =∫ x2+1-1×2+1tan-1 x dx =∫ 1-1×2+1tan-1 x dx =∫ 1II.tan-1I x dx-∫ tan-1 xx2+1dx =tan-1 x∫1 dx-∫ddxtan-1x∫1 dxdx-∫ tan-1 xx2+1dx =tan-1x×x-∫x1+x2dx-∫tan-1 xx2+1dxPutting x2+1=t in the first integral and tan-1 x=p in the second integral⇒2x dx=dt and 11+x2dx=dp⇒x dx=dt2and 11+x2dx=dp∴ I=tan-1 x.x-12∫ dtt-∫p.dp =x tan-1 x-12ln t-p22+C =x tan-1 x-12ln 1+x2-tan-1 x22+C ∵ t=x2+1 and p=tan-1x

Q41.

Answer :

∫ cos-1 4×3-3xdxLet x=cos θ ⇒θ=cos-1 x& dx=-sin θ dθ∴∫ cos-1 4×3-3xdx=∫ cos-1 4 cos3 θ-3 cos θ.-sin θdθ =∫ cos-1 cos 3θ.-sin θdθ ∵cos 3θ=4 cos3 θ-3 cos θ =-3 ∫ θI sin θII dθ =θ∫sin θ dθ-∫ddθθ∫sin θ dθdθ =3 θ -cos θ-∫1.-cos θdθ =3θ cos θ-3 sin θ+C =3 cos-1 x.x-31-x2+C ∵x=cos θ

Q42.

Answer :

Let I=∫ cos-1 1-x21+x2dx =2 ∫ 1II. tan-1I x dx ∵ cos -11-x21+x2=2 tan-1x =2tan-1x∫1 dx-∫ddxtan-1x∫1 dxdx =2tan-1 x. x-∫11+x2×x dx =2 x tan-1 x-∫2×1+x2dxPutting 1+x2=t⇒2x dx=dt∴ I=2x tan-1 x-∫ dtt =2x tan-1 x-ln t+C =2x tan-1 x-ln 1+x2+C ∵ t=1+x2

Q43.

Answer :

Let I=∫ tan-1 2×1-x2dx =2∫ 1II.tan-1 xI dx =2 tan-1x∫1 dx-∫ddxtan-1x∫1 dxdx =2tan-1 x.x-∫11+x2×x dx =2 tan-1 x.x-∫ 2×1+x2dxPutting 1+x2=t⇒2x dx=dt∴ I=2x tan-1 x-∫ dtt =2x tan-1 x-ln t+C =2x tan-1 x-ln 1+x2+C ∵ t= 1+x2

Q44.

Answer :

Let I=∫ tan-1 1-x1+x dxPutting x=cos θ⇒dx=-sin θ dθ& θ=cos-1 x∴ I=∫ tan-1 1-cos θ1+cos θ -sin θ dθ =∫ tan-1 2 sin2 θ22 cos2 θ2 -sin θ dθ =∫ tan-1 tan θ2 -sin θ dθ =-12∫ θI sin θII dθ =-12θ∫ sinθ dθ-∫ddθθ∫sin θ dθdθ =-12 θ-cos θ-∫ 1.-cos θ dθ =-12 -θ cos θ+sin θ+C =-12 -θ.cos θ+1-cos2 θ+C =-12-cos-1 x.x+1-x2+C ∵ θ=cos-1 x =x cos-1 x2-1-x22+C

Page 19.122 Ex.19.20

Q45.

Answer :

Let I=∫ sin-1 xa+x dxPutting x=a tan2 θ⇒xa=tan θ⇒dx=a2 tan θsec2 θ dθ∴ I=∫ sin-1 a tan2 θa+a tan2 θ 2a tan θsec2 θ dθ =∫ sin-1 tan2 θsec2 θ 2a tan θ sec2 θ dθ =2a ∫ sin-1 sin θtan θ sec2 θ dθ

=2a ∫ θ Itan θ sec2 θII dθ =2a θtan2θ2-∫1tan2 θ2dθ=2a θ.tan2 θ2-12∫sec2 θ-1dθ=a θ tan2 θ-a tan θ+aθ+C=axa tan-1 xa-axa+a tan-1 xa+C=x tan-1 xa-ax+a tan-1 xa+C

Q46.

Answer :

∫ elog x+sin x cos x dx=∫ x+sin xcos x dx ∵elog x=x =∫ x cos x+sin x cos x dx=∫ x cos x dx+12∫ 2 sin x cos x dx=∫ xI cos xII dx+12 ∫sin 2x dx=x∫cos x dx-∫ddxx∫cos x dxdx+12 ∫sin 2x dx=x sin x-∫1.sin x dx+12-cos 2×2+C=x sin x–cos x-14cos 2x+C=x sin x+cos x-141-2 sin2 x+C=x sin x+cos x+sin2 x2-14+C=x sin x+cos x+sin2 x2+C’ where C’=C-14

Q47.

Answer :

Let I=∫ x tan-1 x1+x232dxPutting x=tan θ⇒dx=sec2 θ dθ& θ=tan-1 x∴ I=∫ tan θ.θ.sec2 θ dθ1+tan2 θ32 =∫ θ.tan θ sec2 θ dθsec2 θ32 =∫ θ tan θ.sec2 θ dθsec3 θ =∫ θ.tan θsec θ dθ =∫ θI.sin θII dθ =θ∫sin θ dθ-∫ddθθ∫sin dθdθ =θ -cos θ-∫1.-cos θ dθ =-θ cos θ+ sin θ+C =-θsec θ+1cosec θ+C =-θ1+tan2 θ+11+cot2 θ+C =-θ1+tan2 θ+tan θtan2 θ+1+C =-tan-1 x1+x2+xx2+1+C

Q48.

Answer :

Let I=∫ x3 × sin-1 x21-x4dxPutting sin-1 x2=t ⇒x2=sin t⇒ 1×2x dx1-x22=dt⇒x dx1-x4=dt2∴I=∫ x2. sin-1 x21-x4.x dx =∫ sin t.t.dt2 =12∫ tI.sin tII dt =12t∫sin t dt-∫ddtt∫sin t dtdt =12 t.-cos t-∫ 1.-cos t dt =12-t cos t+sin t+C =12 -t1-sin2 t+sin t+C =12 -sin-1 x2 1-x4+x2+C

Q49.

Answer :

∫ x3II.tan-1 xI dx=tan-1 x∫ x3 dx-∫ddxtan-1 x∫x3 dxdx=tan-1 x.x44-∫11+x2×x44 dx=tan-1 x.x44-14 ∫ x4x2+1dx=tan-1 x.x44-14∫ x4-1+1×2+1dx=tan-1 x.x44-14∫ x4-1×2+1dx-14 ∫ 1×2+1dx=tan-1 x.x44-14∫ x2-1 x2+1×2+1dx-14∫ 1×2+1dx=tan-1 x.x44-14×33-x-14 tan-1 x+C=x4-14tan-1x -112×3-3x+C

Q50.

Answer :

∫ x. cos 2x sin x dx=12 ∫ x 2 cos 2x sin x dx ∵2 cos A sin B=sin A+B-sin A-B=12 ∫ x sin 3x-sin x dx=12 ∫ x.sin 3x dx-12 ∫ x sin x dx=12x∫sin 3x dx-∫ddxx∫sin 3x dxdx-12x∫sin x dx-∫ddxx∫sinx dxdx=12 x.-cos 3×3-∫ 1.-cos 3x3dx-12 x.-cos x-∫ 1.-cos x dx=12x.-cos 3×3+19sin 3x -12x.-cos x+sin x=-x cos 3×6+sin 3×18+x cos x2-sin x2+C

Q51.

Answer :

Let I =∫ (tan^-1 x²) x dx
Putting x² = t
⇒ 2x dx = dt
⇒x dx=dt2∴ I=12∫ 1II.tan-1 tI.dt =12tan-1t∫1 dt-∫ddttan-1t∫1 dtdt =12 tan-1 t. t-∫ t1+t2dtNow putting 1+t2=p⇒2t dt=dp⇒t dt=dp2∴ I=12t. tan-1 t-12∫ t dt1+t2 =t.tan-1 t2-12×2 ∫ dpp =t.tan-1 t2-14ln p+C =x2.tan-1 x22-14 ln 1+x4+C ∵ p=1+t2

Q52.

Answer :

∫ x3II.tan-1 xI dx=tan-1x ∫x3 dx-∫ddxtan-1x∫x3 dxdx=tan-1 x.x44-∫11+x2×x44dx=tan-1 x.x44-14∫ x4 dxx2+1=tan-1 x.x44-14∫ x4-1+1×2+1dx=tan-1 x.x44-14∫ x4-1×2+1dx-14∫ 1×2+1dx=tan-1 x.x44-14∫x2-1 x2+1×2+1dx-14∫ 1×2+1dx=tan-1x.x44-14∫ x2-1dx-14 tan-1 x+C=tan-1 x.x44-14×33-x-14tan-1 x+C=x4-14 tan-1x -112×3-3x+C

Q53.

Answer :

Let I=∫ tan-1 x dx =∫ x.tan-1 x dxxLet x=t⇒12xdx=dt=dxx=2dt∴I=2∫ tII.tan-1 tI dt =2 tan-1t∫t dt-∫ddttan-1t∫t dtdt =2 tan-1 t.t22-∫ 11+t2.t22dt =tan-1 t.t2-∫ t21+t2 dt =tan-1 t.t2-∫ 1+t2-11+t2dt =tan-1 t.t2-∫ dt+∫dt1+t2 =tan-1 t.t2-t+tan-1 t+C ∵x=t =tan-1 x.x-x+tan-1 x+C =x+1 tan-1 x-x+C

Q54.

Answer :

∫ 1II.log 1+x2I dx=log1+x2∫1 dx-∫ddxlog1+x2∫1 dxdx=log 1+x2.x-∫ 2×1+x2.x dx=x log 1+x2-2∫ x2+1-11+x2dx=x log 1+x2-2 ∫ dx+2∫ dx1+x2=x log 1+x2-2x+2 tan-1 x+C

Q55.

Answer :

Let I=∫ x2II.tan-1 xI dx =tan-1x∫x2dx-∫ddxtan-1x∫x2dxdx =tan-1 x×x33-∫ 11+x2×x33 dx =tan-1 x.x33-13∫ x2. x1+x2dxLet 1+x2=t⇒2x dx=dt⇒x dx=dt2∴I=tan-1 x.x33-16∫ t-1t.dt =tan-1 x.x33-16∫ dt+16∫ dtt =tan-1x.x33-t6+16log t+C =tan-1 x.x33-1+x26+16log 1+x2+C =tan-1 x.x33-x26+16log 1+x2-16+C =tan-1 x.x33-x26+16log 1+x2+C’ where C’=C-16

Q56.

Answer :

Let I=∫ sin-1 x dx=∫ x.sin-1 xxdxLet x=t⇒12xdx=dt⇒dxx=2dt∴I=∫ tII.sin-1 tI dt =sin-1 t∫t dt-∫ddtsin-1 t∫tdtdt =2 sin-1 t.t22-∫ 11-t2×t22dt =sin-1 t.t2-∫ t21-t2dt =sin-1 t.t2+∫1-t2-11-t2dt =sin-1 t.t2+∫ 1-t2 dt-∫ dt1-t2 =sin-1 t.t2+t21-t2+12sin-1 t-sin-1 t+C =sin-1 t.t2+t21-t2-12sin-1t+C =x.sin-1 x+x2 1-x-12sin-1 x+C ∵x=t =2x-1 sin-1 x2+x-x22+C

Q57.

Answer :

∫ 1II.sec-1 xI dx=sec-1 x∫1 dx-∫ddxsec-1 x∫1 dxdx=sec-1 x.x-∫ 1x 1-x×12x×x dx= x sec-1 x-12 ∫ 1-x-12 dx=x sec-1 x-12 1-x-12+1-12+1 -1+C=x sec-1 x+1-x12+C

Q58.

Answer :

Let I=∫ x2.sin-1 x dx1-x232Putting x=sin θ ⇒dx=cos θ dθ& θ=sin-1 x∴I=∫ sin2 θ.θ.cos θ dθ1-sin2 θ32 =∫ sin2 θ.θ.cos θ dθcos2 θ32 =∫ sin2 θ.θ.cos θ dθcos3 θ =∫ tan2 θ.θ.dθ =∫ sec2 θ-1θ.dθ =∫ θI.sec2 θII dθ-∫ θ.dθ =θ∫sec2 θ dθ-∫ddθθ∫sec2 θ dθdθ-∫ θ.dθ =θ tan θ-∫ 1.tan θ dθ-θ22 =θ.tan θ-ln sec θ-θ22+C =θ.sin θcos θ+ln cos θ-θ22+C =θ.sin θcos θ+ln 1-sin2 θ-θ22+C =θ. sin θ1-sin2 θ+12ln 1-sin2 θ-θ22+C =x sin-1 x1-x2+12ln 1-x2-12sin-1 x2+C ∵ θ=sin-1 x

Q59.

Answer :

∫ x+1II.log xI dx=log x∫x+1dx-∫ddxlog x∫x+1dxdx=log xx22+x-∫ 1xx22+xdx=log xx22+x-∫ x2+1dx=log xx22+x-x24+x+C

Q60.

Answer :

Let I=∫ sin-1 xx2 dxPutting x=sin θ⇒θ=sin-1 x & dx=cos θ dθ∴ I=∫ θ.cos θ sin2 θdθ =∫ θ.cos θsin θ×1sin θ dθ =∫ θI.cosec θII cot θ dθ =θ∫cosec θ cot θ dθ-∫ddθθ∫cosec θ cot θ dθdθ =θ -cosec θ-∫1.-cosec θ dθ =-θ cosec θ+∫ cosec θ dθ =-θ cosec θ+ln cosec θ-cot θ+C =-θsin θ+ln 1-cos θsin θ+C =-θsin θ+ln 1-1-sin2 θsin θ+C =-sin-1 xx+ln 1-1-x2x+C ∵ θ=sin-1 x

Page 19.129 Ex.19.21

Q1.

Answer :

Let I=∫ex cos x- sin x dx let ex cos x=t Diff both sides w.r.t xex·cos x+ex-sin x=dtdx Put ex fx=t⇒ex cos x-sin x dx=dt∴∫ex cos x-sin x dx=∫dt⇒I=t+C=ex cos x+C

Q2.

Answer :

Let I=∫ex 1×2- 2×3 dxAlso let ex×1×2=t Diff both sides w.r.t xex×1×2+ex -2×3=dtdx⇒ex 1×2-2×3 dx=dt∴∫ex 1×2-2×3 dx=∫dt=t+C=exx2+C

Q3.

Answer :

Let I=∫ex 1+sin x1+cos x dx=∫ex 11+cosx +sin x1+cos x dx=∫ex 12 cos2 x2+2 sin x2 cos x22 cos2 x2 dx=∫ex 12 sec2 x2+tan x2 dx Putting ex tan x2=tDiff both sides w.r.t. xex·tan x2+ex×12sec2 x2=dtdx⇒ex tan x2+12 sec2 x2 dx=dt∴∫ex 12 sec2 x2+tan x2 dx=∫dt=t+C=ex tanx2+C

Q4.

Answer :

Let I=∫excotx-cosec2xdxhere f(x)=cotx put exf(x)=t f'(x)=-cosec2xlet excotx=tDiff both sides w.r.t xexcotx+ex-cosec2x=dtdx⇒excotx-cosec2xdx=dt∴∫excotx-cosec2xdx=∫dt=t+C=excot x+C

Q5.

Answer :

Let I=∫exx-12x2dx=12∫ex1x-1x2dxhere 1x=f(x) Put exf(x)=t⇒-1×2=f'(x)let ex1x=tDiff both sides w.r.t xex1x+ex-1×2=dtdx⇒ex1x-1x2dx=dt∴I=12∫dt=t2+C=ex2x+C

Q6.

Answer :

Let I=∫exsecx1+tanxdx=∫exsecx+secx tanxdxHere, f(x)=secx Put exf(x)=t⇒f'(x)=secx tanxlet exsecx=tDiff both sides w.r.t xexsecx+exsecx tanx=dtdx⇒exsecx+tanxdx=dt∴∫exsecx+secx tanxdx=∫dt=t+C=exsecx+C

Q7.

Answer :

Let I=∫extanx-log cosxdxhere f(x)=-log cosx Put exf(x)=t⇒f'(x)=tanxlet -exlog cosx=tDiff both sides w.r.t x-exlogcosx+ex1cosx×-sinx=dtdx⇒-exlogcosx+extanxdx=dt∴∫extanx-log cosxdx=∫dt=t+C=-exlogcosx+C=exlogsec x+C

Q8.

Answer :

Let I=∫exsecx+logsecx+tanxdxHere, f(x)=logsecx+tanx Put exf(x)=t⇒f'(x)=secx let exlogsecx+tanx=tDiff both sides w.r.t xexlogsecx+tanx+ex1secx+tanxsecx+tanx+sec2x=dtdx⇒exlogsecx+tanx+exsecxdx=dt⇒exsecx+logsecx+tanxdx=dt∴∫exsecx+logsecx+tanxdx=∫dt=t+C=exlogsecx+tanx+C

Q9.

Answer :

Let I=∫excotx+log sinxdxHere,f(x)=log sinx Put exf(x)=t⇒f'(x)=cotxlet exlog sinx=tDiff both sides w.r.t xexlog sinx+ex×1sinx×cosx=dtdx⇒exlogsinx+excotxdx=dt⇒excotx+log sinxdx=dt∴∫excotx+log sinxdx=∫dt=t+C=exlog sinx+C

Q10.

Answer :

Let I=∫exx-1x-13dx=∫exx+1-2x+13dx=∫ex1x-12-2x+13dxHere, f(x)=1x+12⇒f'(x)=-2x+12Put exf(x)=tlet ex1x+12=tDiff both sidesex1x+12+ex-2x+13=dtdx⇒ex1x+12-2x+13dx=dt∴∫ex1x+12-2x+13dx=∫dt=t+C=exx+12+C

Q11.

Answer :

Let I=∫exsin4x-41-cos4xdx=∫ex2sin2x cos2x2sin2(2x)-42sin22xdx=∫excot(2x)-2cosec2(2x)dxHere, f(x)=cot(2x)⇒f'(x)=-2cosec2(2x)Put exf(x)=tlet excot(2x)=tDiff both sides w.r.t xexcot(2x)+ex×-2cosec2(2x)=dtdx⇒excot(2x)-2cosec2(2x)dx=dt∴∫excot2x-2cosec22xdx=∫dt⇒I=t+C=excot2x+C

Q12.

Answer :

Let I=∫ex1-x21+x22dx=∫ex1+x2-2×1+x22dx=∫ex11+x2-2×1+x22dxHere, f(x)=11+x2⇒f'(x)=-2×1+x22Put exf(x)=t⇒ex11+x2=tDiff both sides w.r.t xex11+x2+ex-11+x222x=dtdx⇒ex11+x2-2×1+x22dx=dt∴∫ex11+x2-2×1+x22dx=∫dt⇒I=t+C=ex1+x2+C

Q13.

Answer :

Let I=∫ex1+x2+x2dx=∫ex2+x-12+x2dx=∫ex12+x-12+x2dxHere, f(x)=12+x⇒f'(x)=-12+x2Put exf(x)=t⇒ex1x+2=tDiff both sides w.r.t xex1x+2+ex-1x+22=dtdx⇒ex1x+2-1x+22dx=dt∴∫ex12+x-12+x2dx=∫dt⇒I=t+C=ex2+x+C

Q14.

Answer :

Let I=∫1-sinx1+cosxe-x2dx=∫cos2x2+sin2x2-2sinx2cosx22cos2x2e-x2dx=∫cosx2-sinx222cos2x2e-x2dx=∫sinx2-cosx22cos2x2e-x2dx=∫12secx2tanx2-12secx2e-x2dx=12∫secx2tanx2-secx2e-x2dxlet e-x2 secx2=tDiff both sides w.r.t xe-x2secx2tanx22+secx2×e-x2×-12=dtdx⇒e2-x2secx2tanx2-secx2dx=dt∴12∫secx2tanx2-secx2e-x2dx=∫dt⇒I=∫t+C=e-x2secx2+C

Q15.

Answer :

Let I=∫exlogx+1xdxHere, f(x)=logx⇒f'(x)=1xput exf(x)=t⇒ ex logx=tDiff both sides w.r.t xexlogx+ex1x=dtdx⇒exlogx+1xdx=dt∴∫exlogx+1xdx=∫dt⇒I=t+C=exlogx+C

Q16.

Answer :

Let I=∫logx+1x2exdx=∫exlogx+1x-1x+1x2dx=∫exlogx+1xdx+∫ex-1x+1x2dxlet exlogx=tDiff both sidesexlogx+ex1xdx=dtlet ex-1x=pDiff both sidesex-1x+ex1x2dx=dp∴I=∫dt+∫dp=t+p+C=exlogx+ex-1x+C=exlogx-1x+C

Q17.

Answer :

Let I=∫exxxlogx2+2logxdx=∫exlogx2+2logxxdxHere, f(x)=logx2⇒f'(x)=2logxxput exf(x)=t⇒exlogx2=tDiff both sides w.r.t xexlogx2+ex2logxxdx=dt∴I=∫dt=t+C=exlogx2+C

Q18.

Answer :

Let I=∫ex1-x2sin-1x+11-x2dx=∫exsin-1x+11-x2dxHere, f(x)=sin-1x⇒f'(x)=11-x2Put exf(x)=t⇒ exsin-1x=tDiff both sides w.r.t xexsin-1x+ex×11-x2dx=dt∴I=∫dt=t+C=exsin-1x+C

Page 19.130 Ex.19.21

Q19.

Answer :

Let I=∫e2x-sinx+2cosxdxPut e2xcosx=tDiff both sides w.r.t x2e2xcosx+e2x×-sinxdx=dt∴I=∫dt=t+C=e2xcosx+C

Q20.

Answer :

Let I=∫1logx-1logx2dxPut logx=t⇒x=et⇒dx=etdt∴I=∫et1t-1t2dtHere, f(t)=1t⇒ f'(t)=-1t2let et×1t=pDiff both sides w.r.t tet×1t+et×-1t2dt=dp∴I=∫dp=p+C=ett+C=xlogx+C

Q21.

Answer :

Let I=∫exsinx cosx-1sin2xdx=∫excotx-cosec2xdxHere, f(x)=cotx⇒f'(x)=-cosec2xPut exf(x)=t⇒excotx=tDiff both sides w.r.t xexcotx-cosec2xdx=dt∴I=∫dt=t+C=excot x+C

Q22.

Answer :

Let I=∫tanlogx+sec2logxdxPut logx=t⇒x=et⇒dx=etdt∴I=∫tant+sec2tetdtHere, f(t)=tant⇒f'(t)=sec2tlet ettan(t)=pDiff both sides w.r.t tettant+sec2t=dpdt⇒ettant+sec2tdt=dp∴I=∫dp=p+C=ettant+C=x tan(logx)+C

Q23.

Answer :

Let I=∫2-x1-x2exdx=∫1+1-x1-x2exdx=∫11-x2+11-xexdxHere, f(x)=11-x⇒f'(x)=11-x2Put exf(x)=t⇒ex1-x=tDiff both sides w.r.t xex×11-x+ex×11-x2dx=dt∴I=∫dt=t+C=ex1-x+C

Q24.

Answer :

Let I=∫exx-4x-23dx=∫exx-2-2x-23dx=∫ex1x-22-2x-23dxHere, f(x)=1x-22⇒f'(x)=-2x-23Put exf(x)=t⇒ ex1x-22=tDiff both sides w.r.t xex1x-22+ex-2x-23dx=dt∴I=∫dt=t+C=exx-22+C

Q25.

Answer :

Let I=∫extan-1x+11+x2dxHere, f(x)=tan-1(x)⇒f'(x)=11+x2Put exf(x)=t⇒ extan-1(x)=tDiff both sides w.r.t xextan-1x+ex×11+x2dx=dt∴I=∫dt=t+C=extan-1x+C

Q26.

Answer :

Let I=∫exx-3x-13dx=∫exx-1-2x-13dx=∫ex1x-12-2x-13dxHere, f(x)=1x-12⇒f'(x)=-2x-13Put exf(x)=t⇒ ex1x-12=tDiff both sides w.r.t xex1x-12+ex-2x-13dx=dt∴I=∫dt=t+C=exx-12+C

Page 19.135 Ex.19.22

Q1.

Answer :

Let I=∫ eaxcosbxdxConsidering cos bx as first function and eax as second functionI=cosbxeaxa-∫-sinbx×b×eaxadx⇒I=eaxcos bxa+ba∫sin bxeax dx⇒I=eaxacosbx+ba∫eax×sin bxdx⇒I=eaxacosbx+baI1 …..1where I1=∫eaxsinbxdxNow, I1=∫eaxsin bxdxConsidering sin bx as first function eax as second functionI1=sin bxeaxa-∫cos bxbeaxadx⇒I1=sin bxeaxa-ba∫eaxcos bxdx⇒I1=eaxsin bxa-baI …..2From 1 and 2I=eaxacos bx+baeaxsin bxa-baI⇒I=eaxcos bxa+beaxsin bxa2-b2a2I⇒I1+b2a2=eaxa cos bx+b sin bxa2∴I=eaxa cos bx+b sin bxa2+b2+C

Q2.

Answer :

Let I=∫ eaxsin bx+CdxConsidering sin bx+C as first function and eax as second functionI=sin bx+Ceaxa-∫ cos bx+Cbeaxadx⇒I=eaxsin bx+Ca-ba∫ eax cos bx+C dx⇒I=eaxsin bx+Ca-baI1 …1where I1=∫ eaxcos bx+CdxNow, I1=∫ eaxcos bx+CdxConsider cos bx+C as first function eax as second funcitonI1=cos bx+Ceaxa-∫ -sin bx+Cbeaxadx⇒I1=eaxcos bx+Ca+ba∫ eaxsin bx+Cdx⇒I1=eaxcos bx+Ca+baI …..2From 1 & 2I=eaxsin bx+Ca-baeaxcos bx+Ca+baI⇒I=eaxsin bx+Ca-ba2eaxcos bx+C-b2a2I⇒I1+b2a2=eax asin bx+C-beax cos bx+Ca2+C1⇒I=eax a sin bx+C-b cos bx+Ca2+b2+C1Where C1 is integration constant

Q3.

Answer :

Let I=∫ cos log xdxLet log x=t⇒x=et⇒dx=et dtI=∫ etcostdtConsidering cost as first function and et as second functionI=cos t et-∫ -sin tet dt⇒I=cos t et+∫ sin t et dt⇒I=cos t et+I1 …..1where I1=∫etsin t dtI1=∫ etsin t dtCosidering sin t as first function and et as second functionI1=sin t et-∫ cos t etdt⇒I1=sin t et-I …..2From 1 & 2I=cos t et+sin t et-I⇒2I=etsin t+cos t⇒I=etsin t+cos t2+C⇒I=elog x sinlog x+coslog x2+C⇒I=x2sin log x+cos log x+C

Q4.

Answer :

Let I=∫ e2xcos 3x+4dxConsidering cos 3x+4 as first function and e2x as second functionI=cos 3x+4e2x2-∫-sin 3x+4×3e2x2dx⇒I=e2xcos 3x+42+32∫ e2xsin 3x+4dx⇒I=e2xcos 3x+42+32I1 …..1where I1=∫ e2xsin 3x+4dxconsidering sin 3x+4 as first function and e2x as second functionI1=sin 3x+4e2x2-∫ 3 cos 3x+4e2x2dx⇒I1=e2x sin 3x+42-32∫ e2xcos 3x+4dx⇒I1=e2xsin 3x+42-32 I …..2From 1 & 2I=e2xcos 3x+42+34 e2xsin 3x+4-94I⇒I+94I=2e2xcos3x+4+3e2xsin3x+44⇒I=e2x132 cos 3x+4+3 sin 3x+4+C

Q5.

Answer :

Let I=∫ e2xsin x cos x dxI=12∫ e2x 2 sin x cos xdx⇒I=12∫ e2xsin2xdxConsidering sin2x as first function and e2x as second function.I=12sin2xe2x2-∫2cos2xe2x2dx⇒I=e2xsin2x4-12∫ e2xcos2xdx⇒I=e2xsin2x4-12I1 …..1Where I1=∫ e2xcos2xdxConsidering cos2x as first function and e2x as second functionI1=cos2xe2x2-∫-2 sin2xe2x2dx⇒I1=e2xcos2x2+∫ e2xsin2x dx⇒I1=e2xcos2x4+2I …..2⇒I=e2xsin2x4-12e2xcos2x2+2I⇒I=e2xsin2x4-e2xcos2x4-I2×2⇒2I=e2x sin2x-cos2x4+C⇒I=e2x8sin2x-cos2x+C

Q6.

Answer :

Let I=∫ e2xsin x dxConsidering sin x as first function and e2x as second functionI=sin xe2x2-∫ cos xe2x2dx⇒I=sin xe2x2-12∫ cos x e2x dx⇒I=sin x e2x2-12cos xe2x2-∫-sin xe2x2dx⇒I=sin x e2x2-cos x e2x4-12∫e2xsin x2dxI=e2x2 sin x-cos x4-I4⇒5I=e2x 2 sin x-cos x⇒I=e2x 2 sin x-cos x5+C

Q7.

Answer :

Let I=∫ exsin2 x dx=∫ ex 1-cos 2x2dx=12∫ ex dx-12∫ ex cos 2x dx=ex2-12∫ excos 2x dx …..1Let I1=∫ excos 2xdxConsidering cos 2x as first function and ex as second functionI1=cos 2xex-∫-2 sin 2xex dx⇒I1=cos 2xex+2∫ sin 2xex dx⇒I1=cos 2xex+2sin 2xex-∫ 2 cos 2xex dx⇒I1=cos 2xex+2 sin 2xex-4I1⇒5I1=ex cos 2x+2 sin 2x⇒I1=ex5cos 2x+2 sin 2x+C …..2From 1 & 2I=ex2-ex10cos 2x+2 sin 2x+C

Q8.

Answer :

Let I=∫ 1x3sin log xdxPutting log x=t⇒x=et⇒dx=et dt∴I=∫ 1e3tsin t et dt=∫e-2t sin t dtConsidering sin t as first function and e-2t as second functionI=sin te-2t-2-∫ cos te-2t-2dt⇒I=sin t e-2t-2+12∫cos t e-2t dt⇒I=sin t e-2t-2+12cos te-2t-2-∫-sin te-2t-2dt⇒I=sin t e-2t-2-14 cos t e-2t-∫ e-2tsin t dt4⇒I=e-2t -2 sin t-cos t4-I4⇒5I4=e-2t -2 sin t-cos t4⇒I=e-2t5-2 sin t-cos t+C⇒I=-x-252 sin log x+cos log x+C⇒I=-15x2cos log x+2 sin log x+C

Q9.

Answer :

Let I=∫ e2xcos2 x dx=∫ e2x1+cos 2x2dx=12∫ e2xdx +12∫e2xcos 2xdx=e2x4+12I1 …..1Where I1=∫ e2xcos 2x dxConsidering cos 2x as first function and e2x as second function I1=cos 2xe2x2-∫-2 sin 2x×e2x2dx ⇒I1=cos 2xe2x2+∫ e2xsin 2x dxNow considering sin 2x as first function and e2x as second function I1=cos 2xe2x2+sin 2xe2x2-∫ 2 cos 2xe2x2dx⇒ I1=e2xcos 2x+sin 2×2-I1⇒2 I1=e2xcos 2x+sin 2×2⇒ I1=e2xcos 2x+sin 2×4 …..2From 1 & 2I=e2x4+e2x8cos 2x+sin 2x+C

Q10.

Answer :

Let I=∫ e-2xsin x dxConsidering sin x as first function and e-2x as second functionI=sin xe-2x-2-∫cos xe-2x-2dx⇒I=-e-2xsin x2+12∫e-2x cos x dx⇒I=-e-2xsin x2+I12 where …..1Where, I1=∫ e-2xcos x dxConsidering cos x as first and e-2x as second functionI1=cos x e-2x-2-∫-sin xe-2x-2dx⇒I1=e-2xcos x-2-∫sin x e-2x dx2⇒I1=-e-2x cos x2-I2 …..2From 1 & 2I=-e-2x sin x2+12-e-2xcos x2-I2⇒I+I4=-e-2xsin x2-e-2xcos x4⇒5I4=-e-2x 2 sin x+cos x4∴I=e-2×5-2 sin x-cos x+C

Q11.

Answer :

Given integral is,∫ x2ex3cos x3 dxLet x3=t⇒3×2 dx=dt⇒x2 dx=dt3Integral becomes,13∫ etcos t dt=13I …..1Where, I=∫ etcos t dtI=∫ etcos t dtConsidering cos t as first and et as second functionI=cos t et-∫-sin t et dt⇒I=etcos t+∫ sin t et dtAgain considering sin t as first and et as second functionI=etcos t+sin t et-∫ cos t et dt⇒I=et cos t+sin t et-I⇒2I=etsin t+cos t⇒I=et2sin t+cos t∴∫ x2ex3cos x3 dx=13et2sin t+cos t+C From 1=ex36sin x3+cos x3+C

Page 19.139 Ex.19.23

Q1.

Answer :

∫ 3+2x-x2 dx=∫ 3-x2-2x dx=∫ 3-x2-2x+1-1 dx=∫ 4-x-12 dx=∫ 22-x-12 dx ∵∫a2-x2dx=12xa2-x2+12a2sin-1xa+C=x-12 22-x-12+222sin-1 x-12+C=x-123+2x-x2+sin-1 x-12+C

Q2.

Answer :

∫ x2+x+1 dx=∫ x2+x+122-122+1 dx=∫ x+122+322=x+122 x+122+322+38log x+12+12+x2+x+1+C ∵∫x2+a2dx=12xx2+a2+12a2 lnx+x2+a2+C=2x+14 x2+x+1+38log 2x+1+12+x2+x+1+C

Q3.

Answer :

∫ x-x2 dx=∫ -x2-x dx=∫ -x2-x+122-122 dx=∫ 122-x-122 dx=x-122 x-x2 +18sin-1 x-1212+C ∵∫a2-x2dx=12xa2-x2+12a2sin-1xa+C=2x-14 x-x2+18 sin-1 2x-1+C

Q4.

Answer :

∫ 1+x-2×2 dx=∫ 212+x2-x2 dx=2 ∫12-x2-x2 dx=2 ∫ 12-x2-x2+142-142 dx=2 ∫ 12+116-x-142 dx=2 ∫ 342-x-142 dx=2 x-142 342-x-142+932sin-1x-1434+C ∵∫a2-x2dx=12xa2-x2+12a2sin-1xa+C=4x-18 1+x-2×2+9232sin-1 4x-13+C

Q5.

Answer :

Let I=∫ cos x 4-sin2 x dxPutting sin x=t⇒cos x dx=dt∫ 22-t2 dt=t222-t2+222 sin-1 t2+C ∵∫a2-x2 dx=12xa2-x2+12a2sin-1xa+C=sin x2 4-sin2 x+2 sin-1 sin x2+C ∵t= sin x

Q6.

Answer :

Let I=∫ ex e2x+1 dxPutting ex=t⇒ex dx=dt∴I=∫ t2+1dt =t2t2+1+122ln t+t2+1+C ∵∫x2+a2dx=12xx2+a2+12lnx+x2+a2+C =ex2 e2x+1+12ln ex+e2x+1+C ∵ t= ex

Q7.

Answer :

∫ 9-x2 dx=∫ 32-x2 dx=x232-x2+322sin-1 x3+C ∵∫a2-x2 dx=12xa2-x2+12a2 sin-1xa+C=x29-x2+92sin-1 x3+C

Q8.

Answer :

∫ 16×2+25 dx=∫ 16×2+2516dx=4∫ x2+542 dx=4x2x2+542+5422ln x+x2+542+C ∵∫x2+a2 dx=12xx2+a2+12a2 lnx+x2+a2+C=2x x2+2516+258ln x+x2+2516+C

Q9.

Answer :

∫ 4×2-5dx=∫ 4×2-54 dx=2∫ x2-522 dx=2x2x2-54-58ln x+x2-54 +C ∵∫x2-a2 dx=12xx2-a2-12a2 lnx+x2-a2+C=x x2-54-54ln x+x2-54+C

Q10.

Answer :

∫ 2×2+3x+4 dx=2 ∫ x2+32x+2 dx=2 ∫ x2+32x+342-342+2 dx=2 ∫ x+342-916+2 dx=2 ∫ x+342+2342dx=2 x+342 x+342+2342+2332ln x+34+x2+32x+2+C ∵∫x2+a2 dx=12xx2+a2-12a2 lnx+x2+a2+C=2 4x+38 x2+32x+2+2332ln x+34+x2+32x+2+C=4x+38 2×2+3x+4+23232ln x+34+x2+32x+2+C

Q11.

Answer :

Let I=∫3-2x-2x2dx=∫3-2×2+2xdx=∫3-2 x2+xdx=∫3-2 x2+x+14-14dx=∫3-2 x+122+12dx=∫72-2 x+122dx=2∫74-x+122dx=2∫722-x+122dx=2×x+122 722-x+122+2×74×2 sin-1 x+1272+C=2x+14 3-2x-2×2+742 sin-1 2x+17+C

Q12.

Answer :

Let I=∫xx4+1dx=∫xx22+1dxPutting x2=t⇒2x dx=dt⇒x dx=dt2∴I=12∫t2+1dt=12∫t2+12dt=12 t2t2+1+122 log t+t2+1+C=12×22 x4+1+12 log x2+x4+1+C=x24 x4+1+14 log x2+x4+1+C

Q13.

Answer :

Let I=∫x2a6-x6dx=∫x2a32-x32dxPutting x3=t⇒3×2 dx=dt⇒x2 dx=dt3∴I=13∫a32-t2dt=13 t2a32-t2+a322 sin-1 ta3+C=x36 a6-x6+a66 sin-1 x3a3+C

Q14.

Answer :

Let I=∫16+log x2xdxPutting log x=t⇒1x dx=dt∴I=∫16+t2dt=∫42+t2dt=t2 42+t2+422 log t+42+t2+C=log x2 16+log x2+8 log log x+16+log x2+C

Q15.

Answer :

Let I=∫2ax-x2dx=∫a2+2ax-x2-a2dx=∫ a2-x2-2ax+a2dx=∫a2-x-a2dx=x-a2 2ax-x2+a22 sin-1 x-aa+C

Q16.

Answer :

Let I= ∫3-x2dx=∫32-x2dx=x232-x2+322 sin-1 x3+C=x2 3-x2+32 sin-1 x3+C

Page 19.143 Ex.19.24

Q1.

Answer :

Let I= ∫ x+1 x2-x+1 dxAlso, x+1=λddxx2-x+1+μ⇒x+1=λ2x-1+μ⇒x+1=λ2x-1+μ⇒x+1=2λx+μ-λEquating the coefficient of like terms2λ=1⇒λ=12Andμ-λ=1⇒μ-12=1⇒μ=32∴I=∫122x-1+32 x2-x+1dx=12∫2x-1 x2-x+1dx+32∫x2-x+1dx=12∫2x-1 x2-x+1 dx+32∫ x2-x+14-14+1 dx=12∫2x-1 x2-x+1 dx+32∫x-122+34dx=12∫2x-1 x2-x+1 dx+32∫x-122+322dxLet x2-x+1=t⇒2x-1dx=dt∴I=12∫t dt+32∫x-122+322 dx=12×t3232+32x-122 x-122+322+38log x-12+x-122+322+C=13×2-x+132+382x-1 x2-x+1+916log x-12+x2-x+1+C

Q2.

Answer :

Let I=∫ x+1 2×2+3 dxAlso, x+1=λddx2x2+3+μ⇒x+1=λ4x+μEquating coefficient of like terms4λ=1⇒λ=14 and μ=1∴I=∫ 144x+1 2×2+3 dx=14∫ 4x 2×2+3 dx+∫2×2+3 dx=14∫4x2x2+3dx+∫2×2+32dx=14∫4x 2×2+3 dx+2 ∫x2+322 dxLet 2×2+3=t⇒4x dx=dt∴I=14∫ t dt+2x2x2+32+34log x+x2+32=14t3232+2x22x2+32+324 log x+2×2+32+C=162×2+332+x22x2+3+324log 2x+2×2+32+C=162×2+332+x22x2+3+324log 2x+2×2+3-324log 2+C=162×2+332+x22x2+3+324log 2x+2×2+3+C’Where C’=C-324log 2

Q3.

Answer :

Let I=∫ 2x-5 2+3x-x2 dxAlso, 2x-5=λddx2+3x-x2+μ⇒2x-5=λ-2x+3+μ⇒2x-5=-2λx+3λ+μEquating coeffieicents of like terms-2λ=2⇒λ=-1And3λ+μ=-5⇒3-1+μ=-5⇒μ=-5+3⇒μ=-2∴2x-5=-1-2x+3-2Hence, I=∫ –2x+3-2 2+3x-x2 dx=-∫ -2x+3 2+3x-x2dx-2∫2+3x-x2 dx=-I1-2I2 …..1I1=∫-2x+3 2+3x-x2 dxLet 2+3x-x2=t⇒-2x+3dx=dt∴I1=∫ t12 dt=t12+112+1=23t32=232+3x-x232 …..2And I2=∫ 2+3x-x2 dxI2=∫ 2-x2-3x dx=∫ 2-x2-3x+322-322 dx=∫2+94-x-322 dx=∫ 1722-x-322 dx=x-322 1722-x-322+17222sin-1 x-32172=2x-342+3x-x2+178 sin-1 2x-317 …..3From eq 1, 2 and 3 we haveI=-23 2+3x-x232-2x-322+3x-x2-174sin-1 2x-317+C

Q4.

Answer :

Let I=∫ x+2 x2+x+1 dxAlso, x+2=λddx x2+x+1+μ⇒x+2=λ2x+1+μ⇒x+2=2λx+λ+μEquating coefficient of like terms2λ=1 ⇒λ=12And λ+μ=2⇒12+μ=2⇒μ=32∴I=∫ 122x+1+32×2+x+1dx=12∫ 2x+1 x2+x+1 dx+32∫x2+x+1 dx=12∫2x+1 x2+x+1dx+32∫x2+x+122-122+1dx=12∫ 2x+1 x2+x+1 dx+32 ∫x+122+322 dxLet x2+x+1=t⇒2x+1dx=dtThen,I=12∫ t dt+32 ∫x+122+322 dx=12∫ t12 dt+32 x+122 x+122+322+38log x+12+ x+122+322+C=12t3232+382x+1 x2+x+1+916log x+12+x2+x+1+C=13 x2+x+132+38 2x+1 x2+x+1+916log x+12+x2+x+1+C

Q5.

Answer :

Let I=∫ 4x+1 x2-x-2 dxAlso, 4x+1=λddxx2-x-2+μ⇒4x+1=λ2x-1+μ⇒4x+1=2λx+μ-λEquating coefficient of like terms2λ=4⇒λ=2Andμ-λ=1⇒μ-2=1⇒μ=3∴I=∫ 22x-1+3 x2-x-2 dx=2∫2x-1 x2-x-2dx+3∫x2-x-2dx=2∫2x-1 x2-x-2 dx+3 ∫ x2-x+122-122-2 dx=2 ∫ 2x+1 x2-x-2 dx+3 ∫ x-122-2-14 dx=∫ 2x-1 x2-x-2 dx+3 ∫x-122-322 dxLet x2-x-2=t⇒ 2x-1dx=dt∴I=2∫ t dt+3x-122 x-122-322-3222log x-12+x2-x-2=2 t3232+34 2x-1 x2-x-2-278log x-12+x2-x-2+C=43 x2-x-232+34 2x-1 x2-x-2-278log x-12+x2-x-2+C

Q6.

Answer :

Let I=∫ x-2 2×2-6x+5 dxAlso, x-2=λddx2x2-6x+5+μ⇒x-2=4λx+μ-6λEquating the coefficient of like terms4λ=1⇒λ=14Andμ-6λ=-2⇒μ-6×14=-2⇒μ=-2+32=-12∴I=∫ 144x-6-12 2×2-6x+5 dx=14 ∫ 4x-6 2×2-6x+5 dx-12∫2×2-6x+5 dxLet 2×2-6x+5=t⇒4x-6dx=dt∴I=14∫ t12 dt-12∫2×2-3x+52dx=14∫ t12-22∫x2-3x+322-322+52 dx=14t3232-12∫ x-322-94+52 dx=16t32-12 ∫ x-322-9+104 dx=16t32-12∫x-322+122 dx=162×2-6x+532-12 x-322 x-322+122+18log x-32+x2-3x+52+C=16 2×2-6x+532-12 2x-34 x2-3x+52+18log 2x-32+x2-3x+52+C

Q7.

Answer :

Let I=∫ x+1 x2+x+1 dxAlso, x+1=λddx x2+x+1+μ⇒x+1=λ2x+1+μ⇒x+1=2λx+λ+μEquating coefficient of like terms2λ=1 ⇒ λ=12Andλ+μ=1⇒12+μ=1∴μ=12∴I= 12∫ 2x+1 x2+x+1dx+12∫x2+x+1 dx=12∫2x+1 x2+x+1 dx+12∫x2+x+122-122+1 dx=12∫2x+1 x2+x+1 dx+12∫x+122+322 dxLet x2+x+1=t⇒2x+1dx=dtThen,
I=12∫t dt+12x+122x+122+322+38log x+12+x+122+322+C=12×23t32+122x+14 x2+x+1+38log x+12+x2+x+1+C=13×2+x+132+122x+14 x2+x+1+38log x+12+x2+x+1+C

Q8.

Answer :

Let I=∫ 2x+3 x2+4x+3 dxAlso, 2x+3=λddxx2+4x+3+μ⇒2x+3=λ2x+4+μ⇒2x+3=2λx+4λ+μEquating coefficient of like terms.2λ=2 ⇒ λ=1And4λ+μ=3⇒4+μ=3⇒μ=-1∴I=∫ 2x+4-1 x2+4x+3dx=∫ 2x+4 x2+4x+3dx-∫x2+4x+3 dx=∫ 2x+4 x2+4x+3 dx-∫x2+4x+4-1 dx=∫2x+4 x2+4x+3dx-∫x+22-12 dxLet x2+4x+3=t⇒2x+4dx=dtThen,I=∫t dt-∫x+22-12 dx=23t32-x+22x+22-1-122log x+2+x+22-1+C=23 x2+4x+332-12x+2 x2+4x+3-log x+2+ x2+4x+3+C

Q9.

Answer :

Let I=∫ 2x-5 x2-4x+3 dx=∫ 2x-4-1 x2-4x+3 dx=∫2x-4 x2-4x+3 dx-∫x2-4x+3 dx=∫2x-4 x2-4x+3 dx-∫ x2-4x+4-4+3 dx=∫2x-4 x2-4x+3 dx-∫ x-22-12 dxLet x2-4x+3=t⇒2x-4dx=dt∴I=∫t dt-∫x-22-12 dx=23t32-x-22 x-22-12-122logx-2+x-22-1+C=23×2-4x+332-x-22 x2-4x+3+12log x-2+x2-4x+3+C

Q10.

Answer :

Let I=∫ xx2+xdxAlso, x=λddxx2+x+μ⇒x=λ2x+1+μ⇒x=2λx+λ+μEquating coefficient of like terms2λ=1⇒λ=12Andλ+μ=0⇒μ=-12∴I=∫ 122x+1-12 x2+xdx=12∫2x+1 x2+xdx-12∫x2+xdx=12∫ 2x+1 x2+xdx-12∫x2+x+14-14dx=12∫2x+1 x2+x dx-12∫x+122-122dxLet x2+x=t⇒2x+1dx=dtThen,I=12∫t dt-12x+122 x2+x-18logx+12+x2+x+C=12×23t32-2x+18 x2+x+116log x+12+x2+x+C=13×2+x32-2x+18 x2+x+116log x+12+x2+x+C

Page 19.160 Ex.19.25

Q1.

Answer :

∫2x+1x+1x-2 dx Let 2x+1x+1x-2=Ax+1+Bx-2 ….(1)⇒2x+1x+1x-2=Ax-2+Bx+1x+1x-2Then ,2x+1=Ax-2+Bx+1 ….(2)Putting x-2=0 or, x=2 in eq (2) ⇒2×2+1=A×0+B2+1⇒B=53Putting x+1=0 or, x=-1 in eq (2) 2×-1+1+A-1-2+B×0⇒-1=A-3⇒A=13Substituting the values of A and B in eq (1) , we get ∴ 2x+1x+1x-2=13x+1+53x-2∫2x+1dxx+1x-2=13∫1x+1dx+53∫1x-2dx =13 ln x+1+53 ln x-2+C

Q2.

Answer :

∫1xx-2x-4dxLet 1xx-2x-4=Ax+Bx-2+Cx-4⇒1xx-2x-4=Ax-2x-4+Bxx-4+Cx·x-2xx-2x-4⇒1=Ax-2x-4+Bx·x-4+Cx.x-2 …(1)Putting x=0 in eq (1) ⇒1=A0-20-4+B×0+C×0⇒18=APutting x-2=0 or x=2 in eq (1)⇒1=A×0+B22-4+C×2×0⇒B=-14Putting x-4=0 or x=4 in eq (1)⇒1=A×0+B×0+C·44-2⇒C=18∴1xx-2x-4=18x-14x-2+18x-4⇒∫dxxx-2x-4=18∫1xdx-14∫1x-2dx+18∫1x-4dx =18 ln x-14 ln x-2+18 ln x-4+C =18ln x+ln x-4-2 ln x-2+C =18ln xx-4x-22+C

Q3.

Answer :

∫x2+x-1×2+x-6dx=∫x2+x-6+6-1×2+x-6dx=∫x2+x-6×2+x-6dx+5∫1×2+x-6dx=∫dx+5∫1×2+3x-2x-6dx=∫dx+5∫1xx+3-2x+3dx=∫dx+5∫1x-2x+3dx . . . (1)Let1x-2x+3=Ax-2+Bx+3⇒1x-2x+3=Ax+3+Bx-2x-2x+3⇒1=Ax+3+Bx-2 ….(2) Putting x+3=0 or x=-3 in eq (2)⇒1=A×0+B-3-2⇒B=-15Putting x-2=0 or x=2 in eq (2)⇒1=A2+3+B×0⇒A=15

∴1x-2x+3=15x-2-15x+3⇒∫1x-2x+3dx=15∫dxx-2-15∫dxx+3 =15 ln x-2-15 ln x+3+C =15 ln x-2x+3+C . . . (3)From eq (1) & eq (3)∴∫x2+x-1×2+x-6dx=x+55 ln x-2x+3+C =x+ln x-2-ln x+3+C

Q4.

Answer :

∫3+4x-x2x2-x+2x-2dx=∫-x2+4x+3×2+x-2dx-x2+4x+3×2+x-2=-1+5x+1×2+x-2 . . . (1)∴5x+1×2+x-2=5x+1×2+2x-x-2 =5x+1xx+2-1x+2Let5x+1x-1x+2=Ax-1+Bx+2⇒5x+1x-1x+2=Ax+2+Bx-1x-1x+2⇒5x+1=Ax+2+Bx-1 …(2)Putting x+2=0 or x=-2 in eq (2)⇒5x-2+1=A×0+B-2-1⇒B=3Putting x-1=0 or x=1 in eq (2) ⇒5×1+1=A3+B×0⇒A=2∴5x+1x-1x+2=2x-1+3x+2 . . . (3)From (1) & (3)-x2+4x+3×2+x-2=-1+2x-1+3x+2⇒∫-x2+4x+3×2+x-2dx=∫-1 dx+∫2x-1dx+∫3x+2 dx =-x+2 lnx-1+3 lnx+2+C

Q5.

Answer :

∫x2+1×2-1dx=∫x2-1+2×2-1dx=∫dx+2∫1×2-12dx=∫dx+2∫1x-1 x+1dx … 1∴ 1x-1x+1=Ax-1+Bx+1⇒1x-1 x+1=A x+1+Bx-1x-1 x+1⇒1=A x+1 B x-1 ….(2)Putting x+1=0 or x=-1 in eq (2)⇒1=A×0+B -1 -1⇒B=-12Putting x-1=0 or x=1 in eq (2)⇒1=A 1+1+B×0⇒A=12∴ 1x-1x+1=12x-1-12x+1 …(3)From eq (1) & eq (3) ∫x2+1×2-1dx=∫dx+2∫12 x-1-12 x+1dx =∫dx+∫dxx-1-∫dxx+1 =x+ln x-1=-ln x+1+C =x+ln x-1x+1+C

Q6.

Answer :

∫x2 x-1 x-2 x-3dxLetx2x-1 x-2 x-3=Ax-1+Bx-2+Cx-3⇒x2x-1 x-2 x-3=A x-2 x-3+B x-1 x-3+C x-1 x-2x-1 x-2 x-3⇒x2=A x-2 x-3+B x-1 x-3+C x-1 x-2 ….(1)Putting x-1=0 or x=1 in eq (1)⇒1=A 1-2 1-3⇒1=A -1 -2⇒A=12Putting x-2=0 or x=2 in eq (1) ⇒4=B 2-1 2-3⇒B=-4Putting x-3=0 or x=3 in eq (1) ⇒9=C 3-1 3-2⇒C=92∴x2x-1 x-2 x-3=12 x-1-4x-2+92 x-3∫x2 x-1 x-2 x-3dx=12∫1x-1dx-4∫1x-2dx+92∫1x-3dx =12ln x-1-4 ln x-2+92 lnx-3+C

Q7.

Answer :

∫5xx+1 x2-4dxLet5xx+1 x-2 x+2=Ax+1+Bx-2+Cx+2⇒5xx+1 x-2 x+2=A x-2 x+2+B x+1 x+2+C x+1 x-2x+1 x-2 x+2⇒5x=A x-2 x+2+B x+1 x+2+C x+1 x-2 ….(1)Putting x-2=0 or x=2 in eq (1)⇒5×2=B 2+1 2+2⇒B=103×4 =56Putting x+2=0 or x=-2 in eq (1)⇒5×-2=C -2+1 -2-2⇒-10-1×-4 =C⇒C=-52Putting x+1=0 or x=-1 in eq (1) ⇒-5=A -1-2 -1+2⇒-5-3=A⇒A=53∴ 5xx+1 x-2 x+2=53×1x+1+56 x-2- 52 x+2⇒5xx+1 x-2 x+2=56×2x+1+56 x-2-56 3x+2∴ ∫5x x+1 x-2 x+2dx=56∫2x+1 dx+56∫1x-2dx-56∫3x+2 dx =562 ln x+1+ln x-2-3 ln x+2+C =56 ln x+12+ln x-2-ln x+23+C =56 ln x+12 x-2x+23+C

Q8.

Answer :

∫x2+1x x2-1dx=∫x2+1x x-1 x+1dxLetx2+1x x-1 x+1=Ax+Bx-1+Cx+1⇒x2+1x x-1 x+1=A x-1 x+1+B x x+1+C x x-1x x-1 x+1⇒x2+1=A x-1 x+1+B x x+1+C x x-1 ….(1)Putting x-1=0 or x=1 in eq (1) ⇒1+1=A×0+B 1 1+1+C×0⇒B=1Putting x=0 in eq (1) ⇒0+1=A 0-1 0+1⇒A=-1Putting x+1=0 or x=-1 in eq (1) ⇒-12+1=A×0+B×0+C-1 -1-1⇒2=C×2⇒C=1∴x2+1x x2-1=-1x+1x-1+1x+1⇒∫x2+1x x2-1dx=-∫1xdx+∫1x-1dx+∫1x+1dx =-ln x+ln x-1+ln x+1+C =-ln x+ln x2-1+C =ln x2-1x+C

Q9.

Answer :

∫2x-3 x2-1 2x+3dx=∫2x-3 x-1 x+1 2x+3dxLet 2x-3x-1 x+1 2x+3=Ax-1+Bx+1+C2x+3⇒2x-3x-1 x+1 2x+3=A x+1 2x+3+B x+1 2x+3+C x2-1x-1 x+1 2x+3⇒2x-3=A x+1 2x+3+B x-1 2x+3+C x+1 x-1 …(1)Putting x+1=0 or x=-1 in eq (1)⇒-2-3=B -1-1 -2+3⇒-5=B -2 1⇒B=52Putting x-1=0 or x=1 in eq (1)⇒2-3=A 1+1 2+3⇒-1=A 2 5⇒A=-110Putting 2x+3=0 or x=-32in eq (1) ⇒2×-32-3=A×0+B×0+C-32+1 -32-1⇒-6=C -12 -52⇒C=-245∴2x-3x-1 x+1 2x+3=-110 x-1+52 x+1-245 2x+3⇒∫2x-3x-1 x+1 2x+3 dx=-110∫1x-1dx+52∫1x+1dx-245∫12x+3dx =-110 ln x-1+52 ln x+1-245 ln 2x+33+C =-110 ln x-1+52 ln x+1-125 ln 2x+3+C

Q10.

Answer :

∫x3 x-1 x-2x-3dx=∫x3 x-1 x2-5x+6dx=∫x3 x3-5×2+6x-x2+5x-6dx=∫x3 x3-6×2+11x-6dx ∴x3x3-6×2+11x-6=1+6×2+11x+6×2-6×2+11x-6⇒x3x3-6×2+11x-6=1+6×2-11x+6x-1 x-2 x-3∴∫x3x-1 x-2 x-3 dx=∫dx+∫6×2-11x+6x-1 x-2 x-3dx …1

Let6x2-11x+6x-1 x-2 x-3=Ax-1+Bx-2+Cx-3⇒6×2-11x+6x-1 x-2 x-3=A x-2 x-3+B x-1 x-3+C x-1 x-2x-1 x-2 x-3⇒6×2-11x+6=A x-2 x-3+B x-1 x-3+C x-1 x-2 …(2)Putting x-2=0 or x=2 in eq (2) ⇒6×4-22+6=B 2-1 2-3⇒8=B -1⇒B=-8Putting x-3=0 or x=3 in eq (2)⇒6×32-11×3+6=C 3-1 3-2⇒27=C 2 1⇒C=272Putting x-1=0 or x=1 in eq (2) ⇒6×1-11+6=A 1-2 1-3⇒1=A -1 -2⇒A=12∴6×2-11x+6x-1 x-2 x-3=12x-1-8x-2+272x-3 …(3)From eq (2) and eq (3)∴∫x3 dxx-1 x-2 x-3=∫dx+12∫1x-1dx-8∫1x-2dx+272∫1x-3dx =x+12 ln x-1-8 ln x-2+272 ln x-3+C

Q11.

Answer :

We have, I=∫sin 2x dx1+sin x 2+sin x=∫2 sin x cos x dx1+sin x 2+sin xPutting sin x=t⇒cos x dx=dt∴I=∫2t dt1+t 2+t=2∫t dt1+t 2+tLet t1+t 2+t=A1+t+B2+t⇒t1+t 2+t=A 2+t+B 1+t1+t 2+t⇒t=A 2+t+B 1+tPutting 2+t=0⇒t=-2-2=A×0+B -2+1⇒-2=B -1⇒B=2let t+1=0t=-1⇒-1=A -1+2+B×0A=-1∴ I=2∫-1t+1+2t+2dt=2 -log t+1+2 log t+2+C=4 log t+2-2 log t+1+C=log t+24t+12+C=log sin x+24sin x+12+C

Q12.

Answer :

We have,I=∫2x dxx2+1 x2+3Putting x2=t⇒2x dx=dt∴I=∫dtt+1 t+3Let 1t+1 t+3=At+1+Bt+3⇒1t+1 t+3=A t+3+B t+1t+1 t+3⇒1=A t+3+B t+1Putting t+3=0⇒t=-31=A×0+B -3+1⇒B=-12Putting t+1=0⇒t=-11=A -1+3+B -1+1⇒1=A×2+B×0⇒A=12Then,I=12∫dtt+1-12∫dtt+3=12 log t+1-12 log t+3+C=12 log t+1t+3+C=12 log x2+1×2+3+C

Q13.

Answer :

We have,I=∫dxx log x2+log xPutting log x=t⇒1x dx=dt∴I=∫dtt t+2Let 1t t+2=At+Bt+2⇒1t t+2=At+2+Btt t+2⇒1=A t+2+BtPutting t+2=0⇒t=-21=A×0+B -2⇒B=-12Putting t=01=A 0+2+B ×0⇒A=12Then,I=12∫dtt-12∫dtt+2=12 log t-log t+2+C=12 log tt+2+C=12 log log xlogx+2+C

Q14.

Answer :

We have,I=∫dxcos x 5-4 sin x=∫cos x dxcos2x 5-4 sin x=∫cos x dx1-sin2x 5-4 sin x=∫cos x dx1-sin x 1+sin x 5-4 sin xPutting sin x=t⇒cos x dx=dt∴I=∫dt1-t 1+t 5-4tLet 11-t 1+t 5-4t=A1-t+B1+t+C5-4t⇒11-t 1+t 5-4t=A1+t 5-4t+B1-t 5-4t+C1-t 1+t1-t 1+t 5-4t⇒1=A1+t 5-4t+B1-t 5-4t+C1-t 1+tPutting 1+t=0⇒t=-11=B2 5+4B=118Putting 1-t=0⇒t=11=A 2 5-4+B×0+C×0A=12Putting 5-4t=0⇒4t=5⇒t=541=C 1-54 1+54⇒1=C -14 94⇒C=-169∴I=12∫dt1-t+118∫dt1+t-169∫dt5-4t=12 log 1-t-1+118 log 1+t-169× log 5-4t-4+C=118 log 1+t-12 log 1-t+49log 5-4t+C=118 log 1+sin x-12 log 1-sin x+49 log 5-4 sin x+C

Q15.

Answer :

We have,I=∫dxsin x 3+2 cos x=∫sin x dxsin2x 3+2 cos x=∫sin x dx1-cos2x 3+2 cos x=∫sin x dx1-cos x 1+cos x 3+2 cos xPutting cos x=t⇒-sin x dx=dt⇒sin x dx=-dt∴I=∫-dt1-t 1+t 3+2t=∫dtt-1 t+1 3+2tLet 1t-1 t+1 3+2t=At-1+Bt+1+C3+2t⇒1t-1 t+1 3+2t=A t+1 3+2t+B t-1 3+2t+C t+1 t-1t-1 t+1 3+2t⇒1=A t+1 3+2t+B t-1 3+2t+C t+1 t-1Putting t+1=0⇒t=-11=A×0+B -2 3-2+C×0⇒1=B -2⇒B=-12Putting t-1=0⇒t=11=A 2 5+B×0+C×0⇒A=110Putting 3+2t=0⇒t=-321=A×0+B×0+C -32+1 -32-1⇒1=C -12 -52C =45Then,I=110∫dtt-1-12∫dtt+1+45∫dt3+2t=110 log t-1-12 log t+1+45× log 3+2t2+C=110 log t-1-12 log t+1+25log 3+2t+C=110 log cos x-1-12 log cos x+1+25 log 3+2 cos x+C

Q16.

Answer :

We have,I=∫dxsin x +sin 2x =∫dxsin x+2 sin x cos x=∫dxsin x 1+2 cos x=∫sin x dxsin2x 1+2 cos x=∫sin x dx1-cos2x 1+2 cos x=∫sin x dx1-cos x 1+cos x 1+2 cos xPutting cos x=t⇒-sin x dx=dt⇒sin x dx=-dt∴I=∫-dt1-t 1+t 1+2t=∫dtt-1 t+1 1+2tLet 1t-1 t+1 1+2t=At-1+Bt+1+C1+2t⇒1t-1 t+1 1+2t=A t+1 1+2t+B t-1 1+2t+C t-1 t+1t-1 t+1 1+2t⇒1=A t+1 1+2t+B t-1 1+2t+C t-1 t+1Putting t+1=0⇒t=-11=B -1-1 1-2⇒1=B -2 -1⇒B=12Putting t-1=0⇒t=11=A 1+1 1+2⇒1=A23⇒A=16Putting 1+2t=0t=-12⇒1=A×0+B×0+C -12-1 -12+11=C -32 12C=-43Then,I=16∫dtt-1+12∫dtt+1-43∫dt1+2t=16 log t-1+12 log t+1-43×log 1+2t2+C=16 log t-1+12 log t+1-23log 1+2t+C=16 log cos x-1+12 log cos x+1-23 log 1+2 cos x+C

Q16.

Answer :

We have,I=∫dxsin x +sin 2x =∫dxsin x+2 sin x cos x=∫dxsin x 1+2 cos x=∫sin x dxsin2x 1+2 cos x=∫sin x dx1-cos2x 1+2 cos x=∫sin x dx1-cos x 1+cos x 1+2 cos xPutting cos x=t⇒-sin x dx=dt⇒sin x dx=-dt∴I=∫-dt1-t 1+t 1+2t=∫dtt-1 t+1 1+2tLet 1t-1 t+1 1+2t=At-1+Bt+1+C1+2t⇒1t-1 t+1 1+2t=A t+1 1+2t+B t-1 1+2t+C t-1 t+1t-1 t+1 1+2t⇒1=A t+1 1+2t+B t-1 1+2t+C t-1 t+1Putting t+1=0⇒t=-11=B -1-1 1-2⇒1=B -2 -1⇒B=12Putting t-1=0⇒t=11=A 1+1 1+2⇒1=A23⇒A=16Putting 1+2t=0t=-12⇒1=A×0+B×0+C -12-1 -12+11=C -32 12C=-43Then,I=16∫dtt-1+12∫dtt+1-43∫dt1+2t=16 log t-1+12 log t+1-43×log 1+2t2+C=16 log t-1+12 log t+1-23log 1+2t+C=16 log cos x-1+12 log cos x+1-23 log 1+2 cos x+C

Q17.

Answer :

We have,I=∫dxx-1 x+1 x+2Let 1x-1 x+1 x+2=Ax-1+Bx+1+Cx+2⇒1x-1 x+1 x+2=A x+1 x+2+B x-1 x+2+C x-1 x+1x-1 x+1 x+2⇒1=A x+1 x+2+B x-1 x+2+C x-1 x+1Putting x-1=0⇒x=11=A 1+1 1+2+B×0+C×0⇒1=A×6⇒A=16Putting x+1=0⇒x=-11=A×0+B -2 1+C×0⇒B=-12Putting x+2=0⇒x=-21=A×0+B×0+C -2-1 -2+1⇒1=C×3⇒C=13∴I=16∫dxx-1-12∫dxx+1+13∫dxx+2=16 log x-1-12 log x+1+13 log x+2+C=16 log x-1-36 log x+1+26log x+2+C=16 log x-1-3 log x+1+2 log x+2+C=16log x-1 x+22x+13 +C

Q18.

Answer :

We have,I=∫x+1x 1+xex dxI=∫exx+1exx 1+xex dxPut ex=t⇒exx+1dx=dtI=∫dtt 1+t …..1Let,1t 1+t=At +B1+t⇒1=At+1+Bt …..2Putting t=0 in 2, we obtain A=1Putting t=-1 in 2, we obtain B=-1I=∫1t -11+t dtI=logt-logt+1+CI=logtt+1 +CI=logxexxex+1 +C

Q19.

Answer :

We have,I=∫5×2-1 dxx x-1 x+1Let 5×2-1x x-1 x+1=Ax+Bx-1+Cx+1⇒5×2-1x x-1 x+1=A x2-1 +Bx·x+1+C·x· x-1x x-1 x+1⇒5×2-1=A x2-1+B·x x+1+C·x· x-1Putting x=1⇒5-1=A×0+B 1 1+1+C×0⇒4=B 2⇒B=2Putting x=0⇒5×0-1=A 0-1+B×0+C×0⇒-1=A -1⇒A=1Putting x+1=0x=-15-1=A×0+B×0+C -1 -2⇒C=2∴I=∫dxx+2∫dxx-1+2∫dxx+1=log x+2 log x-1+2 log x+1+C=log x+2 log x2-1+C=log xx2-12+C

Q20.

Answer :

We have,I=∫x2+6x-8×3-4xdx=∫x2+6x-8x x2-4dx=∫x2+6x-8x x-2 x+2dxLet x2+6x-8x x-2 x+2=Ax+Bx-2+Cx+2⇒x2+6x-8x x-2 x+2=A x-2 x+2+B x x+2+C x x-2x x-2 x+2⇒x2+6x-8=A x2-4+B x2+2x+C x2-2xPutting x-2=0⇒x=24+6×2-8=A×0+B 4+4⇒8=B×8⇒B=1Putting x=-24-12-8=A×0+B×0+C×8⇒C=-2Putting x=0-8=A -4+B×0+C×0⇒A=2∴I=∫2x+∫dxx-2-2∫dxx+2=2 log x+log x-2-2 log x+2+C=log x2+log x-2-log x+22+C=log x2x-2x+22+C

Q21.

Answer :

We have,I=∫x2+1 dx2x+1 x2-1=∫x2+1 dx2x+1 x-1 x+1Let x2+12x+1 x-1 x+1=A2x+1+Bx-1+Cx+1⇒x2+12x+1 x-1 x+1=A x2-1+B 2x+1 x+1+C 2x+1 x-12x+1 x-1 x+1⇒x2+1=A x2-1+B 2x+1 x+1+C 2x+1 x-1Putting x-1=0⇒x=11+1=A×0+B 2+1 1+1+C×0⇒2=B32⇒B=13Putting x+1=0⇒x=-11+1=A×0+B×0+C -2+1 -1-1⇒2=C -1 -2⇒C=1Putting 2x+1=0⇒x=-12-122+1=A 14-1⇒14+1=A -34⇒54=A -34A=-53∴I=-53∫dx2x+1+13∫dxx-1+∫dxx+1=-53× log 2x+12+13 log x-1+log x+1+C=-56 log 2x+1+13 log x-1+log x+1+C

Q22.

Answer :

We have,I=∫dxx 6 log x2+7 log x+2Putting log x=t⇒1x dx=dt∴I=∫dt6t2+7t+2=∫dt3t+2 2t+1Let 13t+2 2t+1=A3t+2+B2t+1⇒13t+2 2t+1=A 2t+1+B 3t+23t+2 2t+1⇒1=A 2t+1+B 3t+2Putting 2t+1=0⇒t=-121=0+B 3×-12+2⇒1=B 12⇒B=2Putting 3t+2=0⇒t=-231=A 2×-23+1+0⇒1=A -43+1⇒1=A -13⇒A=-3∴I=∫-33t+2+22t+1dt=-3 log 3t+23+2 log 2t+12+C=-log 3t+2+log 2t+1+C=log 2t+13t+2+C=log 2 log x+13 log x+2+C

Page 19.161 Ex.19.25

Q23.

Answer :

We have,I=∫dxx xn+1=∫xn-1dxxn-1x xn+1=∫xn-1 dxxn xn+1Putting xn=t⇒n xn-1 dx=dt⇒xn-1dx=dtn∴I=1n∫dtt t+1Let 1t t+1=At+Bt+1⇒1t t+1=A t+1+Btt t+1⇒1=A t+1+BtPutting t+1=0⇒t=-11=A×0+B -1⇒B=-1Putting t=01=A 0+1+B×0⇒A=1Then,I=1n∫dtt-1n∫dtt+1=1n log t-1nlog t+1+C=1n log tt+1+C=1n log xnxn+1+C

Q24.

Answer :

We have,I=∫x dxx2-a2 x2-b2Putting x2=t⇒2x dx=dt⇒x dx=dt2∴I=12∫dtt-a2 t-b2Let 1t-a2 t-b2=At-a2+Bt-b2⇒1t-a2 t-b2=A t-b2+B t-a2t-a2 t-b2⇒1=A t-b2+B t-a2Putting t=b21=A×0+B b2-a2⇒B=1b2-a2Putting t=a21=A a2-b2+B ×0⇒A=1a2-b2I=12∫dtt-a2 t-b2=12 a2-b2∫dtt-a2+12 b2-a2∫dtt-b2=12 a2-b2 log t-a2+12 b2-a2 log t-b2+C=12 a2-b2 log t-a2-log t-b2+C=12 a2-b2 log t-a2t-b2 +C=12 a2-b2 log x2-a2x2-b2+C

Q25.

Answer :

We have,I=∫ax2+bx+cx-a x-b x-c dxLet ax2+bx+cx-a x-b x-c=Ax-a+Bx-b+Cx-c⇒ax2+bx+c=Ax-b x-c+B x-cx-a+Cx-a x-b⇒ax2+bx+c=Ax2-b+cx+bc+Bx2-c+ax+ca+Cx2-a+bx+ab⇒ax2+bx+c=A+B+Cx2-Ab+c+Bc+a+Ca+bx+Abc+Bca+CabEquating the coefficients on both sides, we geta=A+B+C …..1b=-Ab+c+Bc+a+Ca+b …..2c=Abc+Bca+Cab …..3Solving 1, 2 and 3, we getA=a3+ab+ca-ba-cB=ab2+b2+cb-ab-cC=ac2+bc+cc-ac-b∴I=∫a3+ab+ca-ba-c×1x-a+ab2+b2+cb-ab-c×1x-b+ac2+bc+cc-ac-b×1x-c dx=a3+ab+ca-ba-clog x-a+ab2+b2+cb-ab-clog x-b+ac2+bc+cc-ac-blog x-c+K

Q26.

Answer :

We have,I=∫x3+x+1×2-1dx

As Degree of Numerator is greater than Degree of Denominator we divide numerator by denominator

x2-1×3+x+1x x3-x – + 2x+1∴x3+x+1×2-1=x+2x+1×2-1 …..1⇒x3+x+1×2-1=x+2xx2-1+1×2-1Then,I=∫x dx+∫2xx2-1+∫dxx2-12Putting x2-1=t⇒2x dx=dt∴I=∫x dx+∫dtt+∫dxx2-12=x22+log t+12 log x-1x+1+C=x22+log x2-1+12 log x-1x+1+C

Q27.

Answer :

We have,I=∫3x-2 dxx+12 x+3Let 3x-2x+12 x+3=Ax+1+Bx+12+Cx+3⇒3x-2x+12 x+3=A x+1 x+3+B x+3+C x+12x+12 x+3⇒3x-2=A x+1 x+3+B x+3+C x+12⇒3x-2=A x2+x+3x+3+B x+3+C x2+2x+1⇒3x-2=A+C x2+x 4A+B+2C+3A+3B+CEquating the coefficients of like termsA+C=0 …..14A+B+2C=3 …..23A+3B+C=-2 …..3Solving 1, 2 and 3 we getA=114,B=-52 and C=-114∴3x-2x+12 x+3=114 x+1-52 x+12-114 x+3⇒I=114∫dxx+1-52∫x+1-2 dx -114∫dxx+3=114 log x+1-52 x+1-1-1-114log x+3+C=114log x+1x+3+52 x+1+C

Q28.

Answer :

We have,I=∫2x+1 dxx+2 x-32Let 2x+1x+2 x-32=Ax+2+Bx-3+Cx-32⇒2x+1x+2 x-32=A x-32+B x+2 x-3+C x+2x+2 x-32⇒2x+1=A x2-6x+9+B x2-x-6+C x+2⇒2x+1=A+B x2+-6A-B+C x+9A-6B+2CEquating the coefficients of like termsA+B=0 ….. 1-6A-B+C=2 ….. 29A-6B+2C=1 …..3Solving 1, 2 and 3, we getA=-325, B=325 and C=75∴2x+1 dxx+2 x-32=-325 x+2+325 x-3+75 x-32⇒I=-325∫dxx+2+325∫dxx-3+75∫x-3-2 dx=-325 log x+2+325 log x-3+75x-3-1-1+C=-325log x+2+325 log x-3-75 x-3+C

Q29.

Answer :

We have,I=∫x2+1 dxx-22 x+3Let x2+1x-22 x+3=Ax-2+Bx-22+Cx+3⇒x2+1x-22 x+3=A x-2 x+3+B x+3+C x-22x-22 x+3⇒x2+1=A x2-2x+3x-6+B x+3+C x2-4x+4⇒x2+1=A x2+x-6 +B x+3+C x2-4x+4Equating coefficients of like termsA+C=1 ….. 1A+B-4C=0 ….. 2-6A+3B+4C=1 ….. 3Solving 1, 2 and 3, we getA=35, B=1 and C=25∴I=35∫dxx-2+∫dxx-22+25∫dxx+3=35 log x-2+x-2-2+1-2+1+25 log x+3+C=35log x-2-1x-2+25 log x+3+C

Q30.

Answer :

We have,I=∫x dxx-12 x+2Let xx-12 x+2=Ax-1+Bx-12+Cx+2⇒xx-12 x+2=A x-1 x+2+B x+2+C x-12x-12 x+2⇒x=A x2+2x-x-2+B x+2+C x2-2x+1⇒x=A x2+x-2 +B x+2+C x2-2x+1⇒x=A+C x2+A+B-2C x+-2A+2B+CEquating coefficients of like termsA+C=0 …..1A+B-2C=1 …..2-2A+2B+C=0 …..3Solving 1, 2 and 3, we getA=29, B=13 and C=-29∴xx-12 x+2=29 x-1+13 x-12-29 x+2⇒I=29∫dxx-1+13∫dxx-12-29∫dxx+2=29 log x-1+13×-1x-1-29 log x+2+C=29log x-1x+2-13 x-1+C

Q31.

Answer :

We have,I=∫x2 dxx-1 x+12Let x2x-1 x+12=Ax-1+Bx+1+Cx+12⇒x2x-1 x+12=A x+12+B x+1 x-1+C x-1x+12 x-1⇒x2=A x2+2x+1+B x2-1+C x-1⇒x2=A+B x2+x 2A+C +A-B-CEquating coefficients of like termsA+B=1 …..12A+C=0 …..2A-B-C=0 …..3Solving 1, 2 and 3, we getA=14, B=34 and C=-12∴x2x-1 x+12=14 x-1+34 x+1-12 x+12⇒I=14∫dxx-1+34∫dxx+1-12∫dxx+12=14 log x-1+34 log x+1-12×-1x+1+C=14log x-1+34 log x+1+12 x+1+C

Q32.

Answer :

We have,I=∫x2+x-1 dxx+12 x+2Let x2+x-1x+12 x+2=Ax+1+Bx+12+Cx+2⇒x2+x-1x+12 x+1=A x+1 x+2+B x+2+C x+12x+12 x+2⇒x2+x-1=A x2+3x+2+B x+2+C x2+2x+1Equating coefficients of like termsA+C=1 …..13A+B+2C=1 …..22A+2B+C=-1 …..3Solving 1, 2 and 3, we getA=0 B=-1C=1∴x2+x-1x+12 x+2=-1x+12+1x+2⇒I=∫-dxx+12+∫dxx+2=-∫x+1-2dx+∫dxx+2=-x+1-2+1-2+1 +log x+2+C=1x+1+log x+2+C

Q33.

Answer :

We have,I=∫2×2+7x-3 dxx2 2x+1Let 2×2+7x-3×2 2x+1=Ax+Bx2+C2x+1⇒2×2+7x-3×2 2x+1=A x 2x+1+B 2x+1+Cx2x2 2x+1⇒2×2+7x-3=A 2×2+x+B 2x+1+Cx2⇒2×2+7x-3=2A+C x2+A+2Bx+BEquating coefficients of like terms2A+C=2 …..1A+2B=7 …..2B=-3 …..3Solving 1, 2 and 3, we getA=13B=-3C=-24∴2×2+7x-3×2 2x+1=13x-3×2-242x+1⇒I=13∫dxx-3∫x-2 dx-24∫dx2x+1=13 log x+3x-24 log 2x+12+C=13 log x+3x-12 log 2x+1+C

Q34.

Answer :

We have,I=∫5×2+20x+6×3+2×2+x=∫5×2+20x+6 dxx x2+2x+1=∫5×2+20x+6 dxx x+12Let 5×2+20x+6x x+12=Ax+Bx+1+Cx+12⇒5×2+20x+6x x+12=A x+12 +B x x+1+C xx x+12⇒5×2+20x+6=A x2+2x+1+B x2+x+Cx⇒5×2+20x+6=A+B x2+2A+B+C x+AEquating coefficients of like termsA+B=5 …..12A+B+C=20 …..2 A=6 …..3Solving 1, 2 and 3, we getA=6 B=-1C=9∴5×2+20x+6x x+12=6x-1x+1+9x+12⇒I=6∫dxx-∫dxx+1+9∫dxx+12=6 log x-log x+1-9x+1+C

Q35.

Answer :

We have,I=∫18x+2 x2+4 dxLet 18x+2 x2+4=Ax+2+Bx+Cx2+4⇒18x+2 x2+4=A x2+4+Bx+C x+2x+2 x2+4⇒18=Ax2+4A+Bx2+2Bx+Cx +2C⇒18=A+B x2+x 2B+C+4A+2CEquating coefficients of like termsA+B=0 …..12B+C=0 …..24A+2C=18 …..3Solving 1, 2 and 3, we getA=94B=-94C=92∴18x+2 x2+4=94 x+2+-94x+92×2+4⇒18x+2 x2+4=94 x+2-94 xx2+4+92 x2+4⇒∫18 dxx+2 x2+4=94∫dxx+2-94∫x dxx2+4+92∫dxx2+22let x2+4=t⇒2xdx=dt⇒x dx=dt2∴I=94∫dxx+2-98∫dtt+92∫dxx2+22=94 log x+2-98 log t+92×12 tan-1 x2+C’=94 log x+2-98 log x2+4+94 tan-1 x2+C’

Q36.

Answer :

We have,I=∫5 dxx2+1 x+2 Let 5x+2 x2+1=Ax+2+Bx+Cx2+1⇒5x+2 x2+1=A x2+1+Bx+C x+2x+2 x2+1⇒5=A x2+1+Bx2+2Bx+Cx +2C⇒5=A+B x2+2B+C x+A+2CEquating coefficients of like termsA+B=0 …..12B+C=0 …..2A+2C=5 …..3Solving 1, 2 and 3, we getA=1B=-1C=2∴5x+2 x2+1=1x+2+-x+2×2+1⇒∫5 dxx+2 x2+1=∫dxx+2-∫x dxx2+1+2∫dxx2+1let x2+1=t⇒2xdx=dt⇒x dx=dt2∴I=∫dxx+2-12∫dtt+2∫dxx2+12=log x+2-12 log t+2 tan-1x+C’=log x+2-12 log x2+1+2 tan-1x+C’

Q37.

Answer :

We have,I=∫x dxx+1 x2+1 Let xx+1 x2+1=Ax+1+Bx+Cx2+1⇒xx+1 x2+1=A x2+1+Bx+C x+1x+1 x2+1⇒x=A x2+1+Bx2+Bx+Cx+C⇒x=A+B x2+B+C x+A+CEquating coefficients of like termsA+B=0 …..1B+C=1 …..2A+C=0 …..3Solving 1, 2 and 3, we getA=-12B=12C=12∴xx+1 x2+1=-12 x+1+x2+12×2+1⇒∫x dxx+1 x2+1=-12∫dxx+1+12∫x dxx2+1+12∫dxx2+1let x2+1=t⇒2x dx=dt⇒x dx=dt2∴I=-12∫dxx+1+14∫dtt+12∫dxx2+12=-12 log x+1+14 log t+12 tan-1x+C’=-12 log x+1+14 log x2+1+12 tan-1x+C’

Q38.

Answer :

We have,I=∫dx1+x+x2+x3 =∫dx1+x+x2 1+x=∫dxx+1 x2+1Let 1x+1 x2+1=Ax+1+Bx+Cx2+1⇒1x+1 x2+1=A x2+1+Bx+C x+1x+1 x2+1⇒1=A x2+1+Bx2+Bx+Cx+C⇒1=A+B x2+B+C x+A+CEquating coefficients of like termsA+B=0 …..1B+C=0 …..2A+C=1 …..3Solving 1, 2 and 3, we getA=12B=-12C=12∴I=12∫dxx+1+12∫-x+1×2+1 dx=12∫dxx+1-12∫x dxx2+1+12∫dxx2+12let x2+1=t⇒2x dx=dt⇒x dx=dt2∴I=12∫dxx+1-14∫dtt+12∫dxx2+12=12 log x+1-14 log t+12 tan-1 x+C’=12 log x+1-14 log x2+1+12 tan-1 x+C’

Q39.

Answer :

We have,I=∫dxx+12 x2+1 Let 1x+12 x2+1=Ax+1+Bx+12+Cx+Dx2+1⇒1x+12 x2+1=A x+1 x2+1+B x2+1+Cx+D x+12x+12 x2+1⇒1=A x3+x+x2+1+B x2+1+Cx+Dx2+2x+1⇒1=A x3+x2+x+1+B x2+1+Cx3+2Cx2+Cx+Dx2+2Dx+D⇒1=A+C x3+A+B+2C+Dx2+A+C+2D x+A+B+DEquating coefficients of like termsA+C=0 …..1A+B+2C+D=0 …..2A+C+2D=0 …..3A+B+D=1 …..4A=12, B=12, C=-12 and D=0∴1x+12 x2+1=12 x+1+12 x+12-12 ×xx2+1⇒∫dxx+12 x2+1=12∫dxx+1+12∫dxx+12-12∫x dxx2+1Putting x2+1=t⇒2x dx=dt⇒x dx=dt2∴I=12∫dxx+1+12∫dxx+12-14∫dtt=12 log x+1-12 x+1-14 log t+C’=12 log x+1-12 x+1 -14 log x2+1+C’

Q40.

Answer :

We have,I=∫2x dxx3-1 =∫2x dxx-1 x2+x+1Let 2xx-1 x2+x+1=Ax-1+Bx+Cx2+x+1⇒2xx-1 x2+x+1=A x2+x+1+Bx+C x-1x-1 x2+x+1⇒2x=A x2+x+1+Bx2-Bx+Cx-C⇒2x=A+B x2+A-B+C x+A-CEquating coefficients of like termsA+B=0 …..1A-B+C=2 …..2A-C=0 …..3Solving 1, 2 and 3, we getA=23B=-23C=23∴I=23∫dxx-1+23∫-x+1×2+x+1dxLet -x+1=addx x2+x+1+b⇒-x+1=a 2x+1+b⇒-x+1=2a x+a+bEquating coefficients of like terms 2a=-1⇒a=-12Anda+b=1⇒-12+b=1⇒b=32∴I=23∫dxx-1+23∫-12 2x+1+32×2+x+1dx=23∫dxx-1-13∫2x+1×2+x+1 dx+∫dxx2+x+1=23∫dxx-1-13∫2x+1×2+x+1dx+∫dxx2+x+14-14+1=23∫dxx-1-13∫2x+1 dxx2+x+1+∫dxx+122+322let x2+x+1=t⇒2x+1 dx=dtThen,I=23∫dxx-1-13∫dtt+∫dxx+122+322=23 log x-1-13 log t+23 tan-1 x+1232+C’=23 log x-1-13 log x2+x+1+23 tan-1 2x+13+C’

Q41.

Answer :

We have,I=∫dxx2+1 x2+4 Putting x2=tThen, 1×2+1 x2+4 =1t+1 t+4Let 1t+1 t+4=At+1+Bt+4⇒1=A t+4+B t+1Putting t+4=0⇒t=-4∴1=A×0+B -3⇒B=-13Putting t+1=0⇒t=-1∴1=A -1+4+B×0⇒A=13∴1t+1 t+4=13 t+1-13 t+4⇒1×2+1 x2+4=13 x2+1-13 x2+22⇒∫dxx2+1 x2+4=13∫dxx2+12-13∫dxx2+22=13 tan-1x-13×12 tan-1×2+C=13 tan-1x-16 tan-1 x2+C

Q42.

Answer :

We have,I=∫x2 dxx2+1 3×2+4 Putting x2=tThen, x2x2+1 3×2+4 =tt+1 3t+4Let tt+1 3t+4=At+1+B3t+4⇒tt+1 3t+4=A 3t+4+B t+1 t+1 3t+4⇒t=A 3t+4+B t+1Putting t+1=0⇒t=-1∴-1=A -3+4+0⇒A=-1Putting 3t+4=0⇒t=-43∴-43=0+B -43+1⇒-43=B×-13⇒B=4∴tt+1 3t+4=-1t+1+43t+4⇒x2x2+1 3×2+4=-1×2+1+43×2+4⇒x2x2+1 3×2+4=-1×2+1+43 x2+43⇒∫x2 dxx2+1 3×2+4=-∫dxx2+1+43∫dxx2+232=-tan-1 x+43×32 tan-1 3×2+C=-tan-1 x+23 tan-1 3×2+C

Q43.

Answer :

We have,I=∫ x2+1 x2+2×2+3 x2+4Putting x2=tThen,x2+1 x2+2×2+3 x2+4=t+1 t+2t+3 t+4=t2+3t+2t2+7t+12
Degree of numerator is equal to degree of denominator.
We divide numerator by denominator.
1t2+7t+12 t2+3t+2 t2+7t+12 – – – -4t-10
∴t2+3t+2t2+7t+12=1-4t+10t2+7t+12⇒t2+3t+2t2+7t+12=1-4t+10t+3 t+4 …..1Let 4t+10t+3 t+4=At+3+Bt+4⇒4t+10t+3 t+4=At+4+Bt+3t+3 t+4⇒4t+10=At+4+Bt+3Putting t+4=0⇒t=-4∴-16+10=B-1⇒B=6Putting t+3=0⇒t=-3∴-12+10=A-3+4⇒A=-2∴4t+10t+3 t+4=-2t+3+6t+4 …..2From 1 & 2t2+3t+2t2+7t+12=1+2t+3-6t+4∴∫x2+1 x2+2dxx2+3 x2+4=∫dx+2∫dxx2+32-6∫dxx2+22 =x+23tan-1 x3-62tan-1 x2+C =x+23 tan-1 x3-3 tan-1 x2+C

Q44.

Answer :

We have,I=∫ x3-1dxx3+x
Degree of numerator is equal to degree of denominator.
We divide numerator by denominator.
1×3+x x3-1 x3+x – – -x-1 ∴x3-1×3+x=1-x+1×3+x⇒x3-1×3+x=1-x+1xx2+1 …..1Let x+1xx2+1=Ax+Bx+Cx2+1⇒x+1xx2+1=Ax2+1+Bx+Cxxx2+1⇒x+1=Ax2+A+Bx2+Cx⇒x+1=A+Bx2+Cx+A
Equating coefficient of like terms
A + B = 0
C = 1
A = 1
B = –1
∴x+1xx2+1=1x+-x+1×2+1 …..2Using 1 & 2∫ x3-1dxx3+x=∫ 1-1x+xx2+1-1×2+1dx=∫dx-∫dxx+∫x dxx2+1-∫dxx2+1 Putting x2+1=t⇒2x dx=dt⇒x dx=dt2∴I=∫ dx-∫dxx+12∫dtt-∫dxx2+1=x-log x+12log t-tan-1x+C=x-log x+12log x2+1-tan-1x+C

Q45.

Answer :

We have,I=∫ 4×4+3dxx2+2 x2+3 x2+4Putting x2=tThen,4×4+3×2+2 x2+3 x2+4=4t2+3t+2 t+3 t+4Let 4t2+3t+2 t+3 t+4=At+2+Bt+3+Ct+4⇒4t2+3t+2 t+3 t+4=At+3 t+4+Bt+2 t+4+Ct+2 t+3t+2 t+3 t+4⇒4t2+3=At+3 t+4+Bt+2 t+4+Ct+2 t+3Putting t+3=0⇒t=-3∴4×-32+3=B-3+2 -3+4⇒39=B-1⇒B=-39Putting t+2=0⇒t=-2∴4-22+3=A-2+3 -2+4⇒19=A×1×2⇒A=192Let t+4=0⇒t=-4∴4×-42+3=C-4+2 -4+3⇒67=C-2 -1⇒C=672∴4t2+3t+2 t+3 t+4=192t+2-39t+3+672t+4⇒4×4+3×2+2 x2+3 x2+4=192×2+2-39×2+3+672×2+4∴I=192∫dxx2+22-39∫dxx2+32-672∫dxx2+22=192×12tan-1 x2-393tan-1 x3-672×12tan-1 x2+C=1922 tan-1 x2-393tan-1 x3-674tan-1 x2+C

Q46.

Answer :

We have,I=∫ dxxx4+1=∫x3 dxx4 x4+1Putting x4=t⇒4×3 dx=dt⇒x3 dx=dt4∴I=14∫dttt+1Let 1tt+1=At+Bt+1⇒1tt+1=At+1+Bttt+1⇒1=At+1+BtPutting t+1=0⇒t=-1∴1=A×0+B-1⇒B=-1Putting t=0∴1=A1+B×0⇒A=1∴ I=14∫dtt-14∫dtt+1=14log t-14log t+1+C=14log tt+1+C=14log x4x4+1+C

Q47.

Answer :

We have,I=∫1x x5+1 dx=∫ x4dxx5x5+1Putting x5=t⇒5x4dx=dt⇒x4dx=dt5∴I=15∫dttt+1 …..(1)Let 1tt+1=At+Bt+1 …..(2)⇒1tt+1=At+1+Bttt+1⇒1=At+1+Bt …..(3) Putting t=0 in (3)⇒1=A0+1+B×0⇒A=1Putting t+1=0 in (3)⇒t=-1∴1=A×0+B-1⇒B=-1∴I=15∫dtt-∫dtt+1 From (1) & (2)=15log t-log t+1+C=15log tt+1+C=15log x5x5+1+C

Q48.

Answer :

We have,I=∫ 3 dx1-x 1+x2=3∫dx1-x 1+x2Let 11-x 1+x2=A1-x+Bx+Cx2+1⇒11-x x2+1=Ax2+1+Bx+C 1-x1-x x2+1⇒1=Ax2+A+Bx-Bx2+C-Cx⇒1=A-Bx2+B-Cx+A+CEquating coefficients of like terms.A-B=0 …..1B-C=0 …..2A+C=1 …..3Solving 1, 2 and 3, we getA=12, B=12, C=12∴11-x x2+1=121-x+x2+12×2+1∫ 3 dx1-x x2+1=32∫dx1-x+32∫x dxx2+1+32∫dxx2+1Putting x2+1=t⇒x dx=dt2∴I=32∫dx1-x+34∫dtt+32∫dxx2+1=32log 1-x-1+34log t+32×tan-1x+C=-32log 1-x+34log 1+x2+32tan-1x+C=-34×2 log 1-x+34log 1+x2+342 tan-1 x+C=34log 1+x2-log 1-x2+342 tan-1 x+C=34log 1+x21-x2+2 tan-1 x+C

Q49.

Answer :

We have, I=∫cos x1-sin x3 2+sin x dxLet, sinx=t⇒cosx dx=dtNow, integration becomes,I=∫dt1-t3 2+t =-∫dtt-13 t+2 Let, 1t-13 t+2=At-1+Bt-12+Ct-13+D t+2 …..1⇒1=At-12t+2+Bt-1t+2+Ct+2+Dt-13 …..2

Putting t=1 in 2, we get1=3C⇒C=13Putting t=-2 in 2, we get1=D-2-13⇒1=-27D⇒D=-127Putting t=0 in 2, we get1=2A-2B+2C-D⇒1=2A-2B+23+127⇒2A-2B=827⇒A-B=427Putting t=2 in 2, we get1=4A+4B+4C+D⇒1=4A+4B+43-127⇒A+B=-227Now, A-B=427 and A+B=-227 ⇒A=127 and B=-19
Substituting the values of A, B, C and D in 1, we get1t-13 t+2=127t-1-19t-12+13t-13+-1 27t+2Now, integration becomes I=-∫127t-1-19t-12+13t-13+-1 27t+2dt =-127log t-1+19t-1-16t-12-127log t+2+C =-127log sin x-1-19sin x-1+16sin x-12+127log sin x+2+C =-127log 1-sin x+191-sin x+161-sin x2+127log 2+sin x+C

Q50.

Answer :

We have,I=∫ x4 dxx-1 x2+1=∫ x4-1+1x-1 x2+1dx=∫ x4-1dxx-1 x2+1+∫dxx-1 x2+1=∫x2-1 x2+1 dxx-1 x2+1+∫dxx-1 x2+1=∫ x-1 x+1dxx-1+∫dxx-1 x2+1=∫x+1dx+∫dxx-1 x2+1 …..1Let 1x-1 x2+1=Ax-1+Bx+Cx2+1⇒1x-1 x2+1=Ax2+1+Bx+C x-1x-1 x2+1⇒1=Ax2+A+Bx2-Bx+Cx-C⇒1=A+Bx2+C-Bx+A-CEquating coefficients of like termsA+B=0 …..1C-B=0 …..2A-C=1 …..3Solving 1, 2 and 3, we getB=-12, A=12, C=-12∴1x-1 x2+1=12x-1+-x2-12×2+1⇒1x-1 x2+1=12x-1-12xx2+1-12×2+1 …..2From 1 & 2I=∫x+1dx+12∫dxx-1-12∫x dxx2+1-12∫dxx2+1Putting x2+1=t⇒2x dx=dt⇒x dx=dt2∴I=∫x+1dx+12∫dxx-1-14∫dtt-12∫dxx2+1=x22+x+12log x-1-14log t-12tan-1x+C=x22+x+12log x-1-14log x2+1-12tan-1x+C

Q51.

Answer :

We have,I=∫ cos x dx1-sin x 2-sin xPutting sin x=t⇒cos x dx=dt∴I=∫dt1-t 2-t=∫dtt-1 t-2Let 1t-1 t-2=At-1+Bt-2⇒1t-1 t-2=At-2+Bt-1t-1 t-2⇒1=At-2+Bt-1Putting t-1=0⇒t=1∴1=A1-2+B×0⇒A=-1Putting t-2=0⇒t=2∴1=A×0+B2-1⇒B=1∴I=∫-dtt-1+∫dtt-2=-log t-1+log t-2+C=logt-2t-1+C=log sin x-2sin x-1+C=log 2-sin x1-sin x+C

Q52.

Answer :

We have,I=∫ 2x+1dxx-2 x-3Let 2x+1x-2 x-3=Ax-2+Bx-3⇒2x+1x-2 x-3=Ax-3+Bx-2x-2 x-3⇒2x+1=Ax-3+Bx-2Putting x-3=0⇒x=3∴7=A×0+B×3-2⇒B=7Putting x-2=0⇒x=2∴5=A-1⇒A=-5∴I=-5∫dxx-2+7∫dxx-3=-5 log x-2+7 log x-3+C=log x-37-logx-25+C=log x-37x-25+C

Q53.

Answer :

We have,I=∫ dxx2+1 x2+2Putting x2=tThen,1×2+1 x2+2=1t+1 t+2Let 1t+1 t+2=At+1+Bt+2⇒1t+1 t+2=At+2+Bt+1t+1 t+2⇒1=At+2+Bt+1Putting t+2=0⇒t=-2∴1=A×0+B-1⇒B=-1Putting t+1=0⇒t=-1∴1=A-1+2+B×0⇒A=1∴1t+1 t+2=1t+1-1t+2⇒1×2+1 x2+2=1×2+1-1×2+22∴I=∫ dxx2+12-∫dxx2+22=tan-1 x-12tan-1 x2+C

Q54.

Answer :

We have,I=∫ dxxx4-1=∫ x3 dxx4 x4-1Putting x4=t⇒4×3 dx=dt⇒x3 dx=dt4∴I=14∫dttt-1Let 1tt-1=At+Bt-1⇒1tt-1=At-1+B ttt-1⇒1=At-1+BtPutting t-1=0⇒t=1∴1=A×0+B1⇒B=1Putting t=0∴1=A0-1+B×0⇒A=-1∴I=-14∫dtt+14∫dtt-1=-14log t+14log t-1+C=14log t-1t+C=14log x4-1×4+C

Q55.

Answer :

We have,I=∫dxx4-1=∫dxx2-1 x2+1=∫dxx-1 x+1 x2+1Let 1x-1 x+1 x2+1=Ax-1+Bx+1+Cx+Dx2+1⇒1x-1 x+1 x2+1=Ax2+1 x+1+Bx-1 x2+1 Cx+D x-1 x+1x-1 x+1 x2+1⇒1=Ax2+1 x+1+B x-1 x2+1+Cx+D x2-1⇒1=Ax3+x2+x+1+Bx3+x-x2-1+Cx3-Cx+Dx2-D⇒1=A+B+C x3+x2A-B+D+xA+B-C+A-B-DEquating the coefficients of like terms.A+B+C=0 …..1A-B+D=0 …..2A+B-C=0 …..3A-B-D=1 …..4Solving these four equations we getA=14, B=-14, C=0, D=-12∴1x-1 x+1 x2+1=14x-1-14x+1-12×2+1⇒I=14∫ dxx-1-14∫dxx+1-12∫dxx2+1=14log x-1 -14log x+1-12tan-1 x+C’=14log x-1x+1-12tan-1 x+C’

Q56.

Answer :

We have,I=∫3x+5dxx3-x2-x+1=∫3x+5dxx2x-1-1x-1=∫3x+5dxx2-1 x-1=∫3x+5dxx-1 x+1 x-1=∫3x+5dxx-12 x+1Let 3x+5x-12 x+1=Ax+1+Bx-1+Cx-12⇒3x+5x-12 x+1=Ax-12+Bx+1 x-1+Cx+1x+1 x-12⇒3x+5=Ax2-2x+1+Bx2-1+Cx+C⇒3x+5=A+Bx2+-2A+Cx+A-B+CEquating coefficient of like termsA+B=0 …..1-2A+C=3 …..2A-B+C=5 …..3Solving these three equations we getA=12B=-12C=4∴3x+5x-12 x+1=12x+1-12x-1+4x-12⇒I=12∫dxx+1-12∫dxx-1+4∫x-1-2 dx=12log x+1-12log x-1-4x-1+C’=12log x+1x-1-4x-1+C’

Q57.

Answer :

We have,I=∫x2+x+1x+12 x+2dxLet x2+x+1x+12 x+2=Ax+1+Bx+12+Cx+2⇒x2+x+1x+12 x+2=Ax+1 x+2+Bx+2+Cx+12x+12 x+2⇒x2+x+1=Ax2+x+2x+2+Bx+2B+Cx2+2x+1⇒x2+x+1=A+Cx2+3A+B+2Cx+2A+2B+CEquating coefficient of like terms.A+C=1 …..13A+B+2C=1 …..22A+2B+C=1 …..3Solving these three equations we getA=-2B=1C=3Hence, x2+x+1x+12 x+2=-2x+1+1x+12+3x+2∴I=-2∫dxx+1+∫dx+12+3∫dxx+2=-2 log x+1-1x+1+3 log x+2+C

Page 19.169 Ex.19.26

Q1.

Answer :

We have,I=∫ x2+1dxx4+x2+1Dividing numerator and denominator by x2, we getI=∫ 1+1x2dxx2+1+1×2=∫ 1+1x2dxx2+1×2-2+3=∫ 1+1x2dxx-1×2+3Putting x-1x=t⇒1+1x2dx=dt∴I=∫ dtt2+3=∫dtt2+32=13tan-1 t3+C=13tan-1 x-1×3+C=13 tan-1 x2-13 x+C

Q2.

Answer :

We have,I=∫cot θ dθPutting cot θ=t2⇒-cosec2 θ dθ=2t dt⇒dθ=-2t dtcosec2 θ⇒dθ=-2t dt1+cot2 θ⇒dθ=-2t dt1+t4∴I=∫ t-2t dt1+t4=-∫2t21+t4dt=-∫t2+1+t2-1t4+1dt=-∫t2+1t4+1dt-∫t2-1dtt4+1Dividing numerator and denominator by t2I=-∫1+1t2t2+1t2dt-∫1-1t2t2+1t2dt=-∫1+1t2dtt2+1t2-2+2-∫1-1t2dtt2+1t2+2-2=-∫1+1t2dtt-1t2+22-∫1-1t2dtt+1t2-22Putting t-1t=p⇒1+1t2dt=dpPutting t+1t=q⇒1-1t2dt=dqI=-∫ dpp2+22-∫dqq2-22=-12tan-1 p2-122log q-2q+2+C=-12tan-1 t-1t2-122log t+1t-21+1t+2+C=-12 tan-1 t2-12 t-122log t2+1-2tt2+1+2t+C=-12tan-1 cot θ-12cot θ-122log cot θ+1-2 cot θcot θ+1+2 cot θ+C

Q3.

Answer :

We have,I=∫ x2+9×4+81dxDividing numerator and denominator by x2I=∫1+9x2dxx2+81×2=∫1+9x2dxx2+9×2-2×x×9x+2×x×9x=∫1+9x2dxx-9×2+182Putting x-9x=t⇒1+9x2dx=dt∴I=∫dtt2+182=∫dtt2+322=132tan-1 t32+C=132tan-1 x-9×32+C=132tan-1 x2-932x+C

Q4.

Answer :

We have,I=∫ dxx4+x2+1=12∫ 2 dxx4+x2+1⇒12∫x2+1-x2-1×4+x2+1dx⇒12∫x2+1×4+x2+1dx-12∫x2-1×4+x2+1dxDividing numerator and denominator by x2I=12∫1+1x2x2+1+1x2dx-12∫1-1x2x2+1+1x2dx=12∫1+1x2x2+1×2-2+3dx-12∫1-1x2dxx2+1×2+2-1=12∫1+1x2dxx-1×2+32-12∫1-1x2dxx+1×2-12Putting x-1x=t⇒1+1x2dx=dtPutting x+1x=p⇒1-1x2dx=dp∴I=12∫dtt2+32-12∫dpp2-12=12×13tan-1 t3-12×12×1log p-1p+1+C=123tan-1 x-1×3-14log x+1x-1x+1x+1+C=123tan-1 x2-1×3-14log x2-x+1×2+x+1+C

Q5.

Answer :

We have,I=∫x2-3x+1×4+x2+1dx=∫x2+1dxx4+x2+1-3∫x dxx4+x2+1 …..1=I1-3I2 where I1=∫x2+1dxx4+x2+1, I2=∫x dxx4+x2+1I1=∫x2+1×4+x2+1dxDividing numerator & denominator by x2I1=∫1+1x2dxx2+1×2+1=∫1+1x2dxx2+1×2-2+3=∫1+1x2dxx-1×2+32Let x-1x=t⇒1+1x2dx=dt∴I1=∫dtt2+32I1=13tan-1 t3+C1I1=13tan-1 x-1×3+C1 …..2I2=∫x dxx4+x2+1Putting x2=t⇒2x dx=dt⇒x dx=dt2∴I2=12∫dtt2+t+1=12∫dtt2+t+14+34=12∫dtt+122+322=132×12tan-1 t+1232+C2=13tan-12t+13+C2I2=13tan-1 2×2+13+C2 …3From equating 1, 2 and 3 we haveI=13tan-1 x-1×3+C1-3×13tan-1 2×2+13+C2=13tan-1 x2-13x-3tan-1 2×2+13+C where C=C1+3C2

Q6.

Answer :

We have,I=∫x2+1×4-x2+1dxDividing numerator and denominator by x2I=∫1+1x2x2+1×2-1dx=∫1+1x2dxx2+1×2-2+1=∫1+1x2dxx-1×2+1Putting x-1x=t⇒1+1x2dx=dt∴I=∫dtt2+12=tan-1t+C=tan-1x-1x+C=tan-1 x2-1x+C

Q7.

Answer :

We have,I=∫ x2-1×4+1dxDividing numerator and denominator by x2=∫1-1x2x2+1x2dx=∫1-1x2dxx2+1×2+2-2=∫1-1x2dxx+1×2-22Putting x+1x=t⇒1-1x2dx=dt∴I=∫dtt2-22=122log t-2t+2+C=122log x+1x-2x+1x+2+C=122log x2-2x+1×2+2x+1+C

Q8.

Answer :

We have,I=∫ x2+1×4+7×2+1dxDividing numerator and denominator by x2I=∫1+1x2x2+7+1x2dx=∫1+1x2dxx2+1×2-2+9⇒∫1+1x2dxx-1×2+32Putting x-1x=t⇒1+1x2dx=dt∴I=∫dtt2+32=13tan-1 t3+C=13tan-1 x-1×3+C=13tan-1 x2-13x+C

Q9.

Answer :

We have,I=∫x-12dxx4+x2+1=∫x2-2x+1×4+x2+1dx=∫x2+1×4+x2+1dx-∫2x dxx4+x2+1=I1-I2 where,I1=∫x2+1dxx4+x2+1I2=∫ 2x dxx4+x2+1Now,I1=∫ x2+1×4+x2+1dxDividing numerator and denominator by x2I1=∫1+1x2x2+1×2+1dxI1=∫1+1x2dxx2+1×2-2+3I1=∫1+1x2dxx-1×2+32Putting x-1x=t⇒1+1x2dx=dt∴I1=∫ dtt2+32=13tan-1 t3+C1=13tan-1 x-1×3+C1=13tan-1 x2-13x+C1AndI2=∫2x dxx4+x2+1Putting x2=t⇒2x dx=dtI2=∫ dtt2+t+1=∫dtt2+t+122-122+1=∫dtt+122+322=23tan-1 t+1232+C2=23tan-1 2t+13+C2∴I=13tan-1 x2-13x-23tan-1 2×2+13+C

Q10.

Answer :

We have,I=∫ dxx4+3×2+1=12∫ 2 dxx4+3×2+1=12∫x2+1-x2-1×4+3×2+1dx=12∫x2+1×4+3×2+1dx-12∫x2-1×4+3×2+1dxDividing numerator and denominator by x2=12∫1+1x2x2+1×2+3dx-12∫1-1x2dxx2+1×2+3=12∫1+1x2x2+1×2-2+5dx-12∫1-1x2dxx2+1×2+2+1=12∫1+1x2dxx-1×2+52-12∫1-1x2dxx+1×2+12Putting x-1x=t⇒1+1x2dx=dtPutting x+1x=p⇒1-1x2dx=dp∴I=12∫dtt2+52-12∫dpp2+12=125tan-1 t5-12tan-1 p+C=125tan-1 x-1×5-12tan-1 x+1x+C=125tan-1 x2-15x-12tan-1 x2+1x+C

Page 19.175 Ex.19.27

Q1.

Answer :

We have,I=∫ dxx-1 x+2Putting x+2=t2⇒dx=2t dt∴I=∫2t dtt2-2-1t=∫ 2 dtt2-32=2×123log t-3t+3+C=13log x+2-3x+2+3+C

Q2.

Answer :

We have,I=∫ dxx-1 2x+3Putting 2x+3=t2⇒x=t2-32Diff both sidesdx=t dt∴I=∫ t dtt2-32-1t=∫2 dtt2-3-2=2dtt2-5=2∫dtt2-52=2×125log t-5t+5+C=15log 2x+3-52x+3+5+C

Q3.

Answer :

We have,I=∫ x+1x-1 x+2dxPutting x+2=t2⇒x=t2-2Diff both sidesdx=2t dtI=∫ t2-2+12t dtt2-2-1t=2∫ t2-1t2-3dt=2∫t2-3+2t2-3dt=2∫ t2-3t2-3dt+4∫dtt2-3=2∫dt+4∫dtt2-32=2t+4×123log t-3t+3+C=2x+2+23log x+2-3x+2+3+C

Q4.

Answer :

We have,I=∫ x2x-1 x+2dxPutting x+2=t2x=t2-2Diff both sidesdx=2t dtI=∫ t2-22t2-2-1t2 t dt=2∫ t2-22dtt2-3=2∫ t4-4t2+4t2-3dtDividing numerator by denominator, we get t2-1t2-3 t4-4t2+4 t4-3t2 – + -t2+4 -t2+3 + – 1 ∴I=2∫t2-1+1t2-3dt =2∫ t2 dt-2∫dt+2∫dtt2-32=2t33-2t+2×123log t-3t+3+C=23x+23-2x+2+13log x+2-3x+2+3+C=23x+232-2x+2+13log x+2-3x+2+3+C

Q5.

Answer :

We have,I=∫ x dxx-3 x+1Putting x+1=t2⇒x=t2-1Diff both sidesdx=2t dt∴I=∫ t2-12t dtt2-1-3t=2∫ t2-1t2-4dt=2∫t2-4+3t2-4dt=2∫t2-4t2-4dt+6∫ dtt2-22=2∫ dt+6∫dtt2-22=2t+6×12×2log t-2t+2+C=2x+1+32log t-2t+2+C=2x+1+32log x+1-2x+1+2+C

Q6.

Answer :

We have,I=∫ dxx2+1 xPutting x=t2dx=2t dt∴I=∫ 2t dtt22+1t=2∫ dtt4+1=∫ t2+1-t2-1t4+1dt=∫t2+1t4+1dt-∫t2-1t4+1dtDividing numerator & denominator by t2I=∫1+1t2t2+1t2dt-∫ 1-1t2dtt2+1t2=∫ 1+1t2dtt2+1t2-2+2-∫ 1-1t2dtt2+1t2+2-2=∫ 1+1t2dtt-1t2+22-∫ 1-1t2dtt+1t2-22Putting t-1t=p⇒1+1t2dt=dpPutting t+1t=q⇒1-1t2dt=dq∴I=∫dpp2+22-∫dqq2-22=12tan-1 p2-122log q-2q+2+C=12tan-1 t-1t2-122log t+1t-2t+1t+2+C=12tan-1 t2-12t-122log t2-2t+1t2+2t+1+C=12tan-1 x-12x-122log x-2x+1x+2x+1+C

Q7.

Answer :

We have,I=∫ x dxx2+2x+2 x+1=∫ x dxx+12+1 x+1Putting x+1=t2⇒x=t2-1Diff both sidesdx = 2t dt∴I=∫ t2-12t dtt22+1 t=2∫ t2-1dtt4+1Dividing numerator and denominator by t2I=21-1t2t2+1t2dt

=2∫1-1t2dtt2+1t2+2-2=2∫ 1-1t2dtt+1t2-22Putting t+1t=p⇒1-1t2dt=dpI=2∫ dpp2-22=2×122log p-2p+2+C=12log p-2P+2+C=12log t+1t-2t+1t+2+C=12log t2-2t+1t2+2t+1+C=12log x+1-2x+1+1x+1+2x+1+1+C=12log x+2-2x+1x+2+2x+1+C

Q8.

Answer :

We have,I=∫ dxx-1 x2+1Putting x-1=1t⇒dx=-1t2dt∴I=∫-1t2dt1t 1+1t2+1=∫ -1tdt1+1t2+2t+1=∫ -1tdtt2+1+2t+t2t=∫ -dt2t2+2t+1=-12 ∫ dtt2+t+12=-12∫ dtt2+t+14-14+12=-12 ∫ dtt+122+122=-12log t+12+t+122+14+C where t=1x-1

Q9.

Answer :

We have,I=∫ dxx+1 x2+x+1Putting x+1=1t⇒dx=-1t2dt∴I=∫ -1t2dt1t1t-12+-1+11t=∫ -1t2dt1t1t2-+1+2t1t=∫ -1tdtt2+t-2t+1t=-∫ dtt2-t+1=-∫dtt2-t+14-14+1=-∫dtt-122+322=-log t-12+t-122+322+C=-log t-12+t2-t+1+C=-log 1x+1-12+1x+12-1x+1+1+C=-log 1x+1-12+x+12-x+1+1x+1+C=-log 1x+1-12+x2+x+1x+1+C

Q10.

Answer :

We have,I=∫ dxx2-1 x2+1Putting x=1t⇒dx=-1t2dt∴I=∫ -1t2dt1t2-1 1t2+1=∫ -1t2 dt1-t2t2×1+t2t=∫ -t dt1-t2 1+t2Putting 1+t2=u2⇒t2=u2-1⇒2t dt=2u du⇒t dt=u duI=-∫u du1-u2+1u=-∫ du2-u2=-∫ du22-u2=-122log u+2u-2+C=-122log 1+t2+21+t2-2+C=-122log 1+1×2+21+1×2-2+C=-122log x2+1+2xx2+1-2x+C

Q11.

Answer :

We have,I=∫ x dxx2+4 x2+1Putting x2=t⇒ 2x dx=dt⇒x dx=dt2∴I=12∫ dtt+4 t+1Again Putting t+1=p2⇒t=p2-1⇒dt=2p dpI=12∫ 2p dpp2-1+4p=∫ dpp2+3=∫dpp2+32=13tan-1 p3+C=13tan-1 t+13+C=13 tan-1 x2+13+C

Q12.

Answer :

We have,I=∫ dx1+x2 1-x2Putting x=1t⇒dx=-1t2dt∴I=∫ -1t2dt1+1t2 1-1t2=∫ -1t2dtt2+1t2 t2-1t=-∫ t dtt2+1 t2-1Again Putting t2-1=u2⇒2t dt=2u du⇒t dt=u du∴I=-∫ u duu2+2u=-∫ duu2+22=-12tan-1 u2+C=-12 tan-1 t2-12+C=-12tan-1 1×2-12+C=-12tan-1 1-x22x2+C

Q13.

Answer :

We have,I=∫dx2x2+3 x2-4Putting x=1t⇒dx=-1t2dt∴I=∫-1t2dt2t2+3 1t2-4=∫-1t2 dt2+3t2t2×1-4t2t=-∫t dt2+3t2 1-4t2Again Putting 1-4t2=u2⇒-8t dt=2u du⇒t dt=-u4 du∴I=14∫u du2+3 1-u24 u=14∫4 du8+3-3u2=∫du11-3u2=13∫du113-u2=13∫du1132-u2=13×12×113 log 113+u113-u+C=1233 log 11+3 u11-3 u+C=1233 log 11+3 1-4t211-3 1-4t2+C=1233 log 11+3-12t211-3-12t2+C=1233 log 11+3-12×211-3-12×2+C=1233 log 11x+3×2-1211x-3×2-12+C

Q14.

Answer :

We have,I=∫x dxx2+4 x2+9Putting x2=t⇒2x dx= dt⇒x dx=dt2∴I=12∫dtt+4 t+9Again Putting t+9=u2⇒dt=2u du∴I=12∫2u duu2-9+4 u=∫duu2-5=∫duu2-52=125 log u-5u+5+C=125 log t+9-5t+9+5+C=125 log x2+9- 5×2+9+ 5+C

Page 19.176 (Very Short Answers)

Q1.

Answer :

Let I=∫1+cot xx+log sin xdxLet x +log sin x=t⇒1+1sin x×cos x dx=dt⇒1+cot xdx=dt∴I=∫dtt =log t+C =log x+log sin x+C

Q2.

Answer :

∫e3 log x.x4 dx=∫ elog x3·x4 dx ∵alogx=logxa=∫ x3·x4 dx ∵elog m=m=∫x7·dx=x88+C

Q3.

Answer :

Let I=∫x².sin x³ dx
Let x³ = t
⇒ 3x²dx = dt
⇒x2 dx=dt3∴I=13∫ sin t dt =13-cos t+C =-13cos x3+C ∵ t=x3

Q4.

Answer :

Let I=∫ tan³ x . sec² x dx
Let tan x = t
⇒ sec²x dx = dt
∴I= ∫ t³ . dt
=t44+C=tan4 x4+C ∵ t=tan x

Q5.

Answer :

Let I=∫ e^x (sin x + cos x) dx
Let e^x sin x = t
⇒ (e^x . sin x + e^x cos x) dx = dt
∴ I=∫dt =t+C = exsin x+C ∵t=ex sinx

Q6.

Answer :

Let I=∫ tan⁶ x . sec² x dx
Let tan x = t
sec² x dx = dt
∴I=∫ t⁶ . dt
=t 77+C=tan7 x7+C ∵t=tan x

Q7.

Answer :

Let I=∫ cos x3+2 sin xdxLet 3+2 sin x=t⇒2 cos x dx=dt⇒cos x dx=dt2∴I=12∫dtt =∫12 log t+C =12log 3+2 sin x+C ∵t=3+2 sinx

Q8.
Answer :

Let I=∫ e^x sec x(1 + tan x) dx
=∫ e^x (sec x + sec x tan x) dx
Let e^x sec x = t
⇒ (e^x sec x + e^x sec x tan x)dx = dt
⇒ e^x sec x (1 + tan x) dx = dt
∴I=∫ dt
= t + C
= ex sec x + C ∵t=exsec x

Q9.

Answer :

Let I=∫ log xnx dx =∫ n log xxdx ∵log xa=a logxLet log x=t⇒1xdx=dt∴I=n ∫ t dt =n.t22+C =n.log x22+C ∵t=log x

Q10.

Answer :

Let I=∫log xnxdxLet log x=t⇒1xdx=dt∴I=∫ tn dt =tn+1n+1+C =log xn+1n+1+C ∵t=log x

Q11.

Answer :

Let I=∫ e^{log sin x} . cos x dx
= ∫ sin x × cos x dx ∵elog a=a
Let sin x = t
⇒ cos x dx = dt
∴I=∫ t . dt
=t22+C=sin2 x2+C ∵t= sin x

Q12.

Answer :

Let I=∫ sin³ x . cos x dx
Let sin x = t
⇒ cos x dx = dt
∴ I= ∫ t³ . dt
=t44+C=sin4 x4+C ∵t=sin x

Q13.

Answer :

Let I=∫cos⁴x .sin x dx
Let cos x = t
⇒ –sin x dx = dt
⇒ sin x dx = –dt
∴ I= –∫ t⁴ dt
=-t55+C=-cos5 x5+C ∵t= cos x

Q14.

Answer :

Let I=∫tan x . sec³x dx
=∫sec² x . sec x tan x dx
Let sec x = t
⇒ sec x tan x dx = dt
∴I= ∫ t² dt
=t33+C=sec3 x3+C ∵x=sec x

Q15.

Answer :

Let I=∫dx1+exDividing numerator & denominator by ex⇒I=∫1exdx1ex+1 =∫e-x dxe-x +1Let e-x+1=t-e-x dx=dt⇒e-x dx=-dt∴I=∫-dtt =-log t+C =-log 1+ex+C ∵t=1+ex

Q16.

Answer :

Let I=∫dx1+2exDividing numerator & denominator by ex⇒I=∫1exdx1ex+2 =∫e-x dxe-x+2Let e-x+2=t⇒-e-x dx=dt⇒e-x dx=-dt∴I=-∫dtt =-log t+C =-log e-x+2+C ∵t=e-x+2

Q17.

Answer :

Let I=∫tan-1 x31+x2dxLet tan-1 x=t⇒11+x2dx=dt∴I=∫ t3.dt =t44+C =tan-1 x44+C ∵t=tan-1 x

Q18.

Answer :

Let I=∫sec2 x dx5+tan x4Let 5+tan x=t⇒ sec2 x dx=dt∴I=∫dtt4 =∫ t-4 dt =t-4+1-4+1+C =-13t3+C =-13 5t+tan x3+C ∵t=5+tan x

Q19.

Answer :

∫ sin x+cos x1+sin 2xdx⇒∫ sin x+cos xdxsin2 x+cos2 x+2 sin x cos x⇒∫ sin x+cos xdxsin x+cos x2⇒∫ dx⇒x+C

Q20.

Answer :

∫loge x dx
= ∫ 1II.loge xI dx
=loge x∫1 dx-∫ddxloge x∫1 dxdx
= loge x ∫1 . dx – ∫ 1x×x.dx
= loge x × x – ∫ dx
= x loge x – x + C
= x loge x – x + C
= x (loge x – 1) + C

Q21.
Answer :

∫ a^x . e^x dx
= ∫ (ae)^x dx
=aexln ae+C

Q22.

Answer :

Let I=∫ e2x2+ln xdx=∫ e2x2×eln xdx=∫ e2x2. x dxLet 2×2=t⇒4x dx=dt⇒x dx=dt4∴I=14∫ et dt =14 et+C =14e2x2+C ∵t=2×2

Q23.

Answer :

ex loge a+ea loge xdx⇒∫ elog ax+elog xa dx=∫ ax+xadx=axlog a+xa+1a+1+C

Q24.

Answer :

Let I=∫ cos xsin x ·log sin xdx⇒∫ cot xlog sin xdxLet log sin x=t⇒cot x dx=dt∴I=∫ dtt =log t+C =log log sin x+C

Q25.

Answer :

Let I=∫ sin 2x dxa2 sin2+b2 cos2 xLet a2 sin2 x+b2 cos2 x=t⇒a22 sin x cos x+b22 cos x×-sin xdx=dt⇒a2-b2 sin 2x.dx=dt⇒sin 2x dx=dta2-b2∴I=1a2-b2∫dtt =1a2-b2log t+C =1a2-b2log a2 sin2x+b2 cos2 x+C ∵t=a2 sin2 x+b2 cos2 x

Q26.

Answer :

Let I=∫ ax dx3+axLet 3+ax=t⇒ax.log a dx=dt⇒ax dx=dtlog a∴I=1log a∫dtt =1log alog t+C =1log alog 3+ax+C ∵ t=3+ax

Q27.

Answer :

Let I=∫ 1+log x3+x log xdxLet 3+x log x=t⇒0+x.1x+log xdx=dt⇒1+log xdx=dt∴I=∫ dtt =log t+C =log 3+x log x+C ∵t=3+x log x

Q28.

Answer :

Let I=∫ sin xcos3 xdxLet cos x=t⇒-sin x dx=dt⇒sin x dx=-dt∴I=-∫ dtt3 =-∫ t-3 dt =-t-3+1-3+1+C =12t2+C =12 cos2 x+C ∵t=cos x =12 sec2 x+C

Q29.

Answer :

Let I = ∫sin x+cos x dx1-sin 2x =∫sin x+cos x dxsin2x+cos2x-2 sin x cos x =∫sin x+cos x dxsin x-cos x2 =∫sin x+cos x dxsin x-cos x =±∫sin x+cos xsin x-cos xdxLet sin x-cos x=t⇒cos x+sin xdx=dt∴ I=±∫dtt =±ln t+C =±ln sin x-cos x+C ∵ t=sin x-cos x

Page 19.177 (Very Short Answers)

Q30.

Answer :

Let I=∫ dxxlog xn =∫ log x-n dxxLet log x=t⇒1xdx=dt∴I=∫ t-n.dt =t-n+1-n+1+C =log x-n+1-n+1+C ∵t=log x

Q31.

Answer :

Let I=∫ eax.sin bx dx =sin bx∫eaxdx-∫ddxsin bx∫eaxdxdx =sin bx×eaxa-∫cos bx×b.eaxa =sin bx×eaxa-ba∫eax.cos bx dx =sin bx×eaxa-baI1 …1∴I1=∫ eax ×cos bxdx =cos bx∫eaxdx-∫ddxcos bx∫eaxdxdx =cos bx×eaxa+∫b.sin bx×eaxadx =cos bx.eaxa+baI ….2From 1 & 2∴I=sin bx.eaxa-ba cos bx.eaxa+baI⇒I =sin bx.eaxa-ba2 cos bx eax-b2a2I⇒I+b2a2I=sin bx.eaxa-b cos bx eaxa2⇒a2+b2I=a sin bx-bcos bxeax⇒I=a sin bx-bcos bx eaxa2+b2+C

Q32.

Answer :

Let I=∫ eax.cos bx dx =cos bx∫eaxdx-∫ddxcos bx∫eaxdxdx =cos bx×eaxa-∫-sin bx×b.eaxa =cos bx×eaxa+ba∫eax.sin bx dx =cos bx×eaxa+baI1 …1∴I1=∫ eax ×sin bxdx =sin bx∫eaxdx-∫ddxsin bx∫eaxdxdx =sin bx×eaxa-∫b.cos bx×eaxadx =sin bx.eaxa-baI ….2From 1 & 2∴I=cos bx.eaxa+ba sin bx.eaxa-baI⇒I =cos bx.eaxa+ba2 sin bx eax-b2a2I⇒I+b2a2I=cos bx.eaxa+b sin bx eaxa2⇒a2+b2I=a cos bx+b sin bxeax⇒I=a cos bx+bsin bx eaxa2+b2+C

Q33.

Answer :

Let I =∫ ex 1x-1x2dxAs we know that ∫ ex fx+f’xdx=ex fx+C∴I=exx+C

Q34.

Answer :

Let I =∫ eax a fx+f’xdxLet eax.fx=t⇒eax.a fx+eax.f’xdx=dt∴I=∫ dt =t+C =eax.fx+C ∵t=eax.fx

Q35.

Answer :

∫ 4-x2 dx=∫ 22-x2dx=x222-x2+222sin-1×2+C ∵ a2-x2=x2a2-x2-a22sin-1xa+C=x24-x2+2 sin-1 x2+C

Q36.

Answer :

∫ 9+x2 dx=∫ 32+x2 dx ∵ a2+x2=x2x2+a2+a22ln x+x2+a2=x29+x2+92ln x+9+x2+C

Q37.

Answer :

∫ x2-9 dx=∫ x2-32 dx=x2x2-32-322ln x+x2-3+C ∵ x2-a2=x2x2-a2-a22ln x+x2+a2+C=x2x2-9-92ln x+x2-9+C

Q38.

Answer :

Let I=∫ x2 dx1+x3Putting 1+x3=t⇒3×2 dx=dt⇒x2 dx=dt3∴I=13∫ dtt =13ln t+C =13ln 1+x3+C ∵ t=1+x3

Q39.

Answer :

Let I=∫x2+4xx3+6×2+5 dxLet x3+6×2+5=t⇒3×2+12x dx=dt⇒x2+4x dx=dt3Putting x3+6×2+5=t and x2+4x dx=dt3∴I=13∫dtt =13 ln t+C =13ln x3+6×2+5+C

Q40.

Answer :

Let I =∫sec2xx dxLet x=t⇒dx2x=dt⇒dxx=2 dtPutting x=t and dxx=2 dt∴I=2∫sec2+dt =2 tan t+C =2 tan x+C ∵t=x

Q41.

Answer :

Let I=∫sinxx dxLet x=t⇒12xdx=dt⇒dxx=2 dtPutting x=t and dxx=2 dt , we get ∴I=2∫sin t dt =-2 cos t+C ∵ t=x =-2 cos x+C

Q42.

Answer :

Let I=∫cosxx dxPutting x=t⇒12xdx=dt⇒dxx=2 dt∴ I=2∫cos t dt =2 sin t+C , where t=x =2 sin x+C

Q43.

Answer :

Let ∫1+log x2x dxPutting 1+log x=t⇒1x dx=dt∴I=∫t2·dt =t33+C =1+log x33+C ∵ t=1+log x

Q44.

Answer :

∫sec2 7-4x dx=tan 7-4x-4+C ∵∫sec2 x=tan x+C

Q45.

Answer :

Let I=∫log xx dx& let log x=t⇒1x dx=dt∴I=∫t·dt =t22+C =log x22+C ∵t=log x

Q46.

Answer :

∫2x dx=2xln 2+C ∵∫ax dx=axln a+C

Q47.

Answer :

∫1-sin xcos2x dx=∫1cos2x-sin xcos2xdx ∫1cos2x-sin xcos x×1cos x dx=∫sec2x-sec x tan x dx=tan x-sec x+C

Q48.

Answer :

∫x3-1×2 dx=∫x3x2-1x2dx=∫x-x-2dx=x22-x-2+1-2+1+C=x22+1x+C

Q49.

Answer :

∫x3-x2+x-1x-1 dx=∫x2 x-1+1x-1x-1dx=∫x2+1 x-1x-1dx=∫x2+1 dx=x33+x+C

Q50.

Answer :

Let I=∫etan-1×1+x2 dxLet tan-1x=t⇒dx1+x2=dt∴I=∫etdt =et+C =etan-1x+C

Q51.

Answer :

Let I=∫dx1-x2Let x=sin θ⇒dx=cos θ∴ I=∫cos θ cos θdθ =∫dθ =θ+C =sin-1x+C ∵x=sin θ

Q52.

Answer :

∫sec x sec x+tan x dx=∫sec2x+sec x tan x dx=tan x+sec x+C

Q53.

Answer :

∫1×2+16=∫dxx2+42=14tan-1 x4+C

Page 19.178 (Multiple Choice Questions)

Q1.

Answer :

(b) 14tan-1 x22

Let I=∫x4+x4dx =∫x dx22+x22Putting x2=t⇒2x dx=dt⇒x dx=dt2∴I=12∫dt22+t2 =12×12 tan-1 t2+C ∵∫1a2+x2=1atan-1xa =14 tan-1 x22+C ∵t =x2

Q2.

Answer :

(d) none of these

∫1cos x+ 3 sin xdx=12∫dxcos x×12+sin x×32=12∫dxcos x·cosπ3+sin x·sinπ3=12∫dxcos x-π3=12∫sec x-π3dx=12ln tan π4+12x-π3+C=12ln tan π4+x2-π6+C=12ln tan x2+π12+C

Page 19.179 (Multiple Choice Questions)

Q3.

Answer :

(a) 12 log (sec x² + tan x²) + C

Let I=∫x sec x2 dxPutting x2=t⇒2x dx=dt⇒x dx=dt2∴ I=12∫sec t·dt =12 log sec t+tan t+C =12 log sec x2+tan x2+C ∵ t= x2

Q4.

Answer :

(a) A = 23, B = 53
∫15+4 sin xdx=A tan-1 B tan x2+43+C ….(1)Considering the LHS of eq (1)Putting sin x=2 tan x21+tan2 x2⇒∫15+8 tan x21+tan2 x2dx⇒∫1+tan2 x25 1+tan2 x2+8 tan x2dx⇒∫sec2 x25 tan2 x2+8 tan x2+5 dx …(2) Let tan x2=t⇒sec2 x2×12 dx=dt⇒sec2 x2 dx=2dt∴ Putting tan x2=t and sec2 x2 dx=2dt we get, ∫2dt5t2+8t+5⇒25∫dtt2+85t+1⇒25∫dtt2+85t+452-452+1⇒25∫dtt+452+1-1625⇒25∫dtt+452+352⇒25×53 tan-1 t+4535+C⇒23 tan-1 5t+43+C⇒23 tan-1 53 tan x2+43+C ∵ t= tan x2 …(3)Comparing eq (3) with the RHS of eq (1) we get , ∴ A=23, B=53

Q5.

Answer :

(a) x^{sin x} + C

Let I=∫xsin x sin xx+cos x·log xdxPutting xsin x=t⇒ln xsin x=ln t⇒sin x·ln x=ln t⇒sin x×1x+cos x ln xdx=1tdt∴I=∫t·dtt =t+C =xsin x+C ∵ t= xsin x

Q6.

Answer :

(b) tan-1 loge x+C

We have to integrate 11+logex2 with respect to log ex Let I=∫d loge x1+loge x2Putting loge x=td loge x=dt∴ I=∫dt1+t2 =tan-1 t+C =tan-1 loge x+C ∵ t=logex

Q7.

Answer :

(c) 116

If ∫cos 8x+1tan 2x-cot 2xdx =a cos 8x+C …(1) Considering the LHS of eq (1)∫cos 8x+1tan 2x-cot 2xdx⇒∫2 cos24xsin 2xcos 2x-cos 2xsin 2xdx⇒∫2 cos2 4xsin2 2x-cos2 2x×sin 2x cos 2x⇒∫-cos2 4x×2 sin 2x·cos 2xcos2 2x-sin2 2xdx⇒∫-cos2 4x×sin 4xcos 4xdx ∵cos 2x= cos2x-sin2x⇒12∫-2 sin 4x cos 4x dx ⇒-1 2∫sin 8x dx⇒-12- cos 8×8+C=116cos 8x+C ….(2) Comparing RHS of eq (1) with the eq (2)∴ a=116

Q8.

Answer :

(a) −1/2
If ∫sin8x-cos8x1-2 sin2x cos2xdx=a sin 2x+C ….(1)Considering LHS of eq (1) ⇒∫sin4x-cos4x sin4x+cos4x1-2 sin2xcos2x⇒∫sin2x-cos2x sin2x+cos2x·sin4x+cos4x dxsin2x+cos2x2-2 sin2x cos2x⇒∫sin2x-cos2x·sin4x+cos4xdxsin4x+cos4x+2 sin2x cos2x-2 sin2x cos2x⇒-∫cos2x-sin2x×sin4x+cos4x dxsin4x+cos4x⇒-∫cos 2x dx ∵cos2x-sin2x=cos 2x …(2)Comparing the RHS of eq (1) with eq (2) we get , a=-12

Q9.

Answer :

∫x-1I e-xII dx=x-1∫e-x dx-∫ddxx-1∫e-x dxdx=x-1·e-x -1-∫1·e-x×-1 dx=-x-1 e-x+∫e-x dx=1-x e-x+e-x-1+C⇒1-x-1 e-x+C=-x e-x+C

Q10.

Answer :

(a) -1loge 2

If∫21xx2dx=k·21x+C ….(1) Let 1x=t⇒-1x2dx=dt⇒dxx2=-dtPutting 1x=t and dxx2=-dt in LHS of eq (1) , we get -∫2t·dt⇒-2tln 2+C⇒-21xln 2+C …(2) Comparing RHS of eq (1) with eq (2) we get , ∴ k=-1ln 2 or -1loge2

Q11.

Answer :

Disclaimer : Generally here book is taking loge x as log x . So we are writing ln x or loge x instead log x only .
(d) 12 [x + ln (sin x + cos x)] + C

Let I=∫11+tan xdx =∫11+sin xcos xdx =∫cos x cos x+sin xdx =12∫2 cos x cos x +sin xdx =12∫cos x+sin x+cos x-sin xcos x+sin xdx =12∫cos x +sin xcos x+sin xdx+12∫cos x-sin xcos x+sin xdx =12∫dx+12∫cos x-sin xcos x+ sin xdxPutting sin x+cos x=t⇒cos x-sin x dx=dt∴ I=12∫dx+12∫dtt =x2+12ln t+C =x2+12 ln cos x+sin x+C ∵ t=sin x+cos x =12x+ln sin x+ cos x+C

Q12.

Answer :

(d) none of these

∫x3 dxx=x, x≥0-x, x<0Case 1:When x≥0 ∴∫x3 dx=∫x3 dx=x44+CCase 2:x<0∫x3 dx=-∫x3 dx=-x44+C

Q13.

Answer :

(d) 2 sin x+C

Let I=∫cos xxdxPutting x=t⇒12xdx=dt⇒dxx=2dt∴ I=2∫cos t·dt =2 sin t+C =2 sin x+C ∵ t=x

Q14.

Answer :

(b) −e^x cot x + C

Let I=∫ex1-cot x+cot2xdx =∫excosec2x-cot xdxAs we know that ∫efx+f’xx=exfx+C∴ I=-ex cot x+C

Page 19.180 (Multiple Choice Questions)

Q15.

Answer :

(b) tan7 x7+C

Let I=∫sin6x cos8xdx =∫sin6xcos6x×1cos2xdx =∫tan6x·sec2x dxPutting tan x=t⇒sec2x dx=dt∴I=∫t6·dt =t77+C =tan7x7+C ∵ t= tan x

Q16.

Answer :

(a) 16tan-116tanx2+C

Let I=∫dx7+5 cos xPutting cos x =1-tan2 x21+tan2 x2∴I=∫dx7+5×1-tan2 x21+tan2 x2 =∫1+tan2 x2 dx71+tan2 x2+5-5 tan2 x2 =∫sec2 x2 dx2 tan2 x2+12 =12∫sec2x2dxtan2x2+62Let tan x2=t⇒12 sec2 x2 dx=dt⇒sec2 x2 dx=2 dt∴I=12∫2 dtt2+62 =16 tan-1 t6+C ∵∫1a2+x2=1atan-1xa+C =16 tan-1 tan x26+C ∵ t= tan x2

Q17.

Answer :

(c) log1-cotx2+C
Disclaimer : Here in answer log1-cotx2+C refers to loge1- cotx2+C or ln 1-cotx2+C

Let I=∫dx1-cos x-sin x =∫dx1-1-tan2 x21+tan2 x2-2 tan x21+tan2 x2 =∫1+tan2 x2 dx1+tan2 x2-1-tan2 x2-2 tan x2 =∫sec2 x2 dx2 tan2 x2-2 tan x2 =12∫sec2 x2 dxtan2 x2-tan x2Putting tan x2=t⇒12sec2 x2 dx=dt⇒sec2 x2 dx=2dt∴I=12∫2dtt2-t =∫dtt2-t+122-122 =∫dtt-122-122 =12×12 ln t-12-12t-12+12+C ∵ ∫dxx2-a2=12alnx-ax+a+C =ln t-1t+C =ln 1-1t+C =ln 1-1tan x2+C ∵ t= tan x2 =ln 1-cot x2+C

Q18.

Answer :

(a) exx+4+C

Let I=∫x+3x+42ex dx⇒∫x+4-1x+42 ex dx⇒∫1x+4-1x+42 ex dxAs, we know that ∫exfx+f’x dx=exfx+C∴ I=exx+4+C

Q19.

Answer :

(c) -12 3tan-12 cos x3+C

Let I=∫sin x 3+4 cos2xdxPutting cos x=t⇒-sin x dx=dt∴I=∫-dt3+4t2 =14∫-dtt2+322 =-14×132 tan-1 t×23+C ∵∫1×2+a2=1atan-1xa+C =-123 tan-1 2 t3+C =-123 tan-1 2 cos x3+C ∵ t=cos x

Q20.

Answer :

(b) -ex cotx2+C

Let I =∫ex1-sin x1-cos xdx⇒∫ex11-cos x-sin x1-cos xdx⇒∫ex 12 sin2 x2-2 sin x2 cos x22 sin2 x2dx⇒∫ex 12 cosec2 x2-cot x2dxAs, we know that ∫exfx+f’x dx=exfx+C∴ I=-ex cot x2+C

Q21.

Answer :

(a) -e-xex+e-x+C

Let I =∫2 dxex+e-x2 =∫2 dxex+1ex2 =2∫e2x dxe2x+12Let e2x+1=t⇒e2x·2 dx=dt⇒e2x·dx=dt2∴I=2×12∫dtt2 =-1t+C =-1e2x+1+C ∵ t=e2x+1

Dividing numerator and denominator by ex
⇒I=-1exex+1ex =-e-xex+e-x+C

Q22.

Answer :

(c) tan (xe^x) + C

Let I=∫ex1+xcos2 xexdxPutting xex=t⇒1·ex+x exdx=dt⇒ex1+xdx=dt∴I=∫dtcos2 t =∫sec2t dt =tan t+C =tan x ex+C ∵ t= xex

Page 19.181 (Multiple Choice Questions)

Q23.

Answer :

(c) 13tan3 x+C

Let I =∫sin2x dxcos4x =∫sin2xcos2x×1cos2xdx =∫tan2x·sec2x dxLet tan x=t⇒sec2x dx=dt ∴I=∫t2·dt =t33+C =tan3x3+C ∵ t=tan x

Q24.

Answer :

(a) ax+1xloge a

fx=1-1×2·ax+1x∴∫fxdx=∫1-1×2·ax+1xdxLet x+1x=t⇒1-1x2dx=dt∴∫fxdx=∫at·dt =atloge a+C =ax+1xloge a+C ∵t =x+1x

Q25.

Answer :

(d) log (1 + log x)

Let I=∫dxx+x log x⇒∫dxx 1+log xPutting 1+log x=t⇒1x dx=dt∴ I=∫dtt =ln t+C =ln 1+log x+C

Q26.

Answer :

(d) sin-1 x-x 1-x+C

Let I =∫x1-xdxPutting x=sin θ⇒x=sin2θ⇒dx=2 sin θ cos θ dθ⇒dx=sin 2θ dθ∴ I=∫sin2θ1-sin2θ×sin 2θ·dθ =∫sin θcos θ×2 sin θ·cos θ dθ =∫2 sin2θ·dθ =∫1-cos 2θdθ =θ-sin 2θ2+C =θ-2 sin θ cos θ2+C =θ-sin θ 1-sin2θ+C =sin-1 x-x 1-x+C ∵ θ=sin-1x =sin-1 x-x1-x+C

Q27.

Answer :

(a) e^x f (x) + C

Let I =∫exfx+f’xdxPutting exfx=t⇒ex·fx+exf’xdx=dt∴I=∫dt =t+C =exfx+C ∵t=exfx

Q28.

Answer :

(d) ± log (sin x − cos x) + C

Let I = ∫sin x+cos x dx1-sin 2x =∫sin x+cos x dxsin2x+cos2x-2 sin x cos x =∫sin x+cos x dxsin x-cos x2 =∫sin x+cos x dxsin x-cos x =±∫sin x+cos xsin x-cos xdxLet sin x-cos x=t⇒cos x+sin xdx=dt∴ I=±∫dtt =±ln t+C =±ln sin x-cos x+C ∵ t=sin x-cos x

Q29.

Answer :

(a) sin x + C

∫xI·sin xII dx=-x cos x+α ⇒x∫sin x dx-∫ddxx∫sin x dxdx=-x cos x+α⇒x -cos x-∫1·-cos xdx=-x cos x+α⇒-x cos x+∫cos x dx=-x cos x+α⇒-x cos x +sin x+C=-x cos x+α∴ α=sin x+C

Q30.

Answer :

(c) x − tan x + C

∫cos 2x-1cos 2x+1dx=∫1-2 sin2x-12 cos2x-1+1dx=-∫tan2x dx=-∫sec2x-1dx=∫1-sec2xdx=x-tan x+C

Page 19.182 (Revision Exercise)

Q1.

Answer :

∫1x+1+xdxRationalising the denominator, =∫x+1-xx+1+x x+1-xdx=∫x+1-xx+1-xdx=∫x+1-x dx=∫x+112-x12dx=x+112+112+1-x12+112+1+C=23 x+132-23 x32+C=23x+132- x32+C

Q2.

Answer :

∫1-x41-xdx=∫1-x2 1+x21-xdx=∫1-x 1+x 1+x21-xdx=∫1+x 1+x2dx=∫1+x2+x+x3dx=x+x33+x22+x44+C

Q3.

Answer :

Let ∫x+2x+13dxPutting x+1=t⇒x=t-1⇒dx=dt∴I=∫t-1+2t3dt =∫1t2+1t3dt =∫t-2+t-3dt =t-2+1-2+1+t-3+1-3+1+C =-1t-2t2+C =-1x+1-12x+12+C ∵ t=x+1

Q4.

Answer :

Let I=∫8x+134x+7dxPutting 4x+7=t⇒x=t-74⇒4dx=dt⇒dx=dt4∴ I=14∫8 t-74+13tdt =14∫2t-14+13tdt =14∫2t-1tdt =14∫2ttdt-14∫dtt =12∫t12dt-14∫t-12dt =12t12+112+1-14t-12+1-12+1+C =12×23t32-24 t12+C =13 t32-24t12+C =13 4x+732-12 4x+712+C ∵t=4x+7 =13 4x+732-4x+7+C

Q5.

Answer :

∫1+x+x2x2 1+xdx⇒∫1+xx2 1+xdx+∫x2x2 1+xdx⇒∫dxx2+∫dx1+x⇒∫x-2dx+∫11+xdx⇒x-2+1-2+1+ln 1+x+C⇒-1x+ln 1+x+C

Q6.

Answer :

∫2x+3x26xdx=∫2×2+3×2+2·2x·3x6xdx=∫2x22x·3x+3x22x·3x+2·2x·3x2x·3xdx⇒∫23x+32x+2dx⇒23xln 23+32xln 32+2x+C ∵∫axdx=axln a

Q7.

Answer :

∫sin x1+sin xdxRationalising the denominator ⇒∫sin x1+ sin x×1-sin x1-sin xdx⇒∫sin x-sin2x1-sin2xdx⇒∫sin xcos2x-tan2xdx⇒∫sin xcos x×1cos x-sec2x-1dx⇒∫sec x tan x-sec2x+1dx⇒sec x-tan x+x+C

Q8.

Answer :

∫x4+x 2-1×2+1dx⇒∫x4+x2x2+1dx-∫1×2+1dx⇒∫x2 x2+1×2+1dx-∫1×2+1dx⇒∫x2 dx-∫1×2+1dx⇒x33-tan-1 x+C ∵∫1×2+a2dx=1atan-1xa+C

Q9.

Answer :

∫sec2x·cos2 2xdx=∫sec2 x×2 cos2x-12dx=∫sec2x 4 cos4x-4 cos2x+1dx⇒∫4 cos2x-4+sec2xdx=4∫cos2x dx+∫sec2x dx-4∫dx⇒4∫1+cos 2x2dx+∫sec2x-4∫dx⇒2 x+sin 2×2+tan x-4x+C⇒sin 2x+tan x-2x+C

Q10.

Answer :

∫cosec2x·cos22x dx⇒∫cosec2x 1-2 sin2x2dx⇒∫cosec2x 1+4 sin4x-4 sin2x dx⇒∫cosec2x+4 sin2x-4 dx⇒∫cosec2x dx+4∫1-cos 2x2dx-4∫dx⇒-cot x+2 x-sin 2×2-4x+C⇒-cot x+2x-sin 2x-4x+C⇒-cot x-sin 2x -2x+C

Q11.

Answer :

∫sin4 2x dx⇒∫sin2 2×2 dx⇒∫1-cos 4x22dx⇒14∫1-cos 4×2⇒14∫1+cos2 4x-2 cos 4xdx⇒14∫1+1+cos 8×2-2 cos 4xdx⇒14∫32+cos 8×2-2 cos 4xdx⇒143×2+sin 8×16-2 sin 4×4+C⇒3×8+sin 8×64-sin 4×8+C

Q12.

Answer :

∫cos3 3x dx As we know that cos 3A=4 cos3A-3 cos A ⇒cos 3A+3 cos A4=cos3A⇒∫cos 9x+3 cos 3x4dx⇒14∫cos 9x dx+34∫cos 3x dx⇒14 sin 9×9+34sin 3×3+C⇒sin 9×36+sin 3×4+C⇒1363 sin 3x-4 sin3 3x +sin 3×4+C ∵sin 3x=3 sin x-4 sin3x⇒sin 3×12+sin 3×4-19 sin3 3x+C⇒sin 3×3-sin3 3×9+C

Q13.

Answer :

Let I=∫sin 2xa2+b2 sin2xdxPutting a+b2 sin2x=t⇒b2 2 sin x cos x dx=dt⇒b2×sin 2x dx=dt∴I=1b2∫dtt =1b2ln t+C =1b2ln a2+b2 sin2x+C ∵ t= a+b2 sin2x

Q14.

Answer :

Let I=∫1sin-1x·1-x2dxPutting sin-1 x=t⇒dx1-x2=dt∴ I=∫dtt =ln t+C =ln sin-1x+C ∵ t=sin-1x

Q15.

Answer :

Let I=∫sin-1 x31-x2dxPutting sin-1x=t⇒dx1-x2=dt∴ I=∫t3·dt =t44+C =sin-1 x44+C ∵t= sin-1x

Q16.

Answer :

∫1ex+1dx …(1)
Multiplying numerator and Denominator of eq (1) by ex

⇒∫ex·dxex ex+1Putting ex=t⇒ex dx=dt⇒∫dtt t+1∴1t t+1=At+Bt+11t t+1=A t+1+B tt t+1 …(2)⇒1=A t+1+ B tPutting t+1=0 or, t=-1 in eq (2) we get , ⇒1=A×0+B -1⇒B=-1Now,putting t=0 in eq (2) we get , ⇒1=A 0+1+B×0⇒A=1Putting the values of A and B in eq (2) we get , 1t t+1=1t-1t+1∴∫dtt t+1=∫dtt-∫dtt+1 =ln t-ln t+1+C =ln tt+1+C =ln exex+1+C =ln ex-ln ex+1+C =x-ln ex+1+C

Q17.

Answer :

We have,I=∫ex-1ex+1dx=∫2ex-ex+1ex+1dx=∫2exex+1dx-∫dxPutting ex+1=t⇒exdx=dt∴I=∫2tdt-∫dx=2 log t-x+C=2 log ex+1-x+C=2 log ex+1-x+C

Q18.

Answer :

Let I =∫1ex+e-xdx =∫dxex+1ex =∫ex dxe2x+1 =∫ex dxex2+1Putting ex=t⇒ex dx=dt∴ I=∫dtt2+1 =tan-1t +C ∵ ∫dta2+x2=1atan-1xa+C =tan-1 ex+C ∵t=ex

Q19.

Answer :

Let I=∫cos7x sin xdx =∫cos6x·cos x dxsin x =∫cos2x3·cos xsin xdx =∫1-sin2x3·cos xsin xdxLet sin x=t⇒cos x dx =dt∴ I=∫1-t23tdt =∫1-t6-3t2+3t4tdt =∫1t-t5-3t+3t3dt =ln t-t66-3t22+3t44+C =ln sin x-sin6x6-3 sin2 x2+34 sin4x+C ∵ t=sin x

Q20.

Answer :

∫sin x·sin 2x·sin 3x dx=12∫2 sin 2x·sin x sin 3x dx=12∫cos 2x-x- cos 2x+x sin 3x dx ∵ 2sinAsinB=cos (A-B)-cos (A+B)⇒=12∫cos x-cos 3x sin 3x dx=12∫sin 3x·cos x dx-12∫sin 3x·cos 3x dx=14∫2 sin 3x·cosx dx-14∫2 sin 3x·cos 3x dx =14∫sin 4x+sin 2xdx-14∫sin 6x dx ∵2sin Acos B=sinA+B+sinA-B=14-cos 4×4-cos 2×2-14-cos 6×6+C=-cos 4×16-cos 2×8+cos 6×24+C

Q21.

Answer :

∫cos x·cos 2x·cos 3x dx⇒12∫2 cos 2x·cos x cos 3x dx⇒12∫cos 2x+x+cos 2x-x cos 3x dx ∵2cos Acos B=cos A+B+cosA-B⇒12∫cos3x+cos x cos 3x dx⇒12∫cos2 3x dx+12∫cos 3x·cos x dx⇒12∫1+cos 6x2dx+14∫2 cos 3x·cos x dx ∵cos 2x= cos2 x-1 ⇒14x+sin 6×6+14∫cos 4x+cos 2xdx⇒14x+sin 6×6+14sin 4×4+sin 2×2+C⇒x4+sin 6×24+sin 4×16+sin 2×8+C

Q22.

Answer :

Let I=∫sin x+cos xsin 2 xdxPutting sin x-cos x=t⇒cos x+sin xdx=dtAlso sin x-cos x2=t2⇒sin2x+cos2x-2 sin x cos x=t2⇒1-t2=sin 2x∴I=∫dt1-t2 =sin-1 t+C ∫dta2-x2=sin-1xa+C =sin-1 sin x-cos x+C ∵ t=sin x-cos x

Q23.

Answer :

Let I=∫sin x-cos xsin 2x dxPutting sin x+cos x=t⇒cos x-sin x dx=dt⇒sin x-cos x dx=-dtAlso sin x+cos x=tSquaring both sides, sin x+cos x2=t2⇒sin2x+cos2x+2 sin x cos x =t2⇒1+sin 2x=t2⇒sin 2x=t2-1∴I=∫-dtt2-1 =-ln t+t2-1+C ∵ ∫dtx2-a2=lnx+x2-a2+C =-ln sin x+cos x+sin x+cos x2-1+C ∵t=sin x+cos x =-ln sin x+cos x+sin2x+cos2x+2 sin cos x-1+C =-ln sin x+cos x+sin 2 x+C

Q24.

Answer :

∫1sin x-a·sin x-bdx=1sin b-a∫sin b-asin x-a·sin x-b dx=1sin b-a∫sin x-a-x-bsin x-a·sin x-b dx=1sin b-a∫sin x-a·cos x-b-cos x-a sin x-bsin x-a·sin x-b dx=1sin b-a ∫sin x-a·cos x-bsin x-a·sin x-b – cos x-a sin x-b sin x-a sin x-b dx=1sin b-a∫cot x-b-cot x-a dx=1sin b-a∫cot x-b dx-∫cot x-a dx=1sin b-aln sin x-b-ln sin x-a+C=1sin b-aln sin x-bsin x-a +C=-1sin a-blnsin x-bsin x-a +C=1sin a-b ln sin x-asin x-b+C

Q25.

Answer :

∫1cos x-a·cos x-bdx=1sin a-b∫sin a-bcos x-a·cos x-b dx=1sin a-b∫sin x-b-x-acos x-a·cos x-b dx=1sin a-b∫sin x-b·cos x-a-cos x-b·sin x-acos x-a·cos x-b=1sin a-b∫sin x-b·cos x-acos x-a·cos x-b-cos x-b·sin x-acos x-a·cos x-b dx=1sin a-b∫tan x-b-tan x-a dx=1sin a-b∫tan x-b dx-∫tan x-a dx=1sin a-bln sec x-b-ln sec x-a +C=1sin a-bln cos x-a-ln cos x-b +C=1sin a-bln cos x-acos x-b +C

Q26.

Answer :

We have,I=∫sinx1+sinx dxI=∫2 sinx2cosx2sin2x2+cos2x2+2 sinx2cosx2 dxI=∫2 sinx2cosx2 sinx2+cosx22 dxI=∫2 sinx2cosx2 sinx2+cosx2 dxI=∫1+2sinx2 cosx2-1sinx+cosx dxI=∫sin2x2+cos2x2+2sinx2 cosx2-1sinx+cosx dxI=∫sinx2+cosx22-1sinx2+cosx2 dxI=∫sinx2+cosx22sinx2+cosx2 dx-∫1sinx2+cosx2 dxI=∫sinx2+cosx2 dx-∫1sinx2+cosx2 dxI=2-cosx2+sinx2+C1-12∫112sinx2+cosx2 dxI=2-cosx2+sinx2+C1-12∫1sinx2 cosπ4+cosx2 sinπ4 dxI=2-cosx2+sinx2+C1-12∫1sinx2+π4 dxI=2-cosx2+sinx2+C1-12∫cosecx2+π4 dxI=2-cosx2+sinx2-2logtanx4+π8+C

Q27.

Answer :

Let I=∫sin xcos 2 x dx =∫sin x2 cos2x-1 dx ∵cos 2x=2cos2x-1 Putting cos x=t⇒-sin x dx=dt⇒sin x dx=-dt∴I=∫-dt2t2-1 =12∫-dtt2-12 =-12∫dtt2-122 =-12×12×12 ln t-12t+12+C ∵∫1×2-a2=12alnx-ax+a+C =-122 ln 2t-12t+1+C =-122 ln 2 cosx-12 cos x+1+C ∵t= cos x =122 ln 2 cos x+12 cos x-1+C

Q28.

Answer :

Let I =∫tan3 x dx=∫tan x·tan2x dx=∫tan x sec2x-1dx=∫tan x·sec2x dx-∫tan x dxPutting tan x=t in the Ist integral⇒sec2x dx=dt∴I=∫t·dt-∫tan x dx =t22-ln sec x+C =tan2x2-ln sec x+C ∵ t= tan x

Q29.

Answer :

Let I= ∫tan4x dx =∫tan2x·tan2x dx =∫sec2x-1 tan2x dx =∫sec2x·tan2x dx-∫tan2x dx =∫tan2x·sec2x-∫sec2x-1 dxPutting tan x=t in the Ist integral⇒sec2x dx=dt∴ I=∫t2·dt-∫sec2x-1 dx =t33-tan x+x+C =tan3x3-tan x+x+C ∵ t=tan x

Q30.

Answer :

Let I=∫tan5 x dx =∫tan3x·tan2x dx =∫tan3x sec2x-1 dx =∫tan3x·sec2x dx-∫tan3x dx =∫tan3x·sec2x dx-∫tan x·tan2x dx =∫tan3x·sec2x dx-∫tan x·sec2x-1 dx =∫tan3x·sec2x dx-∫tanx·sec2x dx+∫tan x dxPutting tan x=t in the Ist and IInd integral. ⇒sec2x dx=dt∴ I=∫t3·dt-∫t·dt+∫tan x dx =t44-t22+ln sec x+C =tan4x4-tan2x2+ln sec x+C ∵ t= tan x

Q31.

Answer :

Let I=∫cot4x dx =∫cot2x·cot2x dx =∫cot2 x·cosec2x-1 dx =∫cot2x·cosec2x dx-∫cot2x dx =∫cot2x·cosec2x dx-∫cosec2x-1 dxPutting cot x=t in the Ist integral⇒-cosec2x dx=dt∴ I=-∫t2 dt-∫cosec2x-1 dx =-t33+cot x+x+C =-cot3x3+cot x+x+C ∵t= cot x

Q32.

Answer :

Let I=∫cot5x dx =∫cot2x·cot3x dx =∫cosec2x-1 cot3x dx =∫cot3x·cosec2x dx-∫cot3x dx =∫cot3x·cosec2x dx-∫cot x·cot2x dx =∫cot3x·cosec2x dx-∫cot x cosec2x-1 dx =∫cot3x·cosec2x dx-∫cot x·cosec2x dx+∫cot x dxPutting cot x=t in the Ist and IInd integral ⇒-cosec2x dx=dt⇒cosec2x dx=-dt∴I=-∫t3 dt+∫t·dt+∫cot x dx =-t44+t22+ln sin x+C =-cot4x4+cot2x2+ln sin x+C ∵ t=cot x

Q33.

Answer :

∫x2x-13dx=∫x2-1+1x-13dx=∫x-1 x+1x-13+1x-13dx=∫x+1x-12+1x-13dx=∫x-1+2x-12+1x-13dx=∫1x-1+2x-12+1x-13dx=∫1x-1dx+2∫x-1-2 dx+∫x-1-3 dx=ln x-1+2 x-1-2+1-2+1+x-1-3+1-3+1+C=ln x-1 -2x-1-x-1-22+C=ln x-1 -2x-1-12 x-12+C

Q34.

Answer :

Let I=∫x2x+3 dxPutting 2x+3=t⇒x=t-32⇒2dx=dt⇒dx=dt2∴ I=12∫t-32 t dt =14∫t-3 t dt =14∫t32-3t12dt =14t32+132+1-3 t12+112+1+C =14×25 t52-34×23 t32+C =110 t52-2t32+C =110 2x+352-12 2x+332+C ∵ t=2x+3 =1102x+352-12 2x+332+C

Q35.

Answer :

Let I=∫x31+x22dx =∫x2×x1+x22dxPutting 1+x2=t ⇒x2=t-1⇒2x dx=dt⇒x dx=dt2∴ I=12∫t-1t2dt =12∫1t-1t2dt =12∫dtt-12∫t-2 dt =12 ln t-12t-2+1-2+1+C =12 ln t+12t+C =12 ln 1+x2+12 1+x2+C ∵ t= 1+x2

Q36.

Answer :

Let I=∫x·sin5 x2·cos x2 dxPutting sin x2=t⇒cos x2×2x dx=dt⇒cos x2·x dx=dt2∴ I=12∫t5·dt =12 t66+C =t612+C =sin6 x212+C ∵ t= sin x2

Q37.

Answer :

Let I=∫sin3x·cos4x dx =∫sin2x·sin x·cos4x dx =∫1-cos2x·cos4x·sin x dx =∫cos4x-cos6x·sin x dxPutting cos x=t⇒-sin x dx=dt⇒sin x dx=-dt∴I=-∫t4-t6dt =∫t6-t4dt =t77-t55+C =cos7x7-cos5x5+C ∵ t= cos x

Q38.

Answer :

Let I=∫sin5x dx =∫sin4x·sin x dx =∫sin2x2 sin x dx =∫1-cos2x2 sin x dx =∫cos4x-2 cos2x+1 sin x dxPutting cos x=t⇒-sin x dx=dt⇒sin x dx=-dt∴ I=-∫t4-2t2+1 dt =-∫t4 dt+2∫t2 dt-∫dt =-t55+2t33-t+C =-cos5x5+23 cos3x-cos x+C ∵ t=cos x

Q39.

Answer :

Let I=∫cos5 x dx =∫cos4x·cosx dx =∫cos2x2 cos x dx =∫1-sin2x2 cos x dxPutting sin x=t⇒cos x dx=dt∴ I=∫1-t22·dt =∫t4-2t2+1 dt =∫t4·dt-2∫t2 dt+∫dt =t55-2×t2+12+1+t+C =t55-23 t3+t+C =sin5x5-23 sin3x+sin x+C ∵t= sin x

Q40.

Answer :

Let I=∫sin x·cos3x dx =∫sin x·cos2x·cos x dx =∫sin x 1-sin2x·cos x dxPutting sin x=t⇒cos x dx=dt∴ I=∫t 1-t2·dt =∫t12 dt-∫t12·t2 dt =∫t12dt-∫t52dt =t3232-t7272+C =23t32-27 t72+C =23 sin32x-27 sin72x+C ∵ t= sin x

Q41.

Answer :

Let I=∫sin 2xsin4x +cos4xdx =∫2 sin x ·cos x dxsin4x+cos4xDividing numerator and denominator by cos4x⇒∫2 sin x·cos xcos4 xdx1+tan4x⇒∫2 tan x·sec2x dx1+tan2x2Putting tan2x=t⇒2 tan x·sec2x dx∴ I=∫dt1+t2 =tan-1 t+C =tan-1 tan2x+C ∵ t= tan 2x

Q42.

Answer :

Let I=∫1×2-a2dxPutting x= a sec θ⇒dx=a secθ tanθ dθ∴ I=∫a secθ tanθ dθa2 sec2θ-a2 =∫a secθ·tanθ dθa·tanθ =∫secθ dθ =ln secθ+tanθ+C =ln secθ+sec2θ-1+C =ln xa+xa2-1+C =ln x+x2-a2a+C =ln x+x2-a2-ln a+C =ln x+x2-a2+C’where C’=C-ln a

Q43.

Answer :

Let I=∫dxx2-a2Putting x= a tan θ⇒dx=a sec2θ dθ∴ I=∫a·sec2θ dθa2 tan2θ+a2 =∫a sec2θ·dθa1+tan2θ =∫sec2θ·dθsecθ =∫secθ·dθ =∫secθ·dθ =ln secθ+tanθ+C =ln secθ+sec2θ-1+C =ln xa+x2a2-1+C =ln x+x2-a2-ln a+C =ln x+x2-a2+C’where C’=C-ln a

Q44.

Answer :

Let I=∫dx4x2+4x+1+4 =∫dx2x2+2×2x+1+22 =∫dx2x+12+22Putting 2x+1=t⇒2 dx=dt⇒dx=dt2∴ I=12∫dtt2+22 =12×12 tan-1 t2+C =14 tan-1 2x+12+C ∵ t=2x+1

Q45.

Answer :

∫1×2+4x-5dx=∫1×2+4x+4-4-5dx=∫1×2+4x+4-32dx=∫1x+22-32dx=12×3 ln x+2-3x+2+3+C ∵∫1×2-a2dx=12aln x-ax+a+C=16 ln x-1x+5+C

Q46.

Answer :

We have,I=∫11-x-4x2dx=14∫114–x2x4dx=14∫114-x2+x4dx=14∫114-x2++182-182x4dx=14∫114-x+182+164dx=14∫114+-x+182164dx=14∫116+164-x+182dx=14∫11782-x+182dx=14×12×178 ln 178+x+18178-x-18+C ∵∫1a2-x2dx=12alna+xa-x+C=117 ln 17+18+x17-18-x+C

Q47.

Answer :

∫13×2+13x-10dx=13∫1×2+133x-103dx=13∫1×2+13 x3+1362-1362-103dx=13∫1x+1362-16936-103dx=13∫1x+1362-169-12036dx=13∫1x+1362-1762dx=13×12×176 ln x+136-176x+136+176 ∵∫1×2-a2dx=12aln x-ax+a+C=117 ln x-23x+5+C=117 ln 3x-23x+15+C

Q48.

Answer :

Let I=∫sin x cos2x-2 cos x-3dxPutting cos x=t⇒-sin x dx=dt⇒sin x dx=-dt∴ I=-∫dtt2-2t-3 =-∫dtt2-2t+1-4 =-∫dtt-12-22 =-ln t-1+t-12-4+C ∵ ∫1×2-a2dx=lnx+x2-a2+C =-ln cos x-1+cos2x-2 cos x-3+C ∵t=cos x

Q49.

Answer :

Let I=∫cosec x-1 dx =∫1sin x-1 dx =∫1-sin xsin x dx =∫1-sin x 1+sin xsin x 1+sin xdx =∫1-sin2xsin2x+sin xdx =∫cos xsin2x+sin xdxPutting sin x=t⇒cos x dx=dt∴I=∫dtt2+t =∫dtt2+t+122-122 =∫dtt+122-122 =ln t+12+t+122-122+C ∵ ∫1×2-a2dx=ln x+x2-a2+C =ln t+12+t2+t+C =ln sin x+12+sin2x+sin x+C ∵ t= sin x

Q50.

Answer :

Let I=∫13-2x-x2dx =∫13-x2+2x+1-1dx =∫14-x+12dxPutting x+1=t⇒dx=dt∴ I=∫dt22-t2 =sin-1 t2+C ∵∫ 1a2-x2dx=sin-1xa+C =sin-1 x+12+C ∵ t = x+1

Q51.

Answer :

Let I=∫x+1×2+4x+5dx& let x+1= Addxx2+4x+5+B⇒x+1=A 2x+4+B⇒x+1=2Ax+4A+BEquating the coefficients of like terms2A=1⇒A=12& 4A+B=1⇒4×12+B=1⇒B=-1∴x+1=12 2x+4-1∴I=∫122x+4-1×2+4x+5dx =12∫2x+4×2+4x+5dx-∫1×2+4x+5dxPutting x2+4x+5=t⇒2x+4 dx=dt∴ I=12∫1tdt-∫1×2+4x+4+1dx =12∫dtt-∫1x+22+12dx =12 ln t-tan-1 x+21+C ∵∫1×2+a2dx=1atan-1xa+C =12 ln x2+4x+5-tan-1 x+2+C ∵ t=x2+4x+5

Q52.

Answer :

We have,I=∫5x+7x-5x-4 dx=∫5x+7×2-9x+20 dxLet 5x+7=A ddx x2-9x+20+B⇒5x+7=A 2x-9+BEquating Coefficients of like terms2A=5⇒A=52And-9A+B=7⇒-9×52+B=7⇒B=7+452⇒B=592∴I=∫52 2x-9+592×2-9x+20 dx=52∫2x-9 dxx2-9x+20+592∫dxx2-9x+20Putting x2-9x+20=t⇒2x-9 dx=dtI=52∫dtt+592∫dxx2-9x+922-922+20=52∫t-12 dt+592∫dxx-922-81+804=52 t-12+1-12+1+592 ∫dxx-922-122=52×2t+592 log x-92+x-922-122+C=5t+592 log x-92+x2-9x+20+C=5×2-9x+20+592 log x-92+x2-9x+20+C

Q53.

Answer :

Let I=∫1+xxdx =∫1+xx×1+x1+xdx =∫1+xx2+xdxLet x+1= Addxx2+x+B⇒x+1=A 2x+1+B⇒x+1=2Ax+A+BEquating the coefficients of like terms2A=1⇒A=12& A+B=1⇒12+B=1∴B=12∴ I=∫x+1×2+xdx =∫12 2x+1+12×2+xdx =12∫2x+1×2+xdx+12∫1×2+xdx

Putting x2+x=t⇒2x+1 dx=dt∴ I=12∫1tdt+12∫1×2+x+122-122dx =12∫1tdt+12∫1x+122-122dx =12∫t-12dt+12∫1x+122-122dx =12×2 t+12 ln x+12+x+122-14+C ∵∫1×2-a2dx=ln x+x2-a2+C =t+12 ln x+12+x2+x+C =x2+x+12 ln x+12+x2+x+C ∵ t=x2+x

Q54

Answer :

Let I=∫1-xxdx =∫1-x·1-xx·1-x dx =∫1-xx-x2dxLet 1-x= Addxx-x2+B⇒1-x=A 1-2x+B⇒1-x=-2A x+A+BEquating coefficients of like terms-2A=-1⇒A=12& A+B=1⇒12+B=1∴B=12∴ I=∫12 1-2x+12x-x2dx =12∫1-2xx-x2dx+12∫1x-x2+122-122dx =12∫1-2x x-x2dx+12∫1122-x2-x+122dx =12∫1-2x x-x2dx+12∫1122-x-122dx

Putting x-x2=t in the first integral⇒1-2x dx=dt∴ I=12∫1tdt+12∫1122-x-122dx =12∫t-12dt+12∫dx122-x-122 =12×2 t12 +12×sin-1 x-1212+C ∵ ∫1a2-x2dx=sin-1xa+C =t+12 sin-1 2x-1+C =x-x2+12 sin-1 2x-1+C ∵ t= x-x2

Q55.

Answer :

We have,I=∫a-x1-ax dxI=1a∫1+a-1-ax1-ax dxI=1a∫1-ax1-ax dx+1a∫a-11-ax dxI=1a∫ dx+a-1a∫11-ax dxI=1ax+a-1a∫11-ax dxLet,I1=∫11-ax dxPut ax=z2⇒adx=2zdzI1=1a∫2z1-z dzI1=1a∫2z-2+21-z dzI1=1a∫2z-21-z dz+1a∫21-z dzI1=-2a∫1-z1-z dz+1a∫21-z dzI1=-2a∫ dz+1a∫21-z dzI1=-2az-2alog1-z+C1I1=-2axa-2alog1-ax+C1I=1ax+a-1a-2axa-2alog1-ax+C

Note: The answer in indefinite integration may vary depending on the integral constant.

Q56.

Answer :

Let I=∫1sin x-2 cos x 2 sin x+cos xdx
Dividing numerator and denominator by cos2x we get ,
I= ∫1cos2xtan x-2 2 tan x+1dx =∫sec2xtan x-2 2 tan x+1 dxPutting tan x =t⇒sec2x dx=dt∴ I=∫1t-2 2t+1dt =∫12t2+t-4t-2dt =∫12t2-3t-2dt = 12∫1t2-3t2-1dt =12∫1t2-32t+342-342-1dt =12∫1t-342-916-1dt =12∫1t-342-542dt =12×12×54 ln t-34-54t-34+54+C ∵∫1×2-a2dx=12alnx-ax+a+C =15 ln t-2t+12+C =15 ln 2 t-22t+1+C =15 ln 2 tan x-22 tan x +1+C ∵t=tan x =15 ln tan x-22 tan x+1+15 ln 2+C =15 ln tan x-22 tan x+1+C’where C’=C+15 ln 2

Q57.

Answer :

Let I=∫14 sin2x+4 sin x·cos x+5 cos2xdx

Dividing numerator and denominator by cos2x we get
I=∫sec2x 4 tan2x+4 tan x+5dxPutting tan x =t⇒sec2x dx=dt∴ I=∫dt4t2+4t+5 =14∫dtt2+t+54 =14∫dtt2+t+14-14+54 =14∫dtt+122+12 =14× tan-1 t+12+C ∵∫1×2+a2dx=1atan-1xa+C =14 tan-1 2t+12+C =14 tan-1 2 tan x+12+C ∵ t= tan x =14 tan-1 tan x+12+C

Page 19.183 (Revision Exercise)

Q58.

Answer :

Let I=∫1a+b tan xdx =∫1a+b sin xcos xdx =∫cos x·a cos x+b sin xdxLet cos x=A ddx a cos x+b sin x+B a cos x+b sin x⇒cos x=A -a sin x+b cos x+B a cos x+ b sin x1·cos x=Ab+B·a cos x +sin x-A·a+B·bEquating coefficients of like terms A·b+B·a=1 … 1-A·a+B·b=0 … 2Multiplying equation 1 by a and eq 2 by b and then adding them A·ab+B·a2=a-A·a·b+Bb2=0⇒B=aa2+b2Substituting the value of B in eq 1⇒A·b+a2a2+b2=1⇒A·b=1-a2a2+b2⇒A=ba2+b2∴ I=ba2+b2∫-a sin x+b cos xa cos x+b sin xdx+aa2+b2∫a cos x+b sin xa cos x+b sin xdx =ba2+b2∫-a sin x+b cos xa cos x +b sin xdx+aa2+b2∫dxPutting a cos x +b sin x =t in the Ist integral⇒-a sin x+b cos xdx=dt∴ I=ba2+b2∫dtt+aa2+b2∫dx =ba2+b2 ln t+axa2+b2+C =ba2+b2 ln a cos x+b sin x+axa2+b2+C ∵ t=a cos x +b sin x

Q59.

Answer :

Let I=∫1sin2x+sin 2xdx =∫1sin2x+2 sin x·cos xdx

Dividing numerator and denominator by cos2x, we get
I= ∫1cos2xtan2x+2 tan xdx =∫sec2xtan2x+2 tan x dxPutting tan x =t⇒sec2x dx=dt∴ I=∫1t2+2tdt =∫1t2+2t+1-1dt =∫1t+12-12dt =12 ln t+1-1t+1+1+C =12 ln tt+2+C =12 ln tan xtan x+2+C ∵ t= tan x

Q60.

Answer :

∫sin x+2 cos x2 sin x+cos xdxLet sin x+2 cos x=A ddx 2 sin x+cos x+B 2 sin x+cos x⇒sin x+2 cos x=A 2 cos x-sin x+2 B sin x+B cos x⇒sin x+2 cos x=2 A+B cos x +2 B-A sin xEquating coefficients of like terms⇒2 A+B=2 … 1⇒-A+2B=1 … 2Multiplying eq 2 by 2 and adding it to eq 1 we get,5 B=4⇒B=45Putting B=45 in eq 1 we get,2 A+45=2⇒A=35∴∫sin x+2 cos x2 sin x+cos xdx= ∫35 2 cos x-sin x2 sin x+cos xdx+45∫2 sin x+ cos x2 sin x+cos xdx =35∫2 cos x-sin x2 sin x+cos xdx+45∫dxPutting 2 sin x+cos x=t⇒2 cos x-sin x dx=dt∴ I=35∫dtt+45∫dx =35 ln t+4×5+C =35 ln 2 sin x+cos x+4×5+C ∵ t= 2 sin x + cos x

Q61.

Answer :

Let I=∫x3x8+22dx =∫x3x42+22dxPutting x4=t⇒4x3dx=dt⇒x3·dx=dt4∴ I=14∫1t2+22dt =14 ln t+t2+4+C =14 ln x4+x8+4+C ∵t= x4

Q62.

Answer :

Let I=∫12-3·cos 2xdx =∫12-3 2 cos2x-1dx =∫12-6 cos2x+3dx =∫15-6 cos2xdx
Dividing numerator and denominator by cos2x, we get
I=∫sec2x 5 sec2x-6dx =∫sec2x 5 1+tan2x-6dx =∫sec2x 5 tan2x-1dxPutting tan x =t⇒sec2x dx=dt∴ I=∫15t2-1dt =15∫1t2-152dt =15×12×15 ln t-15t+15+C ∵∫1×2-a2dx=12alnx-ax+a+C =125 ln 5t-15t+1+C =125 ln 5 tan x-15 tan x+1+C ∵ t= tan x

Q63.

Answer :

Let I=∫cos x 14-cos2xdx =∫cos x 14-1-sin2xdx =∫cos x sin2x-34dx =∫cos x sin2x-322dxPutting sin x=t⇒cos x dx=dt∴ I=∫1t2-322dt =12×32 ln t-32t+32+C =13 ln 2t-32t+3+C =13 ln 2 sin x-32 sin x+3+C ∵ t=sin x

Q64.

Answer :

Let I=∫11+2 cos xdx Putting cos x=1-tan2x21+tan2 x2∴ I=∫11+2 1-tan2 x21+tan2x2dx =∫1+tan2 x2 1+tan2 x2+2-2 tan2 x2dx =∫sec2 x2 3-tan2 x2dxPutting tanx2=t⇒12sec2 x2 dx=dt⇒sec2 x2·dx=2dt∴ I=∫23-t2 dt =2∫132-t2dt =2×123 ln 3+t3+t+C ∵∫1a2-x2dx=12alna+xa-x+C =13 ln 3+tanx23-tan x2+C ∵ t= tan x2

Q65.

Answer :

Let I=∫11-2 sin x dx Putting sin x=2 tan x21+tan2 x2∴ I=∫11-2 2 tan x21+tan2x2dx =∫1+tan2 x21+tan2 x2-4 tan x2 dx =∫sec2 x2 1+tan2 x2-4 tan x2dxPutting tan x2=t⇒12 sec2 x2 dx=dt⇒sec2 x2·dx=2dt∴ I=∫2 t2-4t+1dt =2∫1t2-4t+4-4+1dt =2∫1t-22-32dt =2×123 ln t-2-3t-2+3+C ∵ ∫1×2-a2dx=12alnx-ax+a+C =13 ln tanx2-2-3tanx2-2+3+C ∵ t= tan x2

Q66.

Answer :

Let I=∫1sin x 2+3 cos xdx =∫sin x sin2x 2+3 cos xdx =∫sin x1-cos2x 2+3 cos x dx =∫sin x1-cos x 1+cos x 2+3 cos x dxPutting cos x=t⇒-sin x dx=dt∴I=∫-11-t 1+t 2+3tdt =∫1t-1 t+1 3t+2dtLet1t-1 t+1 3t+2=At-1+Bt+1+C3t+2⇒1t-1 t+1 3t+2=A t+1 3t+2+B t-1 3t+2+C t+1 t-1t-1 t+1 3t+2⇒1=A t+1 3t+2+B t-1 3t+2+C t+1 t-1Putting t+1=0 or t=-1⇒1=A×0+B -1 -1 3×-1+2+C×0∴B=12Now, putting t-1=0 or t=1⇒1=A 1+1 3+2+B×0+C×0∴A=110Now, putting 3t+2=0 or t=-2 3⇒1=A×0+B×0+C -23+1 -23-1⇒1=C 13 -53∴ C=-95∴ I=∫110 t-1dt+12∫1t+1dt-95∫13t+2dt =110 ln t-1+12 ln t+1-95 ln 3t+23+C =110 ln t-1+12 log t+1-35 ln 3t+2+C =110+ln cos x-1+12 ln cos x+1 -35 ln 3 cos x+2+C ∵ t= cos x

Q67.

Answer :

Let I=∫1sin x+sin 2xdx =∫1sin x+2 sin x cos xdx =∫1sin x 1+2 cos xdx =∫sin x sin2x 1+2 cos xdx =∫sin x 1-cos2x 1+2 cos xdx =∫sin x dx1-cos x 1+cos x 1+2 cos xPutting cos x=t⇒-sin x dx=dt⇒sin x dx=-dt

∴ I=-∫11-t 1+t 1+2tdt =∫1t-1 t+1 2t+1dt∴1t-1 t+1 2t+1=At-1+Bt+1+C2t+1⇒1=A t+1 2t+1+B t-1 2t+1+C t-1 t+1Putting t+1=0 or t=-1⇒1=A×0+B -1 -1 -2+1+C×0⇒1=B 2∴B=12Now, putting t-1=0 or t=1⇒1=A 2 3+B×0+C×0∴A=16Now, putting 2t+1=0 or t=-12⇒1=A×0+B×0+C -12-1 -12+1⇒1=C -32 12∴C=-43∴I =16∫1t-1dt+12∫1t+1dt-43∫12t+1dt =16 ln t-1+12 log t+1-43 ln 2t+12+C =16 ln t-1+12 ln t+1-23 ln 2t+1+C =16ln cos x-1+12 ln cos x+1 -23 ln 2 cos x+1+C ∵ t= cos x

Q68.

Answer :

We have,I=∫dxsin4x+cos4x
Dividing numerator and denominator by cos4x
I=∫sec4x dxtan4x+1=∫sec2x sec2x dxtan4x+1=∫1+tan2x sec2x dxtan4x+1Putting tan x=t⇒sec2x dx=dt∴I=∫1+t2 dtt4+1=∫1t2+1 dtt2+1t2=∫1+1t2t-1t2+2dtPutting t-1t=p⇒1+1t2dt=dp∴I=∫1p2+22dp=12 tan-1p2+C=12 tan-1t-1t2+C=12 tan-1 t2-12 t+C=12 tan-1 tan2x-12 tan x+C=12 tan-1-2×1-tan2x2 tan x+C=12 tan-1-2tan 2x+C=12 tan-1-2 cot 2x+C

Q69.

Answer :

Let I=∫15-4 sin xdxPutting sin x=2 tan x21+tan2 x2∴ I=∫15-4 ×2 tan x21+tan2 x2dx =∫1+tan2 x2 5 1+tan2 x2-8 tan x2dx =∫sec2 x2 5 tan2 x2-8 tan x2+5dxPutting tan x2=t⇒12 sec2 x2 dx=dt

⇒sec2 x2 dx=2 dt∴ I=2∫15t2-8t+5dt =25∫1t2-85t+1dt =25∫1t2-8t5+452-452+1dt =25∫1t-452-1625+1dt =25∫1t-452+352dt =25×53 tan-1 t-4535+C ∵ ∫1×2+a2dx=1atan-1xa+C =23 tan-1 5t-43+C =23 tan-1 5 tan x2-43+C ∵ t = tan x2

Q70.

Answer :

Let I= ∫sec4x dx =∫sec2x·sec2x dx =∫1+tan2x·sec2x dxPutting tan x=t⇒sec2x dx=dt∴I=∫1+t2 dt =∫dt+∫t2dt =t+t33+C =tan x+13tan3x+C ∵ t= tan x

Q71.

Answer :

Let I= ∫cosec4 2x dx =∫cosec2 2x·cosec2 2x dx =∫1+cot2 2x·cosec2 2x dxPutting cot 2x=t⇒-cosec2 2x·2 dx=dt⇒cosec2 2x·dx=-dt2∴ I=-12∫1+t2·dt =-12 t+t33+C =-12cot 2x+16cot3 2x+C ∵ t= cot 2x

Q72.

Answer :

Let I=∫1+sin x sin x 1+cos xdxPutting sin x=2 tan x21+ tan2 x2and cos x=1-tan2 x21+tan2 x2∴ I=∫1+2 tan x21+tan2 x2 2 tan x21+tan2 x2 1+1-tan2 x21+tan2 x2dx =∫1+tan2 x2+2 tan x2 1+tan2 x2 2 tan x2 1+tan2 x2+1-tan2 x2dx =14∫1+tan2 x2+2 tan x2 sec2 x2tan x2 dxPutting tan x2=t⇒12 sec2 x2 dx=dt⇒sec2 x2 dx=2dt∴ I=14∫1+t2+2t·2 dtt =12∫1t+t+2 dt =12 ln t+t22+2t+C =12 ln tan x2+tan2 x22+2 tan x2+C ∵ t= tan x2 =12 ln tan x2+14tan2x

Q73.

Answer :

Let I=∫12+cos xdxPutting cos x =1-tan2 x21+tan2 x2∴ I=∫12+1-tan2 x21+tan2 x2dx =∫1+tan2 x2 2 1+tan2 x2+1-tan2 x2dx =∫sec2 x2 2+2 tan2 x2+1-tan2 x2dx =sec2 x2 3+tan2 x2dxPutting tan x2=t⇒12 sec2 x2 dx=dt⇒sec2 x2 dx=2 dt∴ I=∫23+t2 dt =2∫1t2+32dt =23 tan-1 t3+C =23 tan-1 tan x23+C ∵ t= tan x2

Q74.

Answer :

Let I=∫a+xxdx =∫a+x a+xx a+x =∫a+xx2+axdx =a∫1×2+axdx+∫x x2+axdx =a∫1×2+ax+a22-a22dx+∫x x2+axdx =a∫1x+a22-a22dx+12∫2x x2+axdx =a∫1x+a22-a22dx+12∫2x+a-ax2+axdx =a∫1x+a22-a22dx+12∫2x+a x2+axdx-a2∫1×2+axdx =a2∫1x+a22-a22dx+12∫2x+a x2+ax dxPutting x2+ax=t in the Ist integral ⇒2x+a dx=dt∴ I=a2∫1x+a22-a22dx+12∫1tdt =a2 ln x+a2+x2+ax+12×2t+C ∵∫1×2-a2dx=lnx+x2-a2+C =a2 ln x+a2+x2+ax+x2+ax+C ∵t=x2+ax

Q75.

Answer :

∫6x+5 dx6+x-2x2Let 6x+5= Addx6+x-2×2+B⇒6x+5=A -4x+1+B⇒6x+5=-4A x+A+BEquating coefficients of like terms-4A=6⇒A=-32 & A+B=5⇒-32+B=5⇒B=5+32⇒B=132Then, 6x+5=-32 -4x+1+132∴∫6x+5 6+x-2x2dx=∫-32-4x+1+1326+x-2x2dx =-32∫-4x+16+x-2×2 dx+132∫16+x-2x2dxPutting 6+x-2×2=t in the Ist integral⇒-4x+1 dx=dt∴ ∫6x+5 6+x-2x2dx=-32∫1tdt+132×2∫13+x2-x2dx =-32∫t-12·dt+1322∫13+x2-x2-142+142dx =-32∫t-12·dt+1322∫13+116-x-142dx =-32∫t-12·dt+1322∫1742-x-142dx =-3 t12+1322×sin-1 x-1474+C ∵∫1a2-x2dx=sin-1 xa+C =-3 6+x-2×2+1322×sin-1 x-1474+C =-36+x-2×2+1322 sin-1 4x-17+C

Q76.

Answer :

Let I=∫sin5x cos4xdx =∫sin4x.sin xcos4xdx =∫sin2x2·sin x cos4xdx =∫1-cos2x2.sin xcos4xdx =∫1+cos4x-2cos2xcos4x sin x dxPutting cos x=t⇒-sin x dx=dt∴ I=-∫1+t4-2t2 dtt4 =-∫t-4 dt-∫dt+2∫t-2 dt =-t-4+1-4+1-t+2 t-2+1-2+1+C =13t3-t-2t+C =13 cos3x-cos x-2cos x+C ∵ t= cos x

Q77.

Answer :

Let I=∫cos5x dxsin x =∫cos4x·cos x dxsin x =∫cos2x2·cos x dxsin x =∫1-sin2x2 cos x dxsin x =∫1+sin4x-2 sin2xsin x cos x dxPutting sin x=t⇒cos x dx=dt∴ I=∫1+t4-2t2tdt =∫dtt+∫t3 dt-2∫t dt =ln t+t44-2t22+C =ln t+t44-t2+C =ln sin x+14sin4x-sin2x+C ∵ t= sin x

Q78.

Answer :

Let I=∫sin6x·cos xdx =∫sin6x·cos xcos2xdx =∫sin6x1-sin2xcos x dxPutting sin x=t⇒cos x dx=dt∴ I=∫t61-t2dt =∫t6-1+11-t2 dt =∫t23-131-t2dt+∫11-t2dt =∫t2-1 1+t2+t41-t2+∫11-t2dt =-∫t4+t2+1dt+∫11-t2dt =-t55+t33+t+12 ln 1+t1-t+C =-15sin5x-13sin3x-sin x+12 ln 1+sin x1-sin x+C ∵ t= sin x

Q79.

Answer :

Let I=∫sin2xcos6xdx =∫sin2xcos2x·cos4xdx =∫tan2x·sec4x dx =∫tan2x sec2x·sec2x dx =∫tan2x 1+tan2x sec2x dxPutting tan x=t⇒sec2x dx=dt∴ I=∫t2 1+t2dt =∫t2+t4dt =t33+t55+C =13tan3x+15tan5x+C ∵ t= tan x

Q80.

Answer :

Let I=∫sec6x dx =∫sec4x·sec2x dx =∫sec2x2·sec2x dx =∫1+tan2x2 sec2x dxPutting tan x=t⇒sec2x dx=dt∴ I=∫ 1+t22·dt =∫1+t4+2t2dt =∫dt+∫t4dt+2∫t2 dt =t+t55+2t33+C =tan x+15tan5 x+23 tan3x+C ∵ t= tan x

Q81.

Answer :

Let I=∫tan5x·sec3x dx =∫tan4x·sec2x·sec x tan x dx =∫sec2x-12·sec2x·sec x tan x dxPutting sec x=t⇒sec x tan x dx=dt∴ I=∫ t2-12·t2·dt =∫t4-2t2+1t2 dt =∫t6-2t4+t2 dt =t77-2t55+t33+C =17sec7x-25sec5x+13sec3x+C ∵t= sec x

Q82.

Answer :

Let I= ∫tan3x·sec4x dx =∫tan3x·sec2x·sec2x dx =∫tan3x 1+tan2x·sec2x dx =∫tan3x+tan5x sec2x dxPutting tan x=t⇒sec2x dx=dt∴ I=∫ t3+t5 dt =t44+t66+C =tan4x4+ tan6x6+C ∵ t= tan x

Q83.

Answer :

We have,I=∫1secx+cosecx dxI=∫11cosx+1sinx dxI=12∫2sinx cosxsinx+cosx dxI=12∫1+2sinx cosx-1sinx+cosx dxI=12∫sin2x+cos2x+2sinx cosx-1sinx+cosx dxI=12∫sinx+cosx2-1sinx+cosx dxI=12∫sinx+cosx2sinx+cosx dx-12∫1sinx+cosx dxI=12∫sinx+cosx dx-12∫1sinx+cosx dxI=12-cosx+sinx+C1-122∫112sinx+cosx dxI=12-cosx+sinx+C1-122∫1sinx cosπ4+cosx sinπ4 dxI=12-cosx+sinx+C1-122∫1sinx+π4 dxI=12-cosx+sinx+C1-122∫cosecx+π4 dxI=12-cosx+sinx-122logtanx2+π8+C

Q84.

Answer :
Let I=∫1·IIa2+x2Idx =a2+x2 ∫1 dx-∫ddxa2+x2 ∫1 dxdx =a2+x2·x-∫1×2×2 a2+x2·x dx =a2+x2·x-∫x2+a2-a2a2+x2dx =xa2+x2-∫a2+x2 dx+a2∫1a2+x2dx =xa2+x2-I+a2∫1a2+x2dx∴2I=xa2+x2+a2 ln x+x2+a2⇒I=x2 a2+x2+a22 ln x+x2+a2+C

Q85.

Answer :

Let I=∫1·IIx2-a2Idx =x2-a2∫1 dx-∫ddxx2-a2∫1 dxdx =x2-a2·x-∫1×2×2 x2-a2·x dx =x2-a2·x-∫x2-a2+a2x2-a2dx =x2-a2·x-∫x2-a2 dx-a2∫dxx2-a2 =xx2-a2-I-a2∫dxx2-a2∴ 2I=xx2-a2-a2 ln x+x2-a2⇒I=x2 x2-a2-a22 ln x+x2-a2+C

Q86.

Answer :

Let I=∫a2-x2 dx =∫1·IIa2-x2Idx =a2-x2∫1 dx-∫ddxa2-x2∫1dxdx =a2-x2·x+∫1×2×2 a2-x2·x dx =a2-x2·x+∫x2-a2+a2a2-x2 dx =xa2-x2-∫a2-x2 dx+a2∫1a2-x2dx =xa2-x2-I+a2∫1a2-x2dx∴2I=xa2-x2+a2 ∫1a2-x2dx⇒I=x2 a2-x2+a22 sin-1 xa+C

Q87.

Answer :

∫3×2+4x+1 dx=3∫x2+43x+13 dx=3∫x2+43x+232-232+13 dx=3∫x+232-49+13 dx=3∫x+232-132 dx=3 12x+23x+232-132 -12×132ln x+23+x+232-132 +C ∵∫ x2-a2 dx=12xx2-a2-12a2ln x+x2-a2+C=163x+23×2+4x+1-318ln x+23+x2+43x+13 +C

Q88.

Answer :

∫1+2x-3x2dx=3∫13+23x-x2dx=3∫13-x2-23xdx=3∫13-x2-23x+132-132dx=3∫13+19-x-132dx=3∫49-x-132dx=3∫232-x-132dx=3 x-132 232-x-132+2322 sin-1 x-1323+C ∵∫a2-x2dx=x2a2-x2+12a2sin-1xa+C=3x-16 1+2x-3×2+239sin-1 3x-12+C

Q89.

Answer :

Let I=∫x1+x-x2dx& let x= Addx1+x-x2+B⇒x=A -2x+1+BBy equating the coefficients of like terms we get,x=-2A x⇒A=-12& A+B=0⇒B=12By substituting the values of A and B in eq (1) we get,I=∫-12 -2x+1+12 1+x-x2 dx =-12∫-2x+1 1+x-x2dx+12 1+x-x2dxPutting 1+x-x2=t⇒-2x+1 dx=dt∴ I=-12∫t·dt+12∫1+x-x2 dx =-12∫t dt+12∫1-x2-x dx =-12∫t12·dt+12∫1-x2-x+14-14dx =-12t12+112+1+12∫1-x-122+14dx =-12×23t32+12∫522-x-122dx =-1 3 t32+12x-122 522-x-122+5222 sin-1 x-1252+C ∵ ∫a2-x2dx=x2a2-x2+12a2sin-1 xa+C =-131+x-x232+122x-14 1+x-x2+58 sin-1 2x-15+C =-1+x-x21+x-x23+2x-18 1+x-x2+516 sin-1 2x-15+C =1+x-x2 -1+x-x23+2x-18+516 sin-1 2x-15+C =1+x-x2 -8-8x+8×2+6x-324+516 sin-1 2x-15+C =1+x-x2 8×2-2x-1124+516 sin-1 2x-15+C

Q90.

Answer :

Let I=∫2x+3 4×2+5x+6 dx& let 2x+3=Addx4x2+5x+6+B⇒2x+3=A 8x+5+B …(1)By equating coefficients of like terms we get,2x=8A x⇒A=14& 5A+B=3⇒54+B=3⇒B=3-54 =74Thus, by substituting the values of A and B in eq (1) we getI=∫ 2x+3 4×2+5x+6 dx =∫148x+5+74 4×2+5x+6 dx =14∫8x+5 4×2+5x+6 dx+74 ∫4×2+5x+6 dxPutting 4×2+5x+6=t in the first integral⇒8x+5 dx=dt∴ I=14∫t·dt+7×24∫x2+5×4+32 dx =14∫t12·dt+72∫x2-5×4+582-582+32dx =14 t12+112+1+72∫x+582-2564+32dx =14×23t32+72∫x+582+-25+9664 =16 t32+72∫x+582+7182 =1 6 4×2+5x+632+72x+582x+582+7182+7164×2 ln x+58+x+582+7182+C ∵∫a2+x2dx=12xa2+x2+12a2lnx+x2+a2+C =16 4×2+5x+632+72 8x+516 x2+54x+32+71×72×128 ln x+58+x2+54x+32+C =16 4×2+5x+632+7×2 8x+54×16 x2+54x+32+497256 ln x+58+x2+54x+32+C =16 4×2+5x+6 4×2+5x+6+764 8x+5 4×2+5x+6+497256 ln x+56+x2+54x+32 +C =4×2+5x+6 4×2+5x+66+764 8x+5+497256 ln x+58+x2+54x+32+C =4×2+5x+6 128×2+328x+297192+ln x+58+x2+54x+32+ C

Q91.

Answer :

Let I=∫1+x2·cos 2x·dx =∫cos 2x dx+∫x2·cos 2x dx =sin 2×2+I1 where I1=∫x2 cos 2x dx … 1I1=∫x2I cos 2xII dx = x2∫cos 2x dx-∫ddxx2∫ cos 2x dxdx =x2·sin 2×2-∫2x×sin 2×2 dx =x2·sin 2×2-∫xI·sin 2xII dx =x2·sin 2×2-x∫ sin 2x dx-∫ddxx∫sin 2x dxdx =x2·sin 2×2-x-cos 2×2-∫1·-cos 2×2 dx =x2·sin 2×2+x·cos 2×2-sin 2×4 … 2From 1 and 2∴I=sin 2×2+x2 sin 2×2+x cos 2×2-sin 2×4+C =x2+1 sin 2×2+x cos 2×2-sin 2×4+C

Q92.

Answer :

∫log10 x dx=∫loge xloge 10 dx=1loge 10∫1II·log xI dx=1loge 10loge x∫1 dx-∫ddxloge x∫1 dxdx=1loge 10loge x·x-∫1x×x dx=1loge 10x loge x-x+C=1loge10×x loge x-1+C=x loge x-1·log10e+C

Q93.

Answer :

Let I=∫log log x dxxPutting log x=t⇒1x dx=dt∴ I=∫1II·log t·dIt =log t∫1 dt-∫ddtlog t∫1 dtdt =log t·t-∫1t×t dt =log t·t-∫dt =log t·t-t+C =t log t-1+C =log x log log x-1+C

Q94.

Answer :

∫xI·sec22x dxII=x∫sec22x dx-∫ddxx∫sec2 2x dxdx=x tan 2×2-∫1·tan 2×2 dx=x tan 2×2-12 ln sec 2×2+C=x tan 2×2-14 ln sec 2x+C

Q95.

Answer :

∫x·sin3x dx =∫x·143 sin x-sin 3x dx sin3A-143 sin A-sin 3A=34∫xI·sin xII dx-14∫x·sin 3x dx=34x∫sin x dx-∫ddxx∫sin x dxdx-14x∫sin 3x dx-∫ddxx∫sin 3x dxdx=34x -cos x-∫1·-cos xdx-14x -cos 3×3-∫1·-cos 3x3dx=-3×4 cos x+34 sin x+x cos 3×12-sin 3×36+C=14-3x cos x+3sin x+x cos 3×3-sin 3×9+C

Q96.

Answer :

∫x+12I exII dx=x+12∫ex dx-∫ddxx+12∫exdxdx=x+12·ex-∫2 x+1·ex dx=x+12 ex-2∫xI exII dx-2∫ex dx=x+12 ex-2 x·ex-∫1·ex dx-2ex=x+12 ex-2x ex+2ex-2ex+C=x+12-2x ex+C=x2+1 ex+C

Q97.

Answer :

Let I=∫1II·log x+x2+a2I dx =log x+x2+a2∫1 dx-∫ddxlog x+x2+a2∫1dx =log x+x2+a2·x-∫1x+x2+a2×1+1×2x2x2+a2·x·dx =log x+x2+a2·x-∫xx2+a2dxPutting x2+a2=t in the second integral⇒2x dx=dt⇒x dx=dt2∴ I=x·log x+x2+a2-12∫1tdt =x·log x+x2+a2-12∫t-12dt =x·log x+x2+a2-12 t-12+1-12+1+C =x·log x+x2+a2-t+C =x·log x+x2+a2-x2-a2+C ∵ t=x2+a2

Q98.

Answer :

∫log xx3dx=∫1x3II log xI dx=log x∫1x3dx-∫ddxlog x∫1x3dxdx=log x∫x-3 dx-∫1x×x-3+1-3+1dx=log x x-3+1-3+1+12∫1x3dx=log x -12×2+12∫x-3 dx=log x -12×2+12 x-3+1-3+1+C=log x -12×2-14×2+C=-14×2 2 log x+1+C

Q99.

Answer :

Let I=∫log 1-xx2dx =∫1x2II log 1-xI dx =log 1-x∫x-2 dx-∫-11-x×x-2+1-2+1 dx =log 1-x x-2+1-2+1+∫-11-x xdx =log 1-x ×-1x+∫1×2-xdx =-log 1-xx+∫1×2-x+122-122dx =-log 1-x x+∫1x-122-122dx =-log 1-xx+12×12 log x-12-12x-12+12+C =-log 1-xx+log x-1x+C =-log 1-xx+log x-1-log x+C =-log 1-xx+log 1-x-log x+C =1-1x log 1-x-log x+C

Q100.

Answer :

∫x3II·log x2I·dx=log x2∫x3 dx-∫2 log xx×x44 dx=log x2×x44-12∫log xI·x3II dx=log x2×x44-12log x∫x3 dx-∫ddxlog x∫x3dxdx=log x2×x44-12 log x·x44-∫1x×x44dx=log x2×x44-12 log x·x44-14∫x3 dx=log x2×x44-12 log x·x44-x416+C=log x2×x44-log x·x48+x432+C

Q101.

Answer :

We have,I=∫dxx 1+xn=∫xn-1 dxxn-1 x1 1+xn=∫xn-1 dxxn 1+xnPutting xn=t⇒n xn-1 dx=dt⇒xn-1 dx=dtn∴I=1n∫dtt 1+tlet 1+t=p2⇒dt=2p dp∴I=1n∫2p dpp2-1 p=2n∫dpp2-12=2n×12 log p-1p+1+C=1n log 1+t-11+t+1+C=1n log 1+xn-11+xn+1+C

Q102.

Answer :

We have,I=∫x21-x dxLet, 1-x=t2Differentiating both sides we get-dx=2t dtNow, integration becomes,I=-∫1-t22t 2tdt=-2∫1-t22dt=-2∫1-2t2+t4 dt=-2t-2t33+t55+C=-215t3t4-10t2+15+C=-2151-x31-x2-101-x+15+C=-2151-x31-2x+x2-101-x+15+C=-2151-x3x2-6x+3-10+10x+15+C=-2151-x3x2+4x+8+C

Q103.

Answer :

We have,I=∫x5 dx1+x3=∫x3x2 dx1+x3Putting x3=t⇒3×2 dx=dt⇒x2 dx=dt3∴I=13∫t dt1+t=13∫1+t-11+t dt=13∫1+t-11+t dt=13 1+t3232-1+t-12+1-12+1+C=29 1+t32-23 1+t12+C=29 1+x332-23 1+x312+C=29 1+x3121+x3-3+C=29 1+x3 x3-2+C

Q104.

Answer :

We have,I=∫1+x21-x2 dx=∫2-1-x21-x2 dx=2∫11-x2 dx-∫1-x21-x2 dx=2∫11-x2 dx-∫1-x2 dx=2 sin-1x-x21-x2+12sin-1x+C=2 sin-1x-x21-x2-12sin-1x+C=32 sin-1x-x21-x2+C

Q105.

Answer :

We have,I=∫x1-x1+x dx⇒I=∫x1-x1-x1+x1-x dx⇒I=∫x1-x1-x2 dx⇒I=∫x-x21-x2 dx⇒I=∫x-x2-1+11-x2 dx⇒I=∫-x2+11-x2 dx+∫x-11-x2 dx⇒I=∫1-x2 dx+∫x1-x2 dx-∫11-x2 dx⇒I=x21-x2+12sin-1x+C1-1-x2+C2-sin-1x+C3 ∵ ∫x1-x2 dx=-1-x2+C2⇒I=1-x2x2-1-12sin-1x+C

Q106.

Answer :

We have,I=∫dxx 1+x3=∫x2dxx31+x3putting x3=t⇒3×2 dx=dt⇒x2dx=dt3∴I=13∫dtt1+tlet 1+t=p2⇒dt=2p dpI=13∫2p dpp2-1×p=23∫dpp2-1=23×12 log p-1p+1+C=13log p-1p+1+C=13log 1+t-11+t+1+C=13log 1+x3-11+x3+1+C

Q107.

Answer :

We have,I=∫sin x+cos xsin4 x+cos4 x dx=∫sin x+cos xsin2 x+cos2 x2-2sin2 x cos2 x dx=∫sin x+cos x1-2sin2 x cos2 x dx=∫sin x+cos x1-122sin x cos x2 dx=∫sin x+cos x1-12sin22xdx

Putting sin x-cos x=t …..1⇒sin x-cos x2=t2⇒sin2x+cos2x-2sin x cos x=t2⇒1-2sin x cos x=t2⇒sin 2x=1-t2Differentiating 1, we getcos x+sin xdx=dt∴I=∫11-121-t22 dt=∫22-1-t22 dt=∫222-1-t22 dt=2∫12+1-t22-1+t2 dt

=222∫12+1-t2+12-1+t2 dt=12∫12+1-t2 dt+12∫12-1+t2 dt=12∫12+12-t2 dt+12∫12-12+t2 dt=12×122+1log2+1+t2+1-t+12×12+1tan-1t2+1+C=12122+1log2+1+t2+1-t+12+1tan-1t2+1+C, where t=sin x-cos x

Q108.

Answer :

We have,I=∫ x2tan-1 x dxConsidering tan-1 x as first function and x2 as second functionI=tan-1xx33-∫11+x2×x33dx=tan-1xx33-13∫x3dx1+x2=tan-1 xx33-13∫x2x1+x2dxPutting 1+x2=t⇒x2=t-1⇒2x dx=dt⇒x dx=dt2∴I=tan-1xx33-16∫t-1tdt=x33tan-1x-16∫dt+16∫dtt=x33tan-1x-16t+16log t+C=x33tan-1x-161+x2+16log 1+x2+C=x33tan-1x-x26+16log x2+1+C’ Where C’=C-16

Q109.

Answer :

We have,I=∫ tan-1 x dxPutting x=tan θ⇒x=tan2 θ⇒dx=2 tan θ sec2 θ dθ∴I=∫ tan-1tan θ 2tan θ sec2 θ dθ=2 ∫ θi tan θ sec2 θii dθ=2θ×tan2 θ2-∫1tan2 θ dθ2 ∵∫ tan θ sec2 θ dθ= tan2 θ2=2θtan2θ2-12∫sec2θ-1dθ=θ tan2 θ-2×tan θ2+2×θ2+C=tan-1x×x-x+tan-1 x+C=x+1tan-1 x -x+C

Q110.

Answer :

We have,I=∫ sin-1 x dxPutting x=sin θ⇒x=sin2 θ⇒dx=2 sin θ cos θ dθ⇒dx=sin2θdθ∴I=∫ θ sin 2θdθ=θ-cos 2θ2-∫1-cos 2θ2dθ=-θ cos 2θ2+12∫cos 2θdθ=-θ cos 2θ 2+12sin 2θ2+C=-sin-1 x 1-2 sin2 θ2+122 sin θ cos θ2+C=-sin x1-2×2+sin θ1-sin2 θ2+C=-sin-1 x 1-2×2+x 1-x2+C=-12sin-1 x 1-2x+12x-x2+C

Q111.

Answer :

We have,I=∫sec-1 x dxPutting x=sec θ⇒x=sec2 θ⇒dx=2 sec θ sec θ tan θ dθ =2 sec2 θ tan θ dθ∴I=2∫θ sec2 θ tan θ dθ=2 ∫ θtan θ sec2 θ dθConsidering θ as first fucction and tan θ sec2 θ as second functionI=2θtan2 θ2-∫1tan2 θ2dθ ∵∫tan θ sec2 θ dθ=tan2 θ2=θ tan2 θ-∫sec2 θ-1dθ=θ tan2θ-tan θ+θ+C=θ1+tan2 θ-tan θ+C=θ sec2 θ-sec2 θ-1+C=sec-1x x-x-1+C=x sec-1x -x-1+C

Q112.

Answer :

We have,I=∫ tan-11-x1+x dxPutting x=cos θ⇒dx=-sin θ dθ∴I=∫ tan-1 1-cos θ1+cos θ -sin θ dθ=∫ tan-1 2 sin2 θ22 cos2 θ2 -sin θdθ=∫ tan-1 tan θ2 -sin θdθ=-12∫θ sin θ dθConsidering θ as first function and sin θ as second functionI=-12θ-cos θ-∫1-cos θdθ=-12θ-cos θ+∫cos θdθ=-12-θ cos θ+sin θ+C=-12-θ cos θ+1-cos2 θ+C=-12-cos-1x ×x+1-x2+C=12x cos-1x-1-x2+C

Q113.

Answer :

We have,I=∫ sin-1 xa+x dxPutting x=a tan2 θ⇒tan θ=xa⇒dx=a2 tan θ sec2 θ dθ∴I=∫ sin-1 a tan2 θa+a tan2 θ 2a tan θ sec2 θdθ=∫ sin-1 tan2 θsec2 θ 2a tan θ sec2 θ dθ=2a ∫ sin-1 sin θ tan θ sec2 θ dθ=2a ∫ θ tan θ sec2 θ dθConsidering θ as first function and tan θ sec2 θ as second functionI=2a θtan2θ2-∫1tan2 θ2dθ=aθ tan2 θ-∫sec2 θ-1dθ=aθ tan2 θ-tan θ+θ+C=aθ×1+tan2 θ-tan θ+C=atan-1xa 1+xa-xa+C=x+atan-1xa-ax+C

Q114.

Answer :

We have,I=∫ sin-1 3x-4x3dxPutting x=sin θ⇒θ=sin-1 x⇒dx=cos θ dθ∴I=∫ sin-1 3 sin θ-4 sin3 θ cos θ dθ=∫ sin-1 sin 3θ cos θ dθ=3∫ θI cos θII dθ=3 θ sin θ-∫1 sin θ dθ=3θ sin θ+cos θ+C=3θ sin θ+1-sin2 θ+C=3 sin-1 x ×x+1-x2+C=3 x sin-1x+1-x2+C

Q115.

Answer :

We have,I=∫sin-1 x3 dxLet, sin-1x=t⇒sin t=x⇒cos t=1-x2Differentiating both sides we getcos t dt=dxNow, integral becomesI=∫sin-1 x3 dx=∫t3 cos t dt=t3sin t-∫3t2 sin t dt Using by parts=t3sin t-3∫t2 sin t dt =t3sin t-3-t2 cos t -∫-2t cos t dt =t3sin t+3t2 cos t-6∫t cos t dt =t3 sin t+3t2 cos t-6tsin t -∫sin t dt =t3 sin t+3t2 cos t-6t sin t+cos t +C=sin-1×3 x+3sin-1×2 1-x2-6sin-1x x-61-x2 +C=xsin-1xsin-1×2-6+3sin-1×2-21-x2 +C

Q116.

Answer :

We have,I=∫ cos-1 1-2x2dxPutting x=sin θ⇒dx=cos θ dθ∴I=∫ cos-11-2 sin2θ cos θ dθ=∫cos-1 cos 2θ cos θ dθ=2∫ θI cos θII dθ=2θ sin θ-∫1 sin θ dθ=2θ sin θ+cos θ+C=2sin-1x×x+1-x2+C=2x sin-1x+1-x2+C

Q117.

Answer :

We have,I=∫x sin-1×1-x232 dxPutting sin-1 x=θ⇒x=sinθ⇒dx=cosθ dθ∴I=∫sinθ θ cosθ dθ1-sin2θ32=∫θ sinθ cosθ dθcos2θ32=∫θsinθcos2θ dθ=∫θ Isec θ tanθII dθ=θ×secθ-∫1×secθ dθ=θ×secθ-∫secθ dθ=θ×secθ-log secθ+tanθ+C=θcosθ-log 1cosθ+sinθcosθ+C=θ1-sin2θ-log 1+sinθcosθ+C=θ1-sin2θ-log 1+sinθ1-sin2θ+C=θ1-sin2θ-log 1+sinθ1-sinθ+C=sin-1×1-x2-log 1+x1-x+C=sin-1×1-x2-12 log 1+x1-x+C

Q118.

Answer :

We have,I=∫ e2x 1+sin 2×1+cos 2xdx=∫ e2x 11+cos 2x+sin 2×1+cos 2xdx=∫ e2x 12 cos2 x+2 sin x cos x2 cos2 xdx=∫ e2xsec2 x2+tan xdxLet e2x tan x=t⇒e2xsec2x+2e2xtan xdx=dt=e2xsec2x2+e2xtan xdx=dt2∴I=∫ dt2=t2+C=e2x tan x2+C

Q119.

Answer :

We have,I=∫1-sin x1+cos xe-x2 dx=∫cos2 x2+sin2 x2-2 sin x2 cos x21+cos xe-x2dx=∫cos x2-sin x22e-x22 cos2 x2 dx=∫cos x2-sin x22 cos2 x2e-x2dx=12∫sec x2-tan x2 sec x2e-x2dxLet e-x2sec x2=t⇒e-x2 sec x2 tan x2×12-e-x2 sec x22 dx=dt⇒12sec x2 tan x2-sec x2e-x2dx=dt∴I=-∫dt=-t+C=-e-x2sec x2+C

Q120.

Answer :

We have,I=∫ex 1-x21+x22 dx=∫ex 1+x2-2×1+x22 dx=∫ex 1+x21+x22-2×1+x22 dx=∫ex 11+x2-2×1+x22 dx=ex1+x2+C ∵∫exfx+f’x dx=exfx+Cwhere, fx=11+x2⇒f’x=-2×1+x22

Q121.

Answer :

We have,I=∫em tan-1 x1+x232dxPutting tan-1x=t⇒x=tan t⇒11+x2 dx=dt⇒dx=1+x2dt⇒dx=1+tan2tdt∴I=∫emt1+tan2 t321+tan2tdt=∫emtdt1+tan2 t=∫eIImtcosI t dt=cos temtm-∫-sin temtm dt=cos temtm+1m∫emtsin t dt=cos temtm+1m I1 …..1Where,I1=∫eIImtsin tI dt=sintemtm-∫costemtmdtI1=sin temtm-1mI …..2from 1 and 2I=cos temtm+1m sin temtm-1mI⇒I=cos temtm+sin t emtm2-1m2 I⇒I+Im2=emt m cos t+sin tm2⇒I=emt m cos t+sin t1+m2+C⇒I=emt1+m2 cos tm1+m2+sin t11+m2+CLet m1+m2=cos θThen, sinθ=11+m2⇒cotθ=m⇒θ=cot-1m∴I=emt1+m2 cos t cos θ+sint sin θ+C=emt1+m2 cos t-θ+C=emt1+m2 cos tan-1x-cot-1m+C

Q122.

Answer :

We have,I=∫x2x-13 x+1 dxLet x2x-13 x+1=Ax-1+Bx-12+Cx-13+Dx+1 …..1⇒x2=Ax-12x+1+Bx-1x+1+C x+1+Dx-13 …..2Putting x=1 in 2, we get1=2C⇒C=12Putting x=-1 in 2, we get1=-8D⇒D=-18

Putting x=2 in 2, we get4=3A+3B+3C+D⇒4=3A+3B+32-18⇒3A+3B=4-32+18⇒3A+3B=32-12+18⇒3A+3B=218⇒A+B=78And putting x=0 in 2, we get0=A-B+C-D⇒0=A-B+12+18 ∵C=12, D=18⇒A-B=-58
Here, A+B=78 and A-B=-58⇒A=18 and B=34Therefore, 1 becomes,x2x-13 x+1=18x-1+34x-12+12x-13-18x+1Now, integral becomesI=∫18x-1+34x-12+12x-13-18x+1dx=18log x-1-34x-1-14x-12-18log x+1+C=18log x-1x+1-34x-1-14x-12+C

Q123.

Answer :

We have,I=∫x dxx3-1=∫x dxx-1 x2+x+1Let xx-1 x2+x+1=Ax-1+Bx+Cx2+x+1⇒xx-1 x2+x+1=A x2+x+1+Bx+C x-1x-1 x2+x+1⇒x=A x2+x+1+Bx2-Bx+Cx-C⇒x=A+B x2+A-B+C x+A-CEquating Coefficient of like termsA+B=0 …..1A-B+C=1 …..2A-C=0 …..3Solving 1, 2 and 3, we getA=13B=-13C=13∴xx-1 x2+x+1=13 x-1+-13x+13×2+x+1=13 x-1+13 -x+1×2+x+1=13 x-1-13 x-1×2+x+1=13 x-1-16 2x-2×2+x+1=13 x-1-16 2x+1×2+x+1-16×-3×2+x+1=13 x-1-16 2x+1×2+x+1+12×1×2+x+1∴I=13∫dxx-1-16∫2x+1 dxx2+x+1+12∫dxx2+x+14-14+1Putting x2+x+1=t⇒2x+1 dx=dt∴I=13 log x-1-16 log t+12∫dxx+122+322=13 log x-1-16 log x2+x+1+1223 tan-1 x+1232+C=13 log x-1-16 log x2+x+1+13 tan-1 2x+13+C

Page 19.184 (Revision Exercise)

Q124.

Answer :

We have,I=∫dx1+x+x2+x3=∫dx1+x+x2 1+x=∫ dx1+x 1+x2Let 1x+1 1+x2=Ax+1+Bx+Cx2+1⇒1x+1 x2+1=A x2+1+Bx+C x+1x+1 x2+1⇒1=A x2+1+Bx2+Bx+Cx+C⇒1=A+B x2+B+C x+A+CEquating Coefficient of like termsA+B=0 …..1B+C=0 …..2A+C=1 …..3Solving 1, 2 and 3, we get,A=12B=-12C=12∴1x+1 x2+1=12 x+1+-x2+12×2+1⇒1x+1 x2+1=12 x+1-12 xx2+1+12 x2+1∴I=12∫dxx+1-12∫x dxx2+1+12∫dxx2+1Putting x2+1=t⇒2x dx=dt⇒x dx=dt2∴I=12∫dxx+1-14∫dtt+12∫dxx2+1=12 log x+1-14 log t+12 tan-1x+C=12 log x+1-14 log x2+1+12 tan-1 x+C=12 log x+1-12 log x2+1+12 tan-1 x+C=12 log x+1×2+1+12 tan-1 x+C

Q125.

Answer :

We have,I=∫dxx2+2 x2+5Putting x2=t∴1×2+2 x2+5=1t+2 t+5Let 1t+2 t+5=At+2+Bt+5⇒1t+2 t+5=A t+5+B t+2t+2 t+5⇒1=A t+5+B t+2Putting t=-5∴1=B -5+2⇒B=-13Putting t=-2∴1=A -2+5+B×0⇒A=13∴I=13∫dxx2+2-13∫dxx2+5=13∫dxx2+22-13∫dxx2+52=132 tan-1 x2-135 tan-1 x5+C

Q126.

Answer :

We have,I=∫x2-2×5-x dx=∫x2-2x x4-1dx=∫x x2-2×2 x2-1 x2+1dxPutting x2=t⇒2x dx=dt⇒x dx=dt2∴I=12∫t-2t t-1 t+1 dtLet t-2t t-1 t+1=At+Bt-1+Ct+1⇒t-2t t-1 t+1=A t-1 t+1+Bt t+1+Ct·t-1t t-1 t+1⇒t-2=A t-1 t+1+B t t+1+C t t-1Putting t=1∴1-2=B×2⇒B=-12Putting t=0∴-2=A -1⇒A=2Putting t=-1∴-3=C -1 -2⇒C=-32∴I=22∫dtt-12×2∫dtt-1-32×2∫dt+1=log t-14 log t-1-34 log t+1+C=log x2-14 log x2-1-34 log x2+1+C=2 log x-14 log x2-1-34 log x2+1+C

Q127.

Answer :

We have,I=∫1-x1+x dxPutting x=cosθ⇒x=cos2θ⇒dx=-2 cosθ sinθ dθ⇒dx=-sin2θ dθ∴I=∫1-cosθ1+cosθ -sin 2θ dθ=∫2 sin2 θ22 cos2 θ2 -2 sinθ cosθ dθ=∫sin θ2cos θ2 -2×2 sin θ2 cos θ2cosθ dθ=-4∫sin2 θ2×cosθ dθ=-4∫1-cosθ2 cosθ dθ=-2∫cosθ-cos2θ dθ=-2∫cosθ-1+cos 2θ2dθ=-2∫cos θ dθ+∫1+cos 2θ dθ=-2sin θ+θ+sin 2θ2+C=-2 1-cos2θ+θ+2 sinθ cosθ2+C=-2 1-cos2θ+θ+sinθ cosθ+C=-21-x+cos-1x+1-xx+C=-21-x+cos-1x+x1-x+C

Q128.

Answer :

We have,I=∫x2+x+1x+12x+2 dxLet x2+x+1x+12x+2=Ax+1+Bx+12+Cx+2 …..1⇒x2+x+1=Ax+1x+2+Bx+2+Cx+12 …..2Putting x=-1 in 2, we get B=1Putting x=-2 in 2, we get C=3Putting x=0 in 2, we get1=2A+2B+C⇒1=2A+2+3⇒-4=2A⇒A=-2Now, 1 becomesx2+x+1x+12x+2=-2x+1+1x+12+3x+2Therefore, integral becomesI=∫-2x+1+1x+12+3x+2dx=-2 log x+1-1x+1+3 log x+2+C

Q129.

Answer :

We have,I=∫sin 4x-21-cos 4x e2x dx=∫2 sin 2x cos 2x-22 sin2 2x e2x dx=∫cot 2x-cosec2 2xe2x dxLet e2x cot 2x=t⇒2e2xcot 2x+e2x-cosec2 2x×2 dx=dt⇒e2xcot 2x-cosec2 2x dx=dt2∴I=12∫dt=t2+C=12 e2xcot 2x+C

Q130.

Answer :

We have,I=∫log log x+1log x2dxPutting log x=t⇒x=et⇒dx=et dt∴I=∫log t+1t2et dt=∫log t+1t-1t+1t2et=∫log t+1tet+∫-1t+1t2et dt=etlog t+et -1t+C ∵∫ex fx+f’x dx=ex fx+C=et log t-1t+C=elog x log log x-1log x+C=xlog log x-1log x+C

Q131.

Answer :

We have,I=∫sin x+cos x dx9+16 sin 2xLet sin x-cos x=t …..1Diff both sidescos x+sin x dx=dtSquaring both sides of 1, we getsin x-cos x2=t2⇒sin2x+cos2x-2 sin x cos x=t2⇒1-t2=2 sin x cos x⇒1-t2=sin 2x∴I=∫dt9+16 1-t2=∫dt25-16 t2=116∫dt2516-t2=116∫dt542-t2=116×12×54 log 54+t54-t+C=140 log 5+4 sin x-cos x5-4 sin x-cos x+C

Q132.

Answer :

We have,I=∫sinx-αsinx+αdx⇒I=∫sinx-αsinx-αsinx+αsinx-αdx⇒I=∫sinx-αsinx+αsinx-αdx⇒I=∫sinx cosα-cosx sinαsin2x-sin2αdx⇒I=∫sinx cosαcos2α-cos2xdx-∫cosx sinαsin2x-sin2αdx⇒I=∫sinx cosαcos2α-cos2xdx-∫cosx sinαsin2x-sin2αdx⇒I=I1-I2where,I1=∫sinx cosαcos2α-cos2xdxI2=∫cosx sinαsin2x-sin2αdxI1=∫sinx cosαcos2α-cos2xdxPutting cosx=t and -sinxdx=dt, we get⇒I1=-∫ cosαcos2α-t2dx⇒I1=-cosα sin-1tcosα+C1⇒I1=-cosα sin-1cosxcosα+C1I2=∫cosx sinαsin2x-sin2αdxPutting sinx=t and cosdx=dt, we get⇒I2=∫ sinαt2-sin2αdt⇒I2=sinα logt+t2-a2+C2⇒I2=sinα logsinx+sin2-a2+C2I=I1-I2I=-cosα sin-1cosxcosα+C1-sinα logsinx+sin2-a2+C2I=-cosα sin-1cosxcosα-sinα logsinx+sin2-a2+C

Q133.

Answer :

We have,I=∫exx3-x+2×2+12 dx=∫exx3+x2-x2-2x+x+2×2+12 dx=∫exx3+x2-x2-2x+x+2×2+12 dx=∫exx3+x2+x+1-x2-2x+1×2+12 dx=∫exx3+x2+x+1×2+12+-x2-2x+1×2+12 dx=∫exx2+1x+1×2+12+-x2-2x+1×2+12 dx=∫exx+1×2+1+-x2-2x+1×2+12 dxAs we know ∫exfx+f’xdx=exfx+CHere,ddxx+1×2+1=-x2-2x+1×2+12Therefore, integral becomesI=exx+1×2+1+C

Q134.

Answer :
We have,I=∫x2xsinx+cosx2 dx⇒I=∫x2cosxxsinx+cosx2cosx dx⇒I=∫xcosxxsinx+cosx2xcosx dx⇒I=xcosx∫xcosxxsinx+cosx2 dx-∫ddxxcosx∫xcosxxsinx+cosx2 Putting xsinx+cosx=t and xcosx dx=dt, we get⇒I=xcosx-1xsinx+cosx -∫cosx+sinxcos2x-1xsinx+cosx  dx⇒I=-xcosxxsinx+cosx+∫sec2x dx⇒I=-xcosxxsinx+cosx+tanx⇒I=-xcosxxsinx+cosx+sinxcosx⇒I=-x+sinxxsinx+cosxcosxxsinx+cosx⇒I=-x+xsin2x+sinxcosxcosxxsinx+cosx⇒I=-x1-sin2x+sinxcosxcosxxsinx+cosx⇒I=-xcosx+sinxcosxcosxxsinx+cosx⇒I=-xcosx+sinxx sinx+cosx

Q135.

Answer :

We have,I=∫cos 2x sin xdx=∫1-tan2x1+tan2x×1sin x dx=∫1-tan2xsec x sin x dx=∫1-tan2xtan xdxPutting 1-tan2x=t2Diff both sides w.r.t x-2 tan x sec2x=2t dtdx⇒dx=-2t dt2 tan x sec2x∴I=∫t×-2t dt2 tan2x sec2x=∫-t2 dt1-t2 1+tan2x=∫-t2 dt1-t2 2-t2Let t2=pThen t2 1-t2 2-t2=p1-p 2-pAnd let p1-p 2-p=A1-p+B2-p⇒p1-p 2-p=A 2-p +B 1-p1-p 2-p⇒p=A 2-p+B 1-pPutting p=2∴2=A×0+B 1-2⇒B=-2Putting p=1∴1=A 2-1+B×0⇒A=1∴ p1-p 2-p=11-p-22-p∴I=-∫dt1-t2+2∫dt2-t2=-∫dt12-t2+2∫dt22-t2=-12 log 1+t1-t+2×122 log 2+t2-t+C=-12 log 1+1-tan2x1-1-tan2x+12 log 2+1-tan2x2-1-tan2x+C=-12 log 1+1-sin2xcos2x1-1-sin2xcos2x+12 log 2+1-sin2xcos2x2-1-sin2xcos2x+C=-12 log cos x+cos2x-sin2x cos x-cos2x-sin2x+12 log 2 cos x+cos2-sin2x2 cos x-cos2-sin2x+C=-12 log cos x+cos 2x cos x-cos 2x+12 log 2 cos x+cos 2×2 cos x-cos 2x+C=-12 log cos x+cos 2x cos x-cos 2x+12 log 2 cos x+cos 2×2 cos x+cos 2×2 cos x-cos 2×2 cos x+cos 2x+C=12 log cos x-cos 2x cos x+cos 2x+12 log 2 cos x+cos 2×22 cos2x-cos 2x+C=12 log cos x-cos 2x cos x+cos 2x+12 log 2 cos x+cos 2×21+C=12 log cos x-cos 2x cos x+cos 2x+12×2 log 2 cos x+cos 2x+C=2 log 2 cos x+cos 2x+12 log cos x-cos 2x cos x+cos 2x+C

Q136.

Answer :

We have,I=∫dxsin x+sec x=∫dxsin x+1cos x=∫cos x dxsin x cos x+1=12∫2 cos x dxsin x cos x+1=12∫cos x+sin xsin x cos x+1 dx+12∫cos x-sin xsin x cos x+1 dxLet sin x-cos x=t …..1⇒sin x-cos x2=t2⇒sin2x+cos2x-2sin x cos x=t2⇒1-t22=sin x cos xDiff both sides of 1cos x+sin x dx=dtLet sin x+cos x=p …..2⇒sin x+cos x2=p2⇒1+2sin x cos x=p2⇒sin cos x=p2-12Diff both sides of 2cos x-sin x dx=dp∴I=12∫dt1-t22+1+12∫dpp2-12+1=12∫2 dt3-t2+12∫2 dpp2+1=∫dt32-t2+∫dpp2+12=123 log 3+t3-t+tan-1p+C=123 log 3+sin x-cos x3-sin x+cos x+tan-1 sin x+cos x+C

Q137.

Answer :

We have I=∫sin xsin 4x dx=∫sin x dx2 sin 2x·cos 2x=∫sin x dx2×2 sin x·cos x·cos 2x=14∫dxcos x·cos 2x=14∫dxcos x2 cos2x-1=14∫cos x·dxcos2x·2 cos2x-1=14∫cos x dx1-sin2x 1-2 sin2xLet sin x=t⇒cos x dx=dt∴I=14∫dt1-t2 1-2t2Let11-t2 1-2t2=A1-t2+B1-2t2Putting t2=p11-p 1-2p=A1-p+B1-2p …..(1)1=A 1-2p+B 1-p …..(2)Let 1-2p=0⇒p=12Substituting p=12in (2)∴1=A×0+B×12⇒B=2Let 1-p=0⇒p=1Substituting p=1 in (2)1=A 1-2+B×0⇒A=-1Therefore,(1) becomes 11-p 1-2p=-11-p+21-2p⇒11-t2 1-2t2=-11-t2+21-2t2 ∵ t2=p …..(3)As I=14∫dt1-t2 1-2t2Therefore, from (3) I=-14∫dt1-t2+24∫dt1-2t2=-14∫dt12-t2+12×12∫dt12-t2=-14∫dt1-t2+14∫dt122-t2=-14×12 log 1+t1-t+14×12 12 log 12+t12-t+C=-18 log 1+sin x1-sin x+142 log 1+2 sin x1-2 sin x+C Since, t= sin x

Q138.

Answer :

We have,I=∫dxx4+x2+1=12∫2 dxx4+x2+1=12∫x2+1-x2-1×4+x2+1dx=12∫x2+1×4+x2+1 dx-12∫x2-1×4+x2+1 dxDividing numerator and Denominator by x2I=12∫1+1×2 dxx2+1+1×2-12∫1-1×2 dxx2+1+1×2=12∫1+1×2 dxx2+1×2-2+3-12∫1-1×2 dxx2+1×2+2-1=12∫1+1×2 dxx-1×2+32-12∫1-1×2 dxx+1×2-1Putting x-1x=t⇒1+1×2 dx=dtPutting x+1x=p⇒1-1×2 dx=dp∴I=12∫dtt2+ 32-12∫dpp2-12=12×13 tan-1t3-12×12×1 log p-1p+1+C=123 tan-1x-1×3-14 log x+1x-1x+1x+1+C=123 tan-1×2-13 x-14 log x2-x+1×2+x+1+C=123 tan-1×2-13 x+14 log x2+x+1×2-x+1+C

Q139.

Answer :

We have,I=∫ex-1 dxPutting ex-1=t⇒ex dx=dt⇒dx=dtex⇒dx=dtt+1∴I=∫ tt+1dtPutting t=p⇒12t dt=dp⇒dtt=2dp⇒dt=2t dp=2p dp∴I=∫p×2p dpp2+1=2∫p2+1-1p2+1 dp=2∫dp-2∫dpp2+1=2p-2×tan-1p+C=2t-2×tan-1t+C=2ex-1-2 tan-1ex-1+C=2ex-1-tan-1ex-1+C

Q140.

Answer :

We have,I=∫cot x+cot3x1+cot3x dx=∫cot x 1+cot2x1+cot3xdx=∫cot x cosec2x1+cot3x dxPutting cot x=t⇒-cosec2x dx=dt⇒cosec2x dx=-dt∴I=-∫t dt1+t3=-∫t dt1+t t2-t+1Let t1+t t2-t+1=At+1+Bt+Ct2-t+1⇒t1+t t2-t+1=A t2-t+1+Bt+C t+1t+1 t2-t+1⇒t=A t2-t+1+Bt2+Bt+Ct+C⇒t=A+B t2+B+C-A t+A+CEquating Coefficients of like termsA+B=0 …..1B+C-A=1 …..2A+C=0 …..3Solving 1, 2 and 3, we getA=-13B=13C=13∴t1+t t2-t+1=-13 t+1+13 t+1t2-t+1⇒t1+t t2-t+1=-13 t+1+16 2t+2t2-t+1⇒t1+t t2-t+1=-13 t+1+16 2t-1+3t2-t+1∴I=–13∫dtt+1+16∫2t-1t2-t+1 dt+12∫dtt2-t+1=+13∫dtt+1-16∫2t-1t2-t+1 dt-12∫dtt2-t+14-14+1=13∫dtt+1-16∫2t-1 dtt2-t+1-12∫dtt-122+322let t2-t+1=p⇒2t-1 dt=dp∴I=13∫dtt+1-16∫dpp-12∫dtt-122+322=13 log t+1-16 log p-12×23 tan-1 t-1232+C=13 log t+1-16 log p-13 tan-1 2t-13+C=13 log cot x+1-16 log cot2x-cot x+1-13 tan-1 2 cot x-13+C

Page 20.15 Ex.20.1

Q1.

Answer :

Let I=∫491x dx. Then,I=2∫4912x dx⇒I=2×49⇒I=23-2⇒I=2

Q2.

Answer :

Let I=∫-231x+7 dx. Then,I=log x+7-23⇒I=log 10-log 5⇒I=log 105 ∵log a- log b=logab⇒I=log 2

Q3.

Answer :

Let I=∫01211-x2 dx. Then,I=sin-1×012⇒I=sin-112-sin-10⇒I=π6-0⇒I=π6

Q4.

Answer :

Let I=∫0111+x2dx. Then,I=tan-1×01⇒I=tan-11-tan-10⇒I=π4-0⇒I=π4

Q5.

Answer :

Let I=∫23xx2+1dx. Then,I=12∫232xx2+1⇒I=12log x2+123⇒I=12log 10-log 5⇒I=12log 105 ∵log a-log b=log ab⇒I=12log 2

Q6.

Answer :

Let I=∫0∞1a2+b2x2 dx. Then,I=1a2∫0∞11+b2x2a2 dx⇒I=1a2∫0∞11+bxa2 dx⇒I=aba2tan-1bxa0∞⇒I=1abtan-1∞-tan-10⇒I=π2ab

Q7.

Answer :

Let I=∫-1111+x2dx. Then,I=tan-1x-11⇒I=tan-11-tan-1-1⇒I=π4–π4⇒I=π2

Q8.

Answer :

Let I=∫0∞e-x dx. Then,I=-e-x0∞⇒I=-e-∞+e0⇒I=0+1⇒I=1

Q9.

Answer :

Let I=∫01xx+1 dx. Then,I=∫011-1x+1 dx⇒I=x-log x+101⇒I=1-log 2-(0-log 1)⇒I=log e-log 2⇒I=log e2

Q10.

Answer :

Let I=∫0π2sin x+cos x dx. Then,I=-cos x+sin x0π2⇒I=0+1–1+0⇒I=2

Q11.

Answer :

Let I=∫π4π2cot x dx. Then,I=-∫π4π2cot x-(cosec x+cot x)cosec x+cot x dx⇒I=-∫π4π2-cosec x cot x-cot2 xcosec x+cot x dx⇒I=-∫π4π2-cosec x cot x-cosec2x+1 cosec x+cot x dx ∵cosec2x=1+cot2x⇒I=-∫π4π2-cosec x cot x-cosec2x cosec x+cot x dx-∫π4π21cosec x+cot xdx⇒I=-∫π4π2-cosec x cot x-cosec2x cosec x+cot x dx-∫π4π2sin x1+cos xdx⇒I=-log cosec x+cot xπ4π2+log 1+cos xπ4π2⇒I=-log 1+∞+log 2+1+log 1+0-log 1+12⇒I=log 2+1-log 2+12⇒I=log 22+12+1⇒I=log2⇒I=12log 2

Q12.

Answer :

Let I=∫0π4sec x dx. Then,I=∫0π4sec x sec x+tan xsec x+tan xdx⇒I=∫0π4sec2 x+sec x tan xsec x+tan xdxPut u=sec x+tan x⇒du=sec2 x+sec x tan x dx∴∫0π4sec2 x+sec x tan xsec x+tan x dx=∫duu⇒I= log u⇒I=log sec x+tan x0π4⇒I=log secπ4+tanπ4-log sec 0+tan 0⇒I=log (2+1)-log 1⇒I=log (2+1)

Q13.

Answer :

Let I=∫π6π4cosec x dx. Then,I=∫π6π4cosec x cosec x-cot xcosec x-cot x dx⇒I=∫π6π4cosec2 x-cosec x cot xcosec x-cot x dx⇒I=log cosec x-cot xπ6π4⇒I=log 2-1-log2-3

Q14.

Answer :

Let I=∫011-x1+x dx. Then,I=∫0111+x-1+x-11+x dxI=∫0111+x-1+x1+x dx⇒I=log 1+x-x+log 1+x01⇒I=log 2-1+log 2-log 1-0+log 1=2 log 2-1

Q15.

Answer :

Let I=∫0π11+sin x dx. Then, I=∫0π1-sin x1+sin x1-sin x dx⇒I=∫0π1-sin x1-sin2 x dx ⇒I=∫0π1-sin xcos2 x dx ∵sin2 x+cos2 x=1⇒I=∫0πsec2 x-sec x tan x dx⇒I=tan x-sec x0π⇒I=tan π-sec π-tan 0-sec 0⇒I=0+1-0-1⇒I=1+1⇒I=2

Q16.

Answer :

Let I=∫-π4π411+sin x dx. Then,I=∫-π4π411+sin x×1-sin x1-sin x dx⇒I=∫-π4π41-sin x1-sin2 x dx⇒I=∫-π4π41-sin xcos2 x dx⇒I=∫-π4π41cos2 x-sin xcos2 x dx⇒I=∫-π4π4sec2 x-sec x tan x dx⇒I=tan x-sec x-π4π4⇒I=1-2–1-2⇒I=2

Q17.

Answer :

Let I=∫0π2cos2 x dx. Then,I=∫0π2cos2 x dx⇒I=12∫0π21+cos 2x dx ∵cos 2x=2 cos2 x-1⇒I=x2+sin 2×40π2⇒I=π4+0-0⇒I=π4

Q18.

Answer :

Let I=∫0π2cos3 x dx. Then,I=∫0π2cos2 x cos x dx⇒I=∫0π21-sin2 x cos x dxLet u= sin x, du= cos x dx⇒I=∫1-u2 du⇒I=u-u33⇒I=sin x-sin3 x30π2⇒I=1-13-0⇒I=23

Q19.

Answer :

Let I=∫0π6cos x cos 2x dx. Then,I=∫0π6cos x cos2 x-sin2 x dx⇒I=∫0π62 cos3 x-cos x dx⇒I=∫0π62 cos x 1-sin2 x – cos x dx⇒I=2sin x-sin3 x3-sin x0π6⇒I=212-124-12-0⇒I=512

Q20.

Answer :

Let I=∫0π2sin x sin 2x dx. Then,I=∫0π22 sin2x cos x dx⇒I=∫0π221-cos2 x cos x dx⇒I=∫0π22 cos x -2 cos3 x dx⇒I=2sinx-2sin x-sin3 x30π2⇒I=2-21-13-0⇒I=23

Page 20.16 Ex.20.1

Q21.

Answer :

Let I=∫π3π4tan x+cot x2 dx. Then,I=∫π3π4tan2 x+cot2 x+2 tan x cot x dx⇒I=∫π3π4tan2 x+cot2 x+2 dx⇒I=∫π3π4sec2 x-1+cosec2 x-1+2 dx⇒I=∫π3π4sec2 x+cosec2 x dx⇒I=tan x-cot xπ3π4⇒I=1-1-3-13⇒I=-23

Q22.

Answer :

Let I=∫0π2cos4 x dx. Then,I=∫0π2cos2 x2 dx⇒I=∫0π21+cos 2×24 dx⇒I=14∫0π21+cos2 2x+2 cos 2x dx⇒I=14∫0π21+2 cos 2x+1+cos 4×2 dx⇒I=14∫0π23+4 cos 2x+cos 4×2 dx⇒I=143×2+2 sin 2×2+sin 4×80π2⇒I=143π4+0⇒I=3π16

Q23.

Answer :

Let I=∫0π2a2 cos2 x+b2 sin2 x dx. Then, I=∫0π2a2 cos2 x+b2 1-cos2 x dx⇒I=∫0π2b2+a2-b2 cos2 x dx⇒I=∫0π2b2+a2-b21+cos 2x2dx⇒I=b2x+a2-b22x+sin 2×20π2⇒I=b2π2+a2-b22π2+0⇒I=π4a2+b2

Q24.

Answer :

Let I=∫0π21+sin x dx. Then,I=∫0π21+sin x×1-sin x1-sin x dx⇒I=∫0π21-sin2 x1-sin x dx⇒I==∫0π2cos x1-sin x dxLet 1-sin x=u⇒-cos x dx=du∴I=∫-duu⇒I==-2u⇒I==-21-sin x0π2⇒I==0+2⇒I==2

Q25.

Answer :

Let I=∫0π21+cos x dx. Then,I=∫0π21+cos x×1-cos x1-cos x dx⇒I=∫0π21-cos2 x1-cos x dx⇒I=∫0π2sin x1-cos x dxLet 1-cos x=u⇒sin x dx=du∴I=∫duu⇒I=2u⇒I=21-cos x0π2⇒I=2-0⇒I=2

Q26.

Answer :

Let I=∫0π2x sin x dxIntegrating by partsI=-x cos x0π2-∫0π21-cos x dx⇒I=-x cos x0π2+sin x0π2⇒I=0+1⇒I=1

Q27.

Answer :

Let I=∫0π2x cos x dx. Then,Integrating by partsI=x sin x0π2-∫0π21 sin x dx⇒I=x sin x0π2+cos x0π2⇒I=π2-1

Q28.

Answer :

Let I=∫0π2×2 cos x dx. Then,Integrating by partsI=x2 sin x0π2-∫0π22x sin x dx⇒I=x2 sin x0π2–2x cos x0π2+∫0π2-2 cos x dx⇒I=x2 sin x0π2+2x cos x0π2-2 sin x0π2⇒I=π24-2

Q29.

Answer :

Let I=∫0π4×2 sin x dx. Then,Integrating by partsI=-x2 cos x0π4-∫0π4-2x cos x dx⇒I=-x2 cos x0π4+2x sin x0π4-∫0π42 sin x dx⇒I=-x2 cos x0π4+2x sin x0π4+2 cos x0π4⇒I=-π2162+π22+22-2⇒I=2+π22-π2162-2

Q30.

Answer :

Let I=∫0π2×2 cos 2x dx. Then,Integrating by partsI=x2 sin 2×20π2-∫0π22x sin 2×2 dx⇒I=x2 sin 2×20π2–x cos 2×20π2+∫0π2-1 cos 2×2 dx⇒I=x2 sin 2×20π2+x cos 2×20π2-sin 2×40π2⇒I=0-π4-0⇒I=-π4

Q31.

Answer :

Let I=∫0π2×2 cos2 x dx. Then,I=∫0π2×2 1+cos 2x2dx⇒I=∫0π2×22+x2 cos 2×2 dx⇒I=x360π2+x2 sin 2×40π2-∫0π2×2 sin 2x dx⇒I=x360π2+x2 sin 2×40π2–x cos 2×40π2+∫0π2-1 cos2x2dx⇒I=x360π2+x2 sin 2×40π2+x cos 2×40π2-sin 2×40π2⇒I=π348-π8

Q32.

Answer :

Let I=∫12log x dx. Then,I=∫121 log x dxIntegrating by parts⇒I=x log x12-∫121x x dx⇒I=x log x12-∫12dx⇒I=x log x12-x12⇒I=2 log 2-2+1⇒I=2 log 2-1

Q33.

Answer :

Let I=∫13log x1+x2 dx. Then,I=-11+x log x13-∫131x-1x+1 dx⇒I=-11+x log x13+∫131xx+1 dx⇒I=-11+x log x13+∫131x-1x+1 dx⇒I=-11+x log x13+log x-log x+113⇒I=-14 log 3+log 3-log 4+log 2⇒I=34 log 3-log 2

Q34.

Answer :

Let I=∫1eexx1+x log x dx. Then,I=∫1eexx+ex log x dx⇒I=∫1eexx dx+∫1eex log x dxIntegrating first term by parts⇒I=log x ex1e-∫1eex log x dx+∫1eex log x dx⇒I=log e ee -0⇒I=ee

Q35.

Answer :

Let I=∫1elog xx dxLet log x=u⇒1x dx=du∴I=∫u du⇒I=u22⇒I=(log x)221e⇒I=12-0⇒I=12

Q36.

Answer :

Let I=∫ee21log x-1log x2 dx. Then,I=∫ee21 1log x dx-∫ee21log x2 dxIntegrating by parts⇒I=xlog xee2-∫ee2-1xlog x2 x dx-∫ee21log x2 dx⇒I=xlog xee2+∫ee21log x2 dx-∫ee21log x2 dx⇒I=xlog xee2+0⇒I=e2log e2-elog e⇒I=e22 log e-elog e⇒I=e22-e

Q37.

Answer :

Let I=∫12x+3xx+2 dx. Then,I=∫12xxx+2+3xx+2 dx⇒I=∫12dxx+2+∫123xx+2 dx⇒I=log x+212+32∫121x-1x+2 dx⇒I=log x+212+32log x-log x+212⇒I=log 4-log 3+32log 2-log 4-0+log 3⇒I=log 4-log 3+32-log 2+log 3⇒I=2 log 2-log 3+32 log 3-32 log 2⇒I=12 log 2+12 log 3⇒I=12log 2+ log 3⇒I=12 log 6

Q38.

Answer :

Let I=∫012x+35×2+1 dx. Then,I=∫012x5x2+1 dx+∫0135×2+1 dx⇒I=15∫0110x5x2+1 dx+3∫0115×2+12 dx⇒I=15log 5×2+101+35tan-15×01⇒I=15 log 6 +35tan-15

Q39.

Answer :

Let I=∫0214+x-x2 dx.Then,I=-∫021×2-x-4 dx⇒I=-∫021×2-x+14-14-4 dx=-∫021x-122-174 dx=-∫021x-122-1722 dx=∫021-2x-122+1722 dx=117log 17+2x-117-2x+102=117log 17+317-3-log 17-117+1=117log 26+6178-log 18-21716=117log 52+121718-217=117log 52+121718-217×18+21718+217⇒I=117 log 1344+32017256⇒I=117 log 21+5174

Q40.

Answer :

Let I=∫0112×2+x+1 dx. Then,I=12∫011×2+x2+12 dxI=12∫011×2+x2+116-116+12 dx⇒I=12∫011x+142+716 dx⇒I=12×47tan-1x+147401⇒I=27tan-157-tan-117

Q41.

Answer :

Let I=∫01×1-x dx. Then,I=∫0114-x-122 dx⇒I=12∫011-x-122 14 dx⇒I=12∫011-x-12 122 dxLet x-1212=sin u⇒2 dx=cos u du∴I=14∫-π2π21-sin2 u cos u du⇒I=14∫-π2π2cos2 u du⇒I=14∫-π2π2cos 2u+12 du⇒I=18sin 2u2+u-π2π2⇒I=18π2+π2⇒I=π8

Q42.

Answer :

Let I=∫0213+2x-x2 dx. Then,I=∫021-x2+2x-1+1+3 dx⇒I=∫021-x-12+4 dx⇒I=sin-1x-1202⇒I=sin-112-sin-1-12⇒I=2 sin-112⇒I=2×π6=π3

Q43.

Answer :

Let I=∫0414x-x2 dx. Then,I=∫0414x-x2-4+4 dx⇒I=∫041-x-22+4 dx⇒I=sin-1x-2204⇒I=sin-11-sin-1(-1)⇒I=2 sin-11⇒I=2 π2=π

Q44.

Answer :

Let I=∫-111×2+2x+5 dx. Then, I=∫-111×2+2x+1+4 dx⇒I=∫-111x+12+22 dx⇒I=12tan-1x+12-11⇒I=12tan-11-tan-10⇒I=12π4⇒I=π8

Q45.

Answer :

Let I=∫14×2+x2x+1 dx. Let 2x+1=u⇒x=u-12⇒dx=du2∴I=∫u-122+u-12u du2⇒I=18∫u2+1-2u+2u-2u du=18∫u2-1u du=18∫u32-u-12 du=182u525-2u121=1822x+1525-22x+112114=1825×243-6-25×93+23⇒I=184565-835⇒I=57-35

Q46.

Answer :

Let I=∫01×1-x5 dx. Then,I=∫01x-1+11-x5 dx⇒I=∫01-1-x6+1-x5 dx⇒I=1-x7701-1-x6601⇒I=-17+16⇒I=142

Q47.

Answer :

Let I=∫12x-1x2ex dx. Then,I=∫12exx-exx2 dx⇒I=∫12exx dx- ∫12exx2 dxIntegrating first term by partsI=exx12-∫12-1x2ex dx- ∫12exx2 dx⇒I=exx12+∫12exx2 dx- ∫12exx2 dx⇒I=exx12⇒I=e22-e

Q48.

Answer :

Let I=∫01x e2x+sin πx2 dx. Then,I=∫01x e2x dx+∫01sin πx2 dxIntegrating first term by partsI=x e2x201-∫011 e2x2 dx+-cos πx2π201⇒I=x e2x201-e2x401-2πcos πx201⇒I=e22-e24+14+2π⇒I=e24+14+2π

Q49.

Answer :

Let I=∫01x ex+cos πx4 dx. Then,I=∫01x ex dx+∫01cosπx4 dxIntegrating first term by partsI=x ex01-∫011 ex dx+sin πx4π401⇒I=x ex01-ex01+sin πx4π401⇒I=e-e+1+4π sin π4⇒I=1+4π2⇒I=1+22π

Disclaimer: The answer given in the book has some error. The solution here is created according to the question given in the book.

Q50.

Answer :

Let I=∫π2πex 1-sin x1-cos x dx. Then,I=∫π2πex1-2 sin x2 cos x22 sin2 x2 dx As, sin A=2 sin A2 cos A2, cos A=1-2 sin2 A2⇒I=∫π2πex 12 cosec2 x2-cot x2 dx⇒I=∫π2π12 ex cosec2 x2 dx-∫π2πex cot x2 dxIntegrating second term by partsI=-ex cot x2π2π-∫π2π12 ex cosec2 x2 dx+∫π2π12 ex cosec2 x2 dx⇒I=-0-eπ2⇒I=eπ2

Q51.

Answer :

Let I=∫02πex2sin x2+π4 dx. Then,Integrating by partsI=-2ex2 cos x2+π402π-∫02π-22 ex2 cos x2+π4 dxAgain, integrating second term by parts⇒I=-2ex2 cos x2+π402π+2ex2sin x2+π402π-∫02π22 ex2sin x2+π4 dx⇒I=-2ex2 cos x2+π402π+2ex2sin x2+π402π-I⇒2I=22 eπ+22-22eπ-22=0⇒I=0

Q52.

Answer :

Let I=∫02πex cosπ4+x2 dx. Then,Integrating by partsI=2ex sin π4+x202π-∫02π2ex sin π4+x2 dxIntegrating second term by partsI=2ex sin π4+x202π+4ex cos π4+x202π+∫02π-4ex cos π4+x2 dx⇒I=2ex sin π4+x202π+4ex cos π4+x202π-4I⇒5I=-2e2π 12-2 12-4e2π 12-4 12⇒5I=-32 e2π-32⇒I=-325e2π+1

Q53.

Answer :

Let I=∫0111+x-x dx. Then,I=∫0111+x-x×1+x+x1+x+x dx⇒I=∫011+x+x1+x-x dx⇒I=∫011+x+x dx⇒I=231+x32+23×3201⇒I=23×22+23-23⇒I=423

Q54.

Answer :

Let I=∫12xx+1x+2 dx. Then,I=∫12-1x+1+2x+2 dx⇒I=-∫121x+1 dx+2∫121x+2 dx⇒I=-log x+1+2 log x+212⇒I=-log 3+2 log 4+log 2-2 log 3⇒I=5 log 2-3 log 3⇒I=log 25- log 33⇒I=log 3227

Q55.

Let I=∫0π2sin3 x dx. Then,I=∫0π2sin x sin2 x dx⇒I=∫0π2sin x 1-cos2 x dxLet u =cos x, du= -sin x dx∴I=∫-1-u2 du⇒I=u33-u⇒I=cos3 x3-cos x0π2⇒I=0-13+1⇒I=23

Q56.

Answer :

Let I=∫0πsin2 x2-cos2 x2 dx. Then,I=-∫0π cos x dx ∵cos A=cos2 A2-sin2 A2⇒I=-sin x0π⇒I=0

Q57.

Answer :

Let I∫12e2x1x-12×2 dx. Then,I=∫12e2x 1x-∫12e2x 12×2 dxIntegrating first term by parts⇒I=e2x2x12-∫12-e2x 12×2-∫12e2x 12×2 dx⇒I=e2x2x12⇒I=e44-e22⇒I=e4-2e24

Q58.

Answer :

We have,∫0k12+8×2 dx=π16⇒18∫0k114+x2 dx=π16⇒14 tan-12x0k=π16⇒tan-12k=π4⇒2k=tanπ4⇒2k=1⇒k=12

Q59.

Answer :

We have,∫0a3x2 dx=8⇒3 x330a=8⇒a3=8⇒a=2

Page 20.32 Ex.20.2

Q1.

Answer :

Let x2=t. Then, 2x dx=dtWhen x=2, t=4 and x=4, t=16. ∴I=∫24xx2+1 dx⇒I=∫41612dtt+1⇒I=12 log t+1416⇒I=12 log 17-12 log 5⇒I=12 log 175

Q2.

Answer :

Let 1+log x=t. Then, 1x dx=dtWhen x=1, t=1 and x=2, t=1+log 2∴I=∫121×1+log x2 dx⇒I=∫11+log 21t2 dt⇒I=-1t11+log 2⇒I=-11+log 2+1⇒I=log 2log 2+log e⇒I=log 2log 2e

Q3.

Answer :

Let x2=t. Then, 2x dx=dtWhen x=1, t=1 and x=2, t=4∴I=∫123x9x2-1 dx⇒I=32∫14dt9t-1⇒I=318log 9t-114⇒I=318log 35-log 8⇒I=log 35-log 86

Q4.

Answer :

Let I=∫0π215 cos x+3 sin x dx. Then,I=∫0π2151-tan2x21+tan2x2+32 tan x21+tan2 x2 dx ∵sin A=2 tan A21+tan2 A2, cos A=1-tan2 A21+tan2 A2⇒I=∫0π21+tan2 x25-5 tan2 x2+6 tan x2 dx⇒I=∫0π2sec2 x25-5 tan2 x2+6 tan x2 dxLet tan x2 =t. Then, 12 sec2 x2 dx=dtAlso, x=0, t=0 and x=π2, t=1∴I=∫012dt5-5t2+6t⇒I=15∫012dt1-t2+65t+36100-36100=25∫01dt-t-6102+136100=25×10136-log t-610-13610t-610+1361001=134-log 4-2344+234+log -6-234-6+234=134 log 6+2346-234×4+2344-234=134 log 160+2034160-2034=134 log 8+348-34

Q5.

Answer :

Let x= a tan t. Then, dx=a sec2 t dtWhen x=0, t=0 and x=a, t=π4∴I=∫0axa2+x2 dx⇒I=∫0π4a tan ta2+a2 tan2 ta sec2 t dt=∫0π4a tan t a sec2 ta sec t dt=∫0π4a tan t sec t dt=asec t0π4=a2-1

Q6.

Answer :

Let ex=t. Then, ex dx=dtWhen x=0, t=1 and x=1, t=e∴I=∫01ex1+e2x dx⇒I=∫1edt1+t2⇒I=tan-1x1e⇒I=tan-1e-tan-11⇒I=tan-1e-π4

Q7.

Answer :

Let I=∫01x ex2dx.Let x2=t. Then, 2x dx=dtWhen x=0, t=0 and x=1, t=1∴ I=12∫01et dt⇒I=12et01⇒I=12e-1

Q8.

Answer :

Let I=∫13cos log xx dx.Let log x=t. Then, 1x dx=dtWhen x=1, t=0 and x=3, t =log 3∴I=∫0log 3cos t dt=sin t0log 3=sin log 3

Q9.

Answer :

Let I=∫012×1+x4 dx.Let x2=t. Then, 2x dx=dtWhen x=0, t=0 and x=1, t=1∴ I=∫012×1+x4 dx⇒I=∫0111+t2 dt⇒I=tan-1t01⇒I=tan-11- tan-10⇒I=π4

Q10.

Answer :

Let I=∫0aa2-x2 dx.Let x= a sin t. Then, dx= a cos t dtWhen x=0, t=0 and x=a, t=π2∴ I=∫0aa2-x2 dx⇒I=∫0π2a2-a2 sin2 t a cos t dt⇒I=∫0π2a2 cos2 t dt⇒I=a2∫0π21+cos 2t2 dt⇒I=a22t+sin 2t20π2⇒I=a22π2-0⇒I=πa24

Q11.

Answer :

∫0π2sin ϕ cos5 ϕ dϕLet sin ϕ=t. Then, cos ϕ dϕ=dtWhen ϕ=0, t=0 and ϕ=π2, t=1Also, cos5 ϕ=cos4 ϕ cos ϕ=1-sin2 ϕ2 cos ϕ∴ I=∫0π2sin ϕ cos5 ϕ dϕ⇒I=∫01t 1-t22 dt⇒I=∫01t1+t4-2t2 dt⇒I=∫01t+t92-2t52 dt⇒I=2t323+2t11211-4t72701⇒I=23+211-47⇒I=64231

Q12.

Answer :

Let I=∫0π2cos x1+sin2 x dx.Let sin x=t. Then, cos x dx=dtWhen x=0, t=0 and x=π2, t=1∴I=∫0π2cos x1+sin2 x dx⇒I=∫0111+t2 dt⇒I=tan-1 t01⇒I=π4

Q13.

Answer :

Let I=∫0π2sin θ1+cos θ dθ.Let cos θ=t. Then, -sin θ dθ=dtWhen θ=0, t=1 and θ=π2, t=0∴ I=∫0π2sin θ1+cos θ dθ=∫10-dt1+t=∫01dt1+t=21+t01=22-1

Q14.

Answer :

Let I=∫0π3cos x3+4 sin x dx.Let sin x =t. Then, cos x dx=dtWhen x=0, t=0 and x=π3, t=32∴ I=∫0π3cos x3+4sin xdx=∫03213+4tdt=14log 3+4t032=14log 3+23-log 3=14 log 3+233

Q15.

Answer :

Let I=∫01tan-1×1+x2 dx.Let tan-1x =t. Then, 11+x2 dx=dtWhen x=0, t=0 and x=1, t=π4∴ I=∫01tan-1×1+x2 dx⇒I=∫0π4t dt⇒I=2t3230π4⇒I=23π432⇒I=112π32

Q16.

Answer :

Let I=∫02xx+2 dx.Let x+2=t2. Then, dx=2t dtWhen x=0, t=2 and x=2, t=2∴ I=∫22t2-2 t 2t dt⇒I=2∫22t4-2t2 dt⇒I=2t55-23t322⇒I=2323-163-425-423⇒I=21615+8215⇒I=16152+2

Q17.

Answer :

∫01tan-12×1-x2dx=∫012tan-1x=2x tan-1×01-2∫01×1+x2dx=2x tan-1×01-log1+x201=2π4-0-log2+0=π2-log2

Q18.

Answer :

Let I=∫0π2sin x cos x1+sin4 x dx.Let sin x=t. Then, cos x dx=dtWhen x=0, t=0 and x=π2, t=1∴ I=∫01t1+t4 dtLet t2=u. Then, 2t dt=duSo, I=∫01t1+t4dt ⇒I=12∫0111+u2du⇒I=12tan-1u01⇒I=π8

Q19.

Answer :

∫0π21acosx+b sinxdx=∫0π21a1-tan2x21+tan2x2+b2tanx21+tan2x2dx=∫0π21+tan2x2a-atan2x2+2b tanx2dx=∫0π2sec2x2a-atan2x2+2b tanx2dxLet tanx2=t, Then, 12sec2x2dx=dtWhen x=0, t=0, x=π2, t=1Therefore the integral becomesI= ∫012dta-at2+2bt=∫012dt-at2-2bta-1=2a∫01dt-t-ba2-1-b2a2=2a∫01dtb2a2+1-t-ba2=2a12a2+b2a2loga2+b2a2+t-baa2+b2a2-t-ba01
1a2+b2loga+b+a2+b2a+b-a2+b2

Q20.

Answer :

Let I=∫0π215+4 sin x dx. Then,I=∫0π215+42 tan x21+tan2 x2 dx⇒I=∫0π21+tan2x251+tan2x2+8 tan x2 dx⇒I=∫0π2sec2 x25 tan2 x2+8 tan x2+5 dxLet tan x2=t. Then, 12sec2 x2 dx=dtWhen x=0, t=0 and x=π2, t=1∴ I=2∫0115t2+8t+5 dt⇒I=2∫0115t2+8t+5+452-452 dt⇒I=2∫0115t+452+95 dt⇒I=23tan-15t+453501⇒I=23tan-13-tan-143⇒I=23tan-13-431+3×43⇒I=23 tan-113

Q21.

Answer :

∫0πsinxsinx+cosxdx=12∫0π2sinxsinx+cosxdx=12∫0πsinx+cosx-cosx-sinxsinx+cosxdx=12∫0πdx-12∫0πcosx-sinxsinx+cosxdx=12×0π-12logsinx+cosx0π=12π-0-12log1-log1=π2

Q22.

Answer :

Let I=∫0π13+2 sin x+cos x dx. Then,I=∫0π13+22 tanx21+tan2x2+1-tan2x21+tan2x2 dx⇒I=∫0π1+tan2 x22 tan2 x2+4 tan x2+4 dxLet tan x2=t. Then, 12 sec2 x2 dx=dtWhen x=0, t=0 and x=π, t=∞∴ I=∫0∞2 dt2t2+4t+4⇒I=∫0∞dtt+12+1⇒I=tan-1t+10∞⇒I=π2-π4⇒I=π4

Q23.

Answer :

Let I=∫01tan-1x dx. Then,I=∫011 tan-1x dxIntegrating by partsI= x tan-1×01-∫01×1+x2 dx⇒I=x tan-1×01-12log x2+101⇒I=π4-0-12 log 2+0⇒I=π4-12 log 2

Q24.

Answer :

Let I=∫0π2sin 2x tan-1sin x dx. Then,I=∫0π22 sin x cos x tan-1sin x dxLet sin x=t. Then, cos x dx=dtWhen x=0, t=0 and x=π2, t=1∴ I=2∫01 t tan-1t dt⇒I=2t22 tan-1t01-2∫01t1+t2 dt⇒I=2t22 tan-1t01-log 1+t201 ⇒I=2π4-1⇒I=π2-1

Q25.

Answer :

Let I=∫01cos-1×2 dx. Then,I=∫011cos-1x2dxIntegrating by parts⇒I=xcos-1×201-∫012x cos-1x -11-x2 dxAgain, integrating second term by parts⇒I=xcos-1×201+21-x2 cos-1×01-2∫0111-x21-x2 dx⇒I=xcos-1×201+21-x2 cos-1×01-2×01⇒I=0+2π2-2⇒I=π-2

Q26.

Answer :

Let I=∫0π4tan3 x1+cos 2x dx. Then,I=∫0π4tan3 x2 cos2 x dx⇒I=12∫0π4tan3 x sec2 x dxLet tan x=t. Then, sec2 x dx=dtWhen x=0, t=0 and x=π4, t=1∴ I=12∫01t3 dt⇒I=12t4401⇒I=1214-0⇒I=18

Q27.

Answer :

Let I=∫0π15+3 cos x dx. Then,I=∫0π15+31-tan2 x21+tan2 x2 dx⇒I=∫0π1+tan2 x25+5 tan2 x2-3 tan2 x2 dx⇒I=∫0πsec2 x25+2 tan2 x2 dxLet tan x2=t. Then, 12 sec2 x2 dx=dtWhen x=0, t=0 and x=π, t=∞∴ I=∫0∞dt5+2t2⇒I=12∫0∞dt52+t2⇒I=12tan-15t20∞⇒I=12π2-0⇒I=π4

Q28.

Answer :

Let I=∫0π21a2 sin2 x+b2 cos2 x dx. Then,Dividing the numerator and denominator by cos2x, we getI=∫0π2sec2xa2 tan2x+b2 dxLet tan x =t. Then, sec2 x dx= dtWhen x=0, t=0 and x=π2 , t=∞∴ I=∫0∞1a2 t2+b2 dt⇒I=1a2∫0∞1t2+b2a2 dt⇒I=1a2×abtan-1atb0∞⇒I=1abπ2⇒I=π2ab

Q29.

Let, I=∫0asin-1xa+x dxLet, x=a tan2θ ⇒θ=tan-1xaWhen, x→x ; θ→0 and x→a ; θ→π4and  dx=2a tanθ sec2θ dθThen,I=∫0π4sin-1a tan2θa+a  tan2θ 2a tanθ sec2θ dθ⇒I= 2a ∫0π4sin-1sinθ tanθ sec2θ dθ⇒I= 2a ∫0π4θ tanθ sec2θ dθLet, tan θ=t⇒θ=tan-1t⇒sec2θ dθ =dtwhen, θ→0 ; t→0 and θ→π4 ; t→1Then, I=2a∫01tan-1t  t dt             =2a∫01tan-1t  t dt             =2atan-1t  t2201-2a2∫01t21+t2   dt             =2aπ4×12-0-a∫011-11+t2   dt             =2aπ8-at-tan-1t01             =πa4-a1-π4             =πa4-a+πa4             =πa2-a            =aπ2-1

Q30.

Answer :

Let I=∫01tan-1×1+x2 dx. Then,Let tan-1x=t. Then, 11+x2 dx=dtWhen x=0, t=0 and x=1, t=π4∴ I=∫0π4 t dt⇒I=t220π4⇒I=π232

Q31.

Answer :

∫π3π21+cosx1-cosx32dx=∫π3π21+cosx1-cosx32×1-cosx1-cosxdx=∫π3π21-cos2x1-cosx2dx=∫π3π2sinx1-cosx2dxLet 1-cosx=t, Then sinx dx=dtWhen x=π3, t=12 and x=π2, t=1Therefore the integral becomes=∫121dtt2=-1t121=-1+2=1

Q32.

Answer :

Let I=∫01x tan-1x dx. Then,Integrating by partsI=x2 tan-1×201-12∫01×21+x2 dx⇒I=x2 tan-1×201-12∫011+x21+x2-11+x2 dx⇒I=x2 tan-1×201-12x-tan-1×01⇒I=π8-0-121-π4-0⇒I=π4-12

Q33.

Answer :

Let, I=∫1-x2x4+x2+1 dx=-∫x2-1×4+x2+1 dx=-∫1-1x2x2+1+1×2 dx=-∫1-1x2x2+2+1×2-1 dx=-∫1-1x2x+1×2-1 dxLet, x+1x=t⇒1-1x2dx=dtThen integral becomes,I=-∫1t2-1 dt=-12logt-1t+1=12logt+1t-1=12logx+1x+1x+1x-1=12logx2+x+1×2-x+1i.e., ∫1-x2x4+x2+1 dx=12logx2+x+1×2-x+1⇒∫011-x2x4+x2+1 dx=12logx2+x+1×2-x+1 01 =12log 3

Page 20.33 Ex.20.2

Q34.

Answer :

Let I=∫0124×31+x24 dx. Then,Let x2 =t. Then, 2x dx=dtWhen x=, t=0 and x=1, t=1∴ I=∫0112t1+t4 dtIntegrating by partsI=12t-31+t301+12∫01131+t3dt⇒I=12t-31+t301-161+t201⇒I=12-124-0-124+16⇒I=12×112⇒I=1

Q35.

Answer :

Let I=∫412xx-413 dx.Let x-4=t. Then, dx=dtWhen x=4, t=0 and x=12, t=8∴ I=∫08t+4t13 dt⇒I=∫08 t43+4t13 dt⇒I=37t73+31t4308⇒I=3847+48⇒I=7207

Q36.

Answer :

Let I=∫0π2×2 sin x dx. Then,Integrating by partsI=-x2 cos x0π2-∫0π2-2x cos x dxAgain, integratting by parts⇒I= -x2 cosx0π2+2x sin x0π2-∫0π21 sin x dx⇒I= -x2 cos x0π2+2x sin x0π2–cos x0π2⇒I=π24 0-0+2π2-0+0-2⇒I=π-2

Q37.

Answer :

Let I=∫011-x1+x dx. Then,I=∫011-x1+x×1-x1-x dx⇒I=∫011-x1-x2 dx⇒I=∫0111-x2 dx-∫01×1-x2 dx⇒I=sin-1×01+12∫01-2×1-x2 dx⇒I=sin-1×01+1221-x201⇒I=π2-0+0-1⇒I=π2-1

Q38.

Answer :

Let I=∫011-x21+x22 dx. Then,I=∫011×2-1x+1×2 dxLet x+1x=t. Then, 1-1×2 dx=dtWhen x=0, t=∞ and x=1, t=2∴I=∫∞2-dtt2⇒I=1t∞2⇒I=12-0⇒I=12

Q39.

Answer :

Let I=∫-115x4x5+1 dx. Then,Let x5+1=t. Then, 5×4 dx=dtWhen x=-1, t=0 and x=1, t=2∴ I=∫02t dt⇒I=23t3202⇒I=238⇒I=423

Q40.

Answer :

Let, I=∫0ax a2-x2a2+x2 dxConsider, x2=a2cos2θ ⇒2x dx=-2a2 sin2θ dθ ⇒x dx=-a2 sin2θ dθWhen, x→0 ; θ→π4 and x→a ;θ→0Now, integral becomes,I=∫π40-a2 sin2θ a2-a2cos2θa2+a2cos2θ dθ =∫π40-a2 sin2θ tanθ dθ =a2 ∫0π42 sinθ cosθ sinθcosθ dθ =a2 ∫0π42sin2θ dθ =a2 ∫0π41-cos 2θ dθ =a2 θ -sin2θ20π4 =a2 π4 -12

Q41.

Answer :

Let, I= ∫-aaa-xa+x dxConsider, x=a cos 2y Then y=12cos-1xa ⇒dx=-2a sin 2y dyWhen, x→-a ;y→π2 and x→a ;y→0Now, integral becomes, I= ∫π20 -2a sin 2ya-a cos 2y a+a cos 2y dy =∫0π2 2a sin 2y tan y dy =2a∫0π2 2sin y cos y sin ycos y dy =2a∫0π2 2sin2 y dy =2a∫0π2 1-cos 2y dy =2a y-sin 2y20π2 =2a π2-sin 2y20π2 =πa

Q42.

Answer :

Let I=∫0π55-4cos θ14 sin θ dθ. Let 5-4 cos θ=t . Then, 4 sin θ dθ=dtWhen θ=0, t=1 and θ=π, t=9∴I=54∫19t14 dt⇒I=544t54519⇒I=93-1

Q43.

Answer :

Let I=∫0π6cos-3 2θ sin 2θ dθ. Then,I=∫0π6sin 2θcos3 2θ dθLet cos 2θ=t. Then, -2 sin 2θ dθ=dtWhen θ=0, t=1 and θ=π6, t=12∴ I=-12∫112dtt3⇒I=1212t2112⇒I=122-12⇒I=34

Q44.

Answer :

Let I=∫0π23x cos2 x32 dx. Then,Let x32=t. Then, 32x dx= dtWhen x=0, t=0 and x=π23, t=π∴ I=23∫0πcos2 t dt⇒I=23∫0π1+cos 2×2 dx⇒I=13x+sin 2×20π⇒I=13π+0⇒I=π3

Q45.

Answer :

Let I=∫121×1+log x2 dx. Then, Let 1+log x=t. Then, 1x dx=dtWhen x=1, t=1 and x=2, t=1+log 2∴ I=∫11+log 21t2 dt⇒I=-1t11+log 2⇒I=-11+log 2+1⇒I=log 21+log 2

Q46.

Answer :

Let I=∫0π2cos5 x dx. Then,I=∫0π2cos4 x cos x dx⇒I=∫0π21-sin2 x2 cos x dx⇒I=∫0π21-2 sin2 x+sin4 x cos x dx Let sin x =t. Then, cos dx= duWhen x=0, t=0 and x=π2, t=1∴ I=∫011-2t2+t4 dt⇒I=t-2t33+t5501⇒I=1-23+15⇒I=815

Q47.

Answer :

Let I=∫49×30-x322 dx. Then,Let 30-x32=t. Then, -32x dx= dtWhen, x=4, t=22 and x=9, t=3∴ I=∫223-231t2 dt⇒I=231t223⇒I=2313-122⇒I=1999

Q48.

Answer :

Let I=∫0π4sin3 2t cos 2t dt. Then,Let sin 2t =u. Then, 2 cos 2t dt=duWhen t=0, u=0 and t=π4, u=1∴ I=12∫01u3 du⇒I=12u4401⇒I=1214-0⇒I=18

Q49.

Answer :

Let I=∫0π2sin x cos xcos2 x+3 cos x+2 dx. Then,Let cos x =t. Then, – sin x dx= dtWhen x=0, t=1 and x =π2, t=0∴ I=-∫10t dtt2+3t+2⇒I=∫10-t dtt+2t+1⇒I=∫101t+1-2t+2 dt⇒I=log t+1-2 log t+210⇒I=log t+1t+2201⇒I=log 14-log 2901⇒I=log 98

Q50.

Answer :

Let I=∫0πsin3 x 1+2 cos x1+cos x2 dx. Then,I=∫0πsin x sin2 x 1+2 cos x1+cos x2 dx⇒I=∫0πsin x 1-cos2 x1+2 cos x1+cos x2 dx⇒I=∫0πsin x 1-cos x1+2 cos x1+cos x3 dxLet cos x =t. Then, – sin x dx = dtWhen x=0, t=1 and x=π, t=-1∴ I=-∫1-11-t1+2t1+t3 dt⇒I=∫-111+t-2t21+t3+3t+3t2 dt⇒I=∫-111+t3+3t+3t2+t+t4+3t2+3t3-2t2-2t5-6t3-6t4 dt⇒I=∫-111+4t+4t2-2t3-5t4-2t5 dt⇒I=t+2t2+4t33-t42-t5-t63-11⇒I=1+2+43-12-1-13+1-2+43+12-1+13⇒I=83

Q51.

Answer :

Let I=∫0π22 sin x cos x tan-1sin x dx. Then,Let sinx =t. Then, cos x dx = dtWhen x=0, t=0 and x=π2, t=1∴ I=∫01 2t tan-1 t dt⇒I=2t2 tan-1 t201-2∫01t21+t2 dt⇒I=2t2 tan-1 t201-2∫011+t21+t2-11+t2 dt⇒I=2t2 tan-1 t201-t-tan-1 t+01⇒I=1 tan-1 1 -0-1+tan-11+0⇒I=π4-1+π4⇒I=π2-1

Q52.

Answer :

Let, I=∫0π/2x+sin x1+cos x dx =∫0π/2x+sin x2 cos2 x2 dx =∫0π/2 x2 cos2 x2 +sin x2 cos2 x2dx =12∫0π/2x sec2 x2dx+∫0π/22sin x2cosx22 cos2 x2dx =12x tanx2120π/2-12∫0π/2 tanx212dx+∫0π/2tanx2dx =x tanx20π/2-∫0π/2 tanx2 dx+∫0π/2tanx2dx =π2 tanπ4 =π2 ×1 =π2

Q53.

Answer :

Let I=∫0π4tan x+cot x dx. Then,I=∫0π4sin xcos x+cos xsin x dx ⇒I=∫0π4sin x+cos xsin x cos x dx⇒I=2∫0π4sin x+cos x2 sin x cos x dx⇒I=2∫0π4sin x +cos x1-sin x-cos x2 dxLet sin x-cos x =t. Then, cos x+ sin x dx= dtWhen x=0, t=1 and x=π4, t=0∴ I=2∫-10dt1-t2⇒I=2 sin-1 t-10⇒I=π2⇒I

Page 20.74 Ex.20.3

Q1.

Answer :

(i) We have,
∫14fx dx, where fx=4x+3, if 1≤x≤23x+5, if 2≤x≤4

I=∫14fx dx⇒I=∫12fx dx+∫24fx dx Additive property⇒I=∫12 4x+3 dx+∫243x+5 dx⇒I=2×2+3×12+3×22+5×24⇒I=8+6-2-3+24+20-6-10⇒I=37

(ii) We have,
∫09fx dx, where fx sin x,0≤x≤π21,π2≤x≤3ex-3,3≤x≤9

I=∫09fx dx⇒I=∫0π2fx dx+ ∫π23fx dx+∫39fx dx Additive property⇒I=∫0π2sin x dx+ ∫π231 dx+∫39ex-3 dx⇒I=-cos x0π2+xπ23+ex-339⇒I=0+1+3-π2+e6-e0⇒I=3-π2+e6

(iii) We have,
∫14fx dx, where fx=7x+3,if 1≤x≤38x,if 3≤x≤4

I=∫14fx dx⇒I=∫13fx dx+∫34fx dx Additive property⇒I=∫137x+3 dx+∫348x dx⇒I=7×22+3×13+4×234⇒I=632+9-72-3+64-36⇒I=562+34⇒I=62

Q2.

Answer :

(i)
∫-44x+2 dxWe know that, x+2=-x+2 , -4≤x≤-2x+2, -2<x≤4∴I=∫-44x+2 dx⇒I=∫-4-2-x+2 dx+∫-24x+2 dx⇒I=-x22-2x-4-2+x22+2x-24⇒I=-2+4-8-8+8+8-2+4⇒I=20

(ii)
I=∫-33x+1 dxWe know that, x+1=-x+1 , -3≤x≤-1x+1, -1<x≤3∴ I=∫-3-1-x+1 dx+∫-13x+1 dx⇒I=-x+122-3-1+x+122-13⇒I=0+2+8-0⇒I=10

(iii)
∫-112x+1 dxWe know that, 2x+1=-2x+1, -1≤x≤-122x+1, -12<x≤1∴I=∫-1-12-2x+1 dx+∫-1212x+1 dx⇒I=-x2+x-1-12+x2+x-121⇒I=-14+12+1-1+1+1-14+12⇒I=52

(iv)
∫-222x+3 dxWe know that, 2x+3=-2x+3, -2≤x≤-322x+3, -32<x≤2∴I=∫-2-32-2x+3 dx+∫-3222x+3 dx⇒I=-x2+3x-2-32+x2+3x-322⇒I=-94+92+4-6+4+6-94+92⇒I=252

(v)
∫02×2-3x+2 dxWe know that, x2-3x+2=-x2-3x+2, x-1x-2≤0 or, 1≤x≤2×2-3x+2, x2-3x+2≥0 or, x∈-∞, 1∪2, ∞∴I=∫02×2-3x+2 dx⇒I=∫01×2-3x+2 dx-∫12×2-3x+2 dx⇒I=x33-3×22+2×01-x33-3×22+2×12⇒I=13-32+2-83-6+4-13+32-2⇒I=13-32+2-83+6-2+13-32⇒I=1

(vi)
∫033x-1 dxWe know that, 3x-1=-3x-1, 0≤x≤133x-1, 13<x≤3∴I==∫013-3x+1 dx+∫1303x+1 dx⇒I=-3×22-x013+3×22+x133⇒I=-16+13-0+272+3-16-13⇒I=656

(vii)
∫-66x+2dxWe know that, x+2=-x+2 , -6≤x≤-2x+2, -2<x≤6∴I=∫-66x+2 dx⇒I=∫-6-2-x+2 dx+∫-26x+2 dx⇒I=-x22-2x-6-2+x22+2x-26⇒I=-2+4+18-12+18+12-2+4⇒I=40

(viii)
∫-22x+1 dxWe know that, x+1=-x+1 , -2≤x≤-1x+1, -1<x≤2∴I=∫-22x+1 dx⇒I=∫-2-1-x+1 dx + ∫-12x+1 dx⇒I=-x22-x-2-1+x22+x-12⇒I=-12+1+2-2+2+2-12+1⇒I=5

(ix)
∫12x-3 dxWe know that, x+1=-x+1 , 1≤x≤3x+1, x>3∴I=∫12x-3 dx⇒I=∫12-x-3 dx⇒I=-x22-3×12⇒I=-2-6+12+3⇒I=32

(x)
∫0π2cos 2x dxWe know that, cos 2x=-cos 2x ,π4 ≤x≤π2cos 2x, 0<x≤π4∴I=∫-22cos 2x dx⇒I=∫0π4cos 2x dx- ∫π4π2 cos 2x dx⇒I=sin 2×20π4-sin 2×2π4π2⇒I=12-0-0+12⇒I=1

(xi)
∫02πsin x dxWe know that, sin x=- sin x ,π ≤x≤2πsin x, 0<x≤π∴I=∫02πsin x dx⇒I=∫0π sin x dx+∫π2π- sin x dx⇒I=-cos x0π+cos xπ2π⇒I=1+1+1–1⇒I=4

(xii)
∫-π4π4sin x dxWe know that, sin x=- sin x ,-π4 ≤x≤0sin x, 0<x≤π4∴I=∫-π4π4sin x dx⇒I=∫-π40-sin x dx +∫0π4 sin x dx⇒I=cos x-π40-cos x0-π4⇒I=1-12-12+1⇒I=2-22⇒I=2-2

(xiii)
∫28x-5 dxWe know that, x-5=-x-5 , 2≤x≤5x-5, 5<x≤8∴I=∫28x-5 dx⇒I=∫25-x-5 dx+∫58 x-5 dx⇒I=-x22-5×25+x22-5×58⇒I=-252+25+2-10+32-40-252+25⇒I=9

(xiv)
∫-π2π2sin x+ cos x dxSince, f-x=sin -x + cos -x=sin x + cos x=fxSo, fx is an even function.∴I=2∫0π2sin x+cos x dx⇒I=2-cos x+sin x0π2⇒I=20+1+1-0⇒I=4

(xv)
∫04x-1 dxWe know that, x-1=-x-1 , 0≤x≤1x-1, 1<x≤4∴I=∫04x-1 dx⇒I=∫01-x-1 dx+∫14x-1 dx⇒I=-x22+x01+x22-x14⇒I=-12+1-0+8-4-12+1⇒I=5

(xvi)
I=∫14x-1+x-2+x-4 dx⇒I=∫14x-1 dx+∫14x-2 dx+∫14x-4 dxWe know that, x-1=-x-1 , x≤1x-1, 1<x≤4x-2=-x-2 , 1≤x≤2x-2, 2<x≤4x-4=-x-4 , 1≤x≤4x-4, x>4∴I=∫14x-1 dx-∫12x-2 dx+∫24x-2 dx-∫14x-4 dx⇒I=x22-x14-x22-2×12+x22-2×24-x22-4×14⇒I=8-4-12+1-2-4-12+2+8-8-2+4-8-16-12+4⇒I=232

(xvii)

I=∫-50x+x+2+x+5 dx⇒I=∫-50x dx+∫-50x+2 dx+∫-50x+5 dxWe know that, x=-x , -5≤x≤0x, x>0x+2=-x+2 , -5≤x≤-2x+2, -2<x≤0x+5=-x+5 , -5≤x≤0x+5, x>-5∴I=-∫-50x dx-∫-5-2x+2 dx+∫-20x+2 dx+∫-50x+5 dx⇒I=-x22-50-x22+2x-5-2+x22+2x-20+x22+5x-50⇒I=252-2-4-252+10-2+4+-252+25⇒I=632

(xviii)
I=∫04x+x-2+x-4 dx⇒I=∫04x dx+∫04x-2 dx+∫04x-4 dxWe know that, x=-x , -5≤x≤0x, x>0x-2=-x-2 , 0≤x≤2x-2, 2<x≤4x-4=-x-4 , 0≤x≤4x-4, x>4∴I=∫04x dx-∫02x-2 dx+∫24x-2 dx-∫04x-4 dx⇒I=x2204-x22-2×02+x22-2×24-x22-4×04⇒I=8-2-4+8-8-2+4-8-16⇒I=20

Q3.

Answer :

Let I=∫0π211+tanxdx … (i)=∫0π211+tanπ2-xdx Using ∫0afxdx= ∫0afa-xdx=∫0π211+cotxdx … (ii)Adding (i) and (ii) 2I =∫0π211+tanx+11+cotxdx =∫0π21+cotx+1+tanx1+tanx1+cotx dx =∫0π22+tanx+cotx1+tanx+cotx+tanxcotxdx =∫0π22+tanx+cotx2+tanx+cotxdx =∫0π2 dx =x0π2 2I =π2∴ I =π4

Q4.

Answer :

Let I=∫0π211+cotxdx … (i)= ∫0π211+cotπ2-xdx Using ∫0afxdx=∫0afa-xdx=∫0π211+tanxdx … (ii) Adding (i) and (ii)2I=∫0π211+cotx+11+tanx dx =∫0π21+tanx+ 1+cotx1+cotx1+tanx dx =∫0π22+tanx+ cotx1+tanx +cotx + tanx cotxdx =∫0π22+tanx+ cotx2+tanx+ cotx dx =∫0π2dx = x0π2=π2Hence , I=π4

Q5.

Answer :

Let I=∫0π2cotxcotx+tanxdx …(i) =∫0 π2cotπ2-xcotπ2-x+tanπ2-xdx Using ∫0afx dx=∫0afa-x dx= ∫0π2tanxtanx+cotx dx …(ii) Adding (i) and (ii)2I=∫0π2cotxcotx+tanx+tanxtanx+cotx dx =∫0π2dx =x0π2 =π2Hence, I =π4

Q6.

Answer :

Let I=∫0π2sinnxsinnx+ cosnxdx … (i)= ∫0π2sinnπ2-xsinnπ2-x+ cosnπ2-xdx Using ∫0afx dx=∫0afa-x dx= ∫0π2cosnxcosnx+ sinnx dx = ∫0π2cosnxsinnx+ cosnx dx … (ii)Adding (i) and (ii) we get2I =∫0π2sinnxsinnx+ cosnx+cosnxsinnx+ cosnx dx =∫0π2sinnx+ cosnxsinnx+ cosnx dx= ∫0π2 dx =x0π2=π2Hence I=π4i.e.,∫0π2sinnxsinnx+ cosnxdx=π4∴∫0π2sin3/2xsin3/2x+ cos3/2xdx=π4

Q7.

Answer :

Let I =∫0π2sinnxsinnx+ cosnxdx … (i) =∫0π2sinnπ2-xsinnπ2-x +cosnπ2-x dx=∫0π2 cosnx cosnx+sinnx dx =∫0π2 cosnx sinnx+ cosnx dx … (ii)Adding (i) and (ii) we get2I =∫0π2sinnxsinnx+ cosnx+ cosnx sinnx+ cosnxdx =∫0π2sinnx + cosnx sinnx+ cosnx dx =∫0π21 dx =∫0π2 dx=x0π2=π2Hence I =π4

Page 20.75 Ex.20.3

Q8.

Answer :

Let I = ∫0π211+tanxdx …(i)= ∫0π211+tanπ2-xdx Using ∫0afx dx=∫0afa-x dx=∫0π211+cotxdx …(ii) Adding (i) and (ii) we get2I = ∫0π211+tanx+11+cotxdx =∫0π21+cotx+1+tanx1+tanx 1+cotx dx =∫0π21+cotx+1+tanx1+cotx+tanx+tanx cotx dx =∫0π22+cotx+tanx 2+cotx+tanx dx= ∫0π2 dx =x0π2=π2Hence I = π4

Q9.

Answer :

We have, I=∫0a1x+a2-x2dxPutting x=a sin θ⇒dx=a cos θ dθWhen x→0; θ→0 And x→a; θ→π2∴I=∫0π2a cos θ a sin θ+a2-a sin θ2dθ=∫0π2a cos θ a sin θ+a cos θdθI=∫0π2cos θ sin θ+cos θdθ …..1⇒I=∫0π2cos π2-θ sin π2-θ +cos π2-θ dθ=∫0π2sin θcos θ+sin θdθI=∫0π2sin θ sin θ+cos θdθ …..2By adding 1 and 2, we get2I=∫0π2cos θ +sin θsin θ+cos θdθ ⇒2I=∫0π2dθ ⇒2I=θ0π2⇒2I=π2⇒I=π4

Q10.

Answer :

We have,I=∫0∞log x1+x2 dxPutting x=tan θ⇒dx=sec2θ dθWhen x→0 ; θ→0and x→∞ ; θ→π2Now, integral becomes,

I=∫0π2log tan θ1+tan2 θ sec2θ dθ⇒I=∫0π2log tan θ dθ …..1⇒I=∫0π2log tan π2-θ dθ ∵∫0afxdx=∫0afa-xdx⇒I=∫0π2log cot θ dθ …..2Adding 1 and 2, we get

2I=∫0π2log tan θ dθ+∫0π2log cot θ dθ=∫0π2log tan θ+log cot θ dθ=∫0π2log tan θ×cot θ dθ=∫0π2log 1 dθ=∫0π20 dθ⇒2I=0⇒I=0∴∫0∞log x1+x2 dx=0

Q11.

Answer :

We have,I=∫01log 1+x1+x2 dxPutting x=tan θ⇒dx=sec2 θ dθWhen x→0 ; θ→0and x→1 ; θ→π4Now, integral becomes

I=∫0π4log 1+tan θsec2 θ sec2 θ dθ⇒I=∫0π4log 1+tan θ dθ …..1⇒I=∫0π4log1+tan π4-θ dθ ∵∫0afxdx=∫0afa-xdx=∫0π4log1+tanπ4-tan θ1+tanπ4 tan θ dθ=∫0π4log1+1-tan θ1+tan θ dθ=∫0π4log21+tan θ dθI=∫0π4log 2-log 1+tan θ dθ …..2

Adding 1 and 2, we get2I=∫0π4log 2 dθ⇒2I=log 2θ0π4⇒2I=π4log 2⇒I=π8log 2∴∫01log1+x1+x2dx=π8log 2

Q12.

Answer :

We have,I=∫0∞x1+x1+x2 dxPutting x=tan θ⇒dx=sec2θ dθWhen x→0 ; θ→0and x→∞ ; θ→π2Now, integral becomes

I=∫0π2tan θ1+tan θ sec2θ sec2θ dθ=∫0π2tan θ1+tan θ dθ=∫0π2sin θcos θ1+sin θcos θdθ⇒I=∫0π2sin θsin θ+cos θdθ …..1⇒I=∫0π2sinπ2-θsinπ2-θ+cosπ2-θdθ ∵∫0afxdx=∫0afa-xdx⇒I=∫0π2cos θcos θ+sin θdθ⇒I=∫0π2cosθsinθ+cosθdθ …..2

Adding 1 and 2, we get2I=∫0π2sinθ+cosθsinθ+cosθ dθ⇒2I=∫0π2dθ⇒2I=π2⇒I=π4∴∫0∞x1+x1+x2 dx=π4

Q13.

Answer :

Let I =∫0πx tanxsecx cosecxdx …(i)=∫0ππ-x tanπ-xsecπ-x cosecπ-xdx Using ∫0afxdx=∫0afa-xdx=∫0π-π-xtanx-secx cosecxdx=∫0ππ-xtanxsecx cosecxdx …(ii)Adding (i) and (ii)2I=∫0πx tanxsecx cosecx+π-xtanxsecx cosecxdx =∫0πx+π-x tanxsecx cosecxdx =∫0π πtanxsecx cosecxdx =∫0ππsin2x dx =π∫0π1- cos2x dx =π x0π-π2∫0π1+cos2x dx =π2×0π-π2sin2x20π =π22Hence, I = π24

Q14.

Answer :

Let I =∫0πx sinx cos4x dx …(i) =∫0ππ-x sinπ-x cos4π-x dx =∫0ππ-x sinx cos4x dx …(ii) Adding (i) and (ii) we get2I= ∫0πx+π-x sinx cos4x dx = π∫0π sinx cos4x dx Let cos x= t, Then -sinx dx = dt, When x=0, t=1, x=π, t=-1Therefore, 2I=-π∫1-1t4 dt =π∫-11t4 dt =πt55-11 =π5+π5 =2π5Hence I =π5

Q15.

Answer :

Let I=∫0πx sin3x dx …(i) =∫0ππ-x sin3π-x dx =∫0ππ-x sin3x dx …(ii)Adding (i) and (ii) we get2I=∫0πx+π-xsin3x dx =∫0ππ sin3x dx =∫0ππ 3 sin x -sin 3×4 dx =π4∫0π 3 sin x -sin 3x dx =π4-3 cos x+cos 3×30π =π4-3 cos π+3cos 0+cos 3π3-cos 03 =π43+3+-13-13 =π23-13 =π2×83 =4π3∴I = 2π3

Q16.

Answer :

∫0πx log sinxdxLet I =∫0πx logsinxdx …..(i) I=∫0ππ-x log sinπ-x dx I=∫0ππ-x logsin x dx …..(ii)Adding (i) and (ii)2I=π∫0π log sinx dx =2π∫0π2 log sinx dx I=π∫0π2log sinx dx …..(iii)Let∫0π2log sinx dx= I2I2=∫0π2log sinπ2-x dx =∫0π2log cosx dx2I2=∫0π2log sinx+log cosx dx=∫0π2logsinx cosx dx=∫0π2logsin2x dx-∫0π2log 2 dxLet 2x=t2dx=dtwhen,x=0⇒t=0x=0 ⇒ t=π2I2=12∫0πlog sint dt-π2log 22I2=22∫0π2log sint dt-π2log 22I2=I2-π2log 2I2=-π2log 2From iii, I=π∫0π2log sinx dx=π I2I=π-π2log 2I=-π2 log 22

Q17.

Answer :

Let I =∫0πx sinx1+ sinxdx … (i) =∫0ππ-xsinπ-x1+sinπ-x dx =∫0ππ-x sinx1+ sinxdx … (ii)Adding (i) and (ii) we get 2I=∫0πx+π-x sinx1+ sinxdx =∫0ππ sinx1+ sinxdx =π∫0π1+sinx-11+sinxdx =π∫0πdx-π∫0π11+sinxdx =π∫0πdx-π∫0π1-sinx1+sinx1-sinxdx =π∫0πdx-π∫0π1-sinx1-sin2xdx =π∫0πdx-π∫0π1-sinxcos2xdx =π∫0πdx-π∫0πsec2x-secx tanxdx =πx0π-πtanx-secx0π =π2-π0+1-0+1 =π2-2πHence I=ππ2-1

Q18.

Answer :

We have, I =∫0πx1+cosα sinxdx …..1 =∫0ππ-x1+cosα sinπ-xdx =∫0ππ-x1+cosα sinxdx …..2Adding 1 and 2 we get,2I=∫0πx+π-x1+cosα sinxdx⇒I=π2∫0π11+cosα sinx dx

= π2∫0π11+cosα sinx=π2∫0π11+cosα 2tanx21+tan2x2dx=π2∫0π1+tan2x21+tan2x2+2cosα tan x2dx=π2∫0πsec2x21+tan2x2+2cosα tan x2dx
Putting tanx2=t⇒12sec2x dx=dtWhen x→0; t→0and x→π; t→∞∴I=π2∫0∞21+t2+2cosα tdt=π2∫0∞2t+cosα2-cos2α+1dt=π∫0∞1t+cosα2+sin2αdt=π1sin αtan-1t+cos αsin α01=πsinαtan-1∞-tan-1cotα=πsinαπ2-tan-1tanπ2-α=παsinα

Q19.

Answer :

Let I=∫0πx cos2x dx … (i) =∫0ππ-x cos2π-x dx =∫0ππ-x cos2x dx … (ii)Adding (i) and (ii) we get2I=∫0πx+π-x cos2x dx =∫0ππ cos2x dx =π∫0π1+cos2x2 dx =π2∫0π1+cos2x dx =π2x+sin2x20π =π2π-0 Hence I=π

Q20.

Answer :

Let I =∫0π22 log cosx-logsin2xdx =∫0π22 log cosx-log2sinx cosxdx =∫0π22logcosx-log2-logsinx-logcosxdx =∫0π2logcosx-log2-logsinxdx =∫0π2logcosx dx-∫0π2log2 dx -∫0π2logsinx dx =∫0π2logcosx dx-∫0π2log2 dx -∫0π2logsinπ2-x dx Using∫0afx dx=∫0afa-x dx =∫0π2logcosx dx-∫0π2log2 dx-∫0π2logcosx dx =-log2 x0π2 =-π2 log2

Q21.

Answer :

Let I=∫0πsin2x cos3x dx =∫0πsin2x cos2x cos x dx =∫0π sin2x1-sin2x cos x dx =∫0πsin2x -sin4x cos x dxLet, sin x=t ⇒cos x dx = dtWhen x→0 ; t→0 and x→π ; t→0∴ I=∫00t2 -t4 dt =0 ∵ ∫aaf(x)dx = 0

Q22.

Answer :

Let I=∫0π2x sinx cosxsin4x+cos4xdx …(i) =∫0π2π2-x sinπ2-x cosπ2-xsin4π2-x+cos4π2-xdx =∫0π2π2-xcosx sinxcos4x+sin4x dx =∫0π2π2-xsinx cosxsin4x+cos4x dx … (ii)Adding (i) and (ii) we get2I =∫0π2x+π2-xsinx cosxsin4x+cos4xdx =π2∫0π2sinx cosxsin4x+cos4xdxLet sin2x=t, Then 2 sinx cosx dx=dtWhen x=0, t=0, x=π2, t=1Therefore2I =π4∫01dtt2+1-t2 =π8∫01dtt-122+14 =π8×2tan-12t-101 =π4π4+π4Hence I=π216

Q23.

Answer :

Let I=∫-π2π2sin3x dx =∫-π2π2 sinx1-cos2xdx =∫-π2π2sinx dx-∫-π2π2sinx cos2x dx =-cosx-π2π2+cos3x3-π2π2 =0 +0 =0

Q24.

Answer :

Let I= ∫-π2π2sin4x dx =∫-π2π2sin2x2 dx =∫-π2π21-cos2x22 dx =14∫-π2π21-2cos2x+cos22x dx =14∫-π2π2dx-12∫-π2π2cos2x dx+18∫-π2π21+cos4x dx =14∫-π2π2dx-12∫-π2π2cos2x dx+18∫-π2π2dx+18∫-π2π2cos4x dx =38∫-π2π2dx-12∫-π2π2cos2x dx+18∫-π2π2cos4x dx =38x-π2π2-14sin2x-π2π2+132sin4x-π2π2 =38π2+π2-140-0+1320-0Hence I=3π8

Q25.

Answer :

Let I=∫-11log2-x2+xdxHere fx=log2-x2+xf-x=log2+x2-x=-log2-x2+x=-fxHence fx is an odd function,Therefore,I=0

Q26.

Answer :

Let I=∫-π4π4sin2x dxHere fx = sin2xf-x=sin2-x=sin2x =fxHence sin2x is an even functionTherefore,I=2∫0π4sin2x dx =2∫0π41-cos2x2dx =∫0π41-cos2x dx =x-sin2x20π4 =π4-12→

Q27.

Answer :

Let, I=∫0πlog1-cosx dx =∫0πlog2sin2x2 dx =∫0πlog2 dx+ 2∫0πlog sinx2 dx Let, t =x2 in the secong integral. then dt= 12dxWhen x→0 ; t→0 and x→π ; t→π2I=log2x0π+4∫0π2log sint dt =πlog2+4×-π2log2 Where, ∫0π2log sint dt=-π2log2 =-π log2

Q28.

Answer :

Let I=∫-π2π2log2-sinx2+sinxdxHere, fx=log2-sinx2+sinxf-x=log2-sin-x2+sin-x=log2+sinx2-sinx=-log2-sinx2+sinx=-fxHence fx is an odd function∴ I =0

Q29.

Answer :

Let I=∫0axx+a-xdx … (i) ⇒ I=∫0aa-xa-x+xdx Using, ∫0afx dx=∫0afa-x dx ⇒ I=∫0aa-xx+a-xdx … (ii)Adding (i) and (ii)2I=∫0ax+a-xx+a-x dx =∫0a dx=x0a=aHence I=a2

Q30.

Answer :

Let I=∫05x+44x+44-9-x4dx … (i)I=∫059-x49-x4-x+44dx Using ∫0afxdx=∫0afa-xdxI=-∫059-x4x+44-9-x4dx … (ii)Adding (i) and (ii)2I=∫05x+44x+44-9-x4-9-x4x+44-9-x4dx =∫05x+44-9-x4x+44-9-x4dx =∫05dx =x05 =5Hence I= 52

Q31.

Answer :

Let I=∫07x3x3+7-x3dx … (i) =∫077-x37-x3+x3 dx Using ∫0afx dx=∫0afa-x dx =∫077-x3x3+7-x3 dx … (ii) Adding (i) and (ii) we get2I=∫07×3+7-x3x3+7-x3dx = ∫07dx =x07=7Hence I =72

Q32.

Answer :

Let I =∫π6π311+tanxdx … (i) =∫π6π311+tanπ3+π6-xdx =∫π6π311+cotxdx … (ii)Adding (i) and (ii)2I=∫π6π311+tanx+11+cotx dx =∫π6π31+cotx+1+tanx1+cotx+tanx+tanx cotx dx =∫π6π32+cotx+tanx2+cotx+tanx dx =∫π6π3 dx =xπ6π3 =π3-π6∴2I=π6Hence I=π12

Q33.

Answer :

Let I=∫abfxfa+b-x+fxdx … (i) =∫abfa+b-xfx+fa+b-xdx … (ii)Adding (i) and (ii) we get2I=∫abfx+fa+b-xfa+b-x+fxdx =∫ab dx =xab=b-aHence I=b-a2

Q34.

Answer :

Let I=∫02×2-xdx =∫022-x2-2+xdx =∫022-xxdx =∫022x-xx dx =∫022×12-x32 dx =2×3232-x525202 =43×32-25×5202 =823-825=16215

Q35.

Answer :

Let I =∫01log1x-1dx … (i) =∫01log11-x-1dx Using ∫0af(x) dx = ∫0af(a-x) dx I =∫01logx1-x dx … (ii)Adding (i) and (ii)2I=∫01log1-xx+logx1-x dx =∫01 log1-xx×x1-x dx =∫01log1 dx =0Hence I=0

Q36.

Answer :

Let I=∫02afxdxBy Additive propertyI=∫0afxdx+∫a2afxdxConsider the integral ∫a2afxdxLet x=2a-t, then dx=-dtWhen x=a, t=a, x=2x, t=0Hence ∫a2afxdx=-∫a0f2a-tdt =∫0af2a-tdt =∫0af2a-xdx Changeing the variableTherefore,I=∫0afxdx+∫0af2a-xdx =∫0afxdx+∫0afxdx Given ∫0afxdx=∫0af2a-xdx =2∫0afxdx

Hence Proved

Q37.

Answer :

Let I=∫02afxdxUsing additive propertyI=∫0afxdx+∫a2afxdxConsider the integral ∫a2afxdxLet x=2a-t, Then dx=-dtWhen x=a, t=a and x=2a, t=0Therefore,∫a2afxdx=-∫a0f2a-tdt =∫0af2a-tdt =∫0af2a-xdx changing the variableWe have f2a-x=-fxTherefore,I=∫0afxdx-∫0afxdx =0

Q38.

Answer :

(i)
I=∫-aafx2dxHere gx=f(x2)⇒g-x=f-x2=f(x2)=gx i.e, gx is even ThereforeI=2∫0afx2dx Using ∫-aagxdx=2∫0agxdx when gx is even
(ii)
I=∫-aaxfx2dxLet gx=xfx2⇒g-x=-xf-x2=-xfx2=-gx i.e, gx is odd ThereforeI=0 Using ∫-aagxdx=0 when gx is odd

Q39.

Answer :

Let I=∫02afxdxBy Additive propertyI=∫0afxdx+∫a2afxdxConsider the integral ∫a2afxdxLet x=2a-t, then dx=-dtWhen x=a. t=a, x=2a, t=0Hence, ∫a2afxdx=-∫a0f2a-tdt =∫0af2a-tdt=∫0af2a-xdx ThereforeI=∫0afxdx+∫0af2a-xdx =∫0afx+f2a-x dxHence, proved.

Page 20.76 Ex.20.3

Q40.

Answer :

Let I=∫-aafxdxBy Additive propertyI=∫-a0fxdx+∫0afxdxLet x=-t, then dx = -dt,When x=-a, t=a, x=0, t=0Hence ∫-a0fxdx=-∫a0f-tdt =∫0af-tdt =∫0af-xdx Changing the variableTherefore,I=∫0af-xdx+∫0afxdx =∫0afx+f-x dxHence, proved.

Page 20.90 Ex.20.4

Q1.

Answer :
∫abfxdx=limh→0 hfa+fa+h+fa+2h…………+fa+n-1h, where, h=b-an

Here, a=0, b=3, fx=x+4, h=3-0n=3nTherefore, I=∫03x+4dx =limh→0 hf0+f0+h+…….+f0+n-1h =limh→0 h0+4+h+4+…….+n-1h+4 =limh→0 h4n+h1+2+…….+n-1 =limh→0 h4n+hnn-12 =limn→∞ 3n4n+3n×nn-12 =limn→∞12+921-1n =12+92=332

Q2.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x+3, h=2-0n=2nTherefore,I=∫02x+3dx =limh→0 hf0+f0+h+………………..+f0+n-1h =limh→0 h0+3+0+h+3+……………+0+n-1h+3 =limh→0 h3n+h1+2+3………+n-1 =limh→0 h3n+hnn-12 =limn→∞ 2n3n+n-1 =limn→∞24-1n =8

Q3.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an
Here a=1, b=3, fx=3x-2, h=3-1n=2nTherefore,I=∫133x-2dx=limh→0 hf1+f1+h+………………..+f1+n-1h=limh→0 h3-2+3+3h-2+3+6h-2……………+3n-1h+3-2=limh→0 hn+3h1+2+3………+n-1=limh→0 hn+3hnn-12=limn→∞ 2nn+3n-3=limn→∞24-3n =8

Q4.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=-1, b=1, fx=x+3, h=1+1n=2nTherefore,I=∫-11x+3dx=limh→0 hf-1+f-1+h+………………..+f-1+n-1h=limh→0 h-1+3+-1+h+3+……………+-1+n-1h+3=limh→0 h2n+h1+2+3………+n-1=limh→0 h2n+hnn-12=limn→∞ 2n2n+n-1=limn→∞23-1n=6

Page 20.91 Ex.20.4

Q5.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here, a=0, b=5, fx=x+1, h=5-0n=5nTherefore,I=∫05x+1dx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+1+h+1+……………+n-1h+1=limh→0 hn+h1+2+3………+n-1=limh→0 hn+hnn-12=limn→∞ 5nn+5n-52=limn→∞572-5n=352

Q6.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here, a=1, b=3, fx=2x+3, h=3-1n=2nTherefore,I=∫132x+3dx=limh→0 hf1+f1+h+………………..+f1+n-1h=limh→0 h2+3+2+2h+3+……………+2+2n-1h+3=limh→0 h5n+2h1+2+3………+n-1=limh→0 h5n+2hnn-12=limn→∞ 2n5n+2n-2=limn→∞27-2n=14

Q7.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h+…+fa+n-1hwhere h=b-an

Here a=3, b=5, fx=2-x, h=5-3n=2nTherefore,I=∫352-xdx =limh→0 hf2+f2+h+…+f2+n-1h =limh→0 h2-2+2-h-2+…+2-n-1h-2 =limh→0 h-h1+2+3+…+n-1 =limh→0 h-2hnn-12 =limn→∞ 2n-2n+2 =limn→∞2-2+2n =-4

Q8.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+1, h=2-0n=2nTherefore,I=∫02×2+1dx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+1+h2+1+……………+n-12h2+1=limh→0 hn+h212+22+32………+n-12=limh→0 hn+h2nn-12n-16=limn→∞ 2nn+2n-12n-13n=limn→∞21+231-1n2-1n=2+83=143

Q9.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=1, b=2, fx=x2, h=2-1n=1nTherefore,I=∫12x2dx=limh→0 hf1+f1+h+………………..+f1+n-1h=limh→0 h1+h+12+……………+n-1h+12=limh→0 hn+h212+22+32………+n-12+2h1+2+3+………..+n-1=limh→0 hn+h2nn-12n-16+2hnn-12=limn→∞ 1nn+n-12n-16n+n-1=limn→∞2+161-1n2-1n-1n=2+13=73

Q10.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=2, b=3, fx=2×2+1, h=3-2n=1nTherefore,I=∫232×2+1dx=limh→0 hf2+f2+h+………………..+f2+n-1h=limh→0 h22.22+1+22+h2+1+……………+22+n-1h2+1=limh→0 hn+222+2+h2+………….2+n-1h2=limh→0 hn+8n+2h212+22+32………+n-12+8h1+2+…….+n-1=limh→0 h9n+h22nn-12n-16+8hnn-12=limn→∞ 1n9n+n-12n-13n+4n-4=limn→∞13+131-1n2-1n-4n =13+23=413

Q11.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=1, b=2, fx=x2-1, h=2-1n=1nTherefore,I=∫12×2-1dx=limh→0 hf1+f1+h+………………..+f1+n-1h=limh→0 h1-1+h2-1+……………+n-12h2-1=limh→0 hn-1+h212+22+32………+n-12=limh→0 hn-1+h2nn-12n-16=limn→∞ 1nn-1+n-12n-16n=limn→∞1-1n+161-1n2-1n=1+13=43

Q12.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+4, h=2-0n=2nTherefore,I=∫02×2+4dx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+4+h2+4+……………+n-12h2+4=limh→0 h4n+h212+22+32………+n-12=limh→0 h4n+h2nn-12n-16=limn→∞ 2n4n+2n-12n-13n=limn→∞24+231-1n2-1n=8+83=323

Q13.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=1, b=4, fx=x2-x, h=4-1n=3nTherefore,I=∫14×2-xdx=limh→0 hf1+f1+h+………………..+f1+n-1h=limh→0 h1-1+1+h2-1+h+……………+1+n-1h2-1+n-1h=limh→0 hh212+22+32………+n-12+1+2h1+2+……+n-1-n-h1+2+…..+n-1=limh→0 hh2nn-12n-16+hn-12=limn→∞ 3n9n-12n-16n+3n-12=limn→∞3321-1n2-1n+321-1n=9+92=272

Q14.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=1, fx=3×2+5x, h=1-0n=1nTherefore,I=∫013×2+5xdx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+0+3h2+5h+……………+3n-12h2+5n-1h=limh→0 h5h1+2+………+n+3h212+22+32………+n-12=limh→0 h5hnn-12+h23nn-12n-16=limn→∞ 1n5n-12+n-12n-12n=limn→∞521-1n+121-1n2-1n=52+1=72

Q15.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=ex, h=2-0n=2nTherefore,I=∫02exdx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 he0+eh+e2h+…….+en-1h=limh→0 hehn-1eh-1=limh→0 e2-1eh-1h=e2-11=e2-1

Q16.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=a, b=b, fx=ex, h=b-anTherefore,I=∫abexdx=limh→0 hfa+fa+h+………………..+fa+n-1h=limh→0 hea+ea+h+…………+ea+n-1h=limh→0 heaehn-1eh-1=limh→0 heaeb-a-1eh-1=limh→0eb-eaeh-1h=eb-ea1=eb-ea

Q17.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h+…+fa+n-1hwhere h=b-an
Here a=a, b=b, fx=cos x, h=b-anTherefore,I=∫abcos x dx=limh→0 hfa+fa+h+…+fa+n-1h=limh→0 hcosa+cosa+h+…+cosa+n-1h=limh→0 hcosa+n-1h2sinnh2sinh2=limh→0h2sinh22cosa+b-a2-h2 sinb-a2 Using nh=b-a=limh→0h2sinh2×limh→02cosa+b2-h2sinb-a2=2cosa+b2sinb-a2=sin b-sin a Since, 2cosA sinB=sinA+B-sinA-B

Q18.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h+…+fa+n-1hwhere h=b-an
Here a=0, b=π2, fx=sinx, h=π2-0n=π2nTherefore,I=∫0π2sinxdx =limh→0 hf0+f0+h+…+f0+n-1h =limh→0 hsin0+sinh+sin2h+…+sinn-1h =limh→0 hsinn-1h2sinnh2sinh2 =limh→0 h2sinh2×2sinπ4-h2sinπ4 Using nh=π2 =limh→0h2sinh2×limh→02sinπ4-h2sinπ4 =2sinπ4sinπ4=2×12×12=1

Q19.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h+…+fa+n-1hwhere h=b-an
Here a=0, b=π2, fx=cosx, h=π2-0n=π2nTherefore,I=∫0π2cosx dx =limh→0 hf0+f0+h+…+f0+n-1h =limh→0 hcos0+cosh+cos2h+…+cosn-1h =limh→0 hcosn-1h2sinnh2sinh2 =limh→0 hcosπ4-h2sinπ4sinh2 Using , nh=π2 =limh→0h2sinh2×2cosπ4-h2sinπ4 =limh→0h2sinh2×limh→02cosπ4-h2sinπ4 =2cosπ4 sinπ4=2×12×12=1

Q20.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=1, b=4, fx=3×2+2x, h=4-1n=3nTherefore,I=∫143×2+2xdx=limh→0 hf1+f1+h+………………..+f1+n-1h=limh→0 h3.12+2×1+31+h2+21+h+……………+31+n-1h2+21+n-1h=limh→0 h312+1+h2+1+2h2+………..+1+n-1h2+21+1+h+……….+1+n-1h=limh→0 h3n+3h212+22+32………+n-12+6h1+2+………n-1h+2n+2h1+2+……….+n-1h=limh→0 h5n+3h2nn-12n-16+8hnn-12=limn→∞ 3n5n+9n-12n-12n+12n-12=limn→∞317-12n+921-1n2-1n=51+27=78

Q21.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=3×2-2, h=2-0n=2nTherefore,I=∫023×2-2dx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0-2+3h2-2+……………+3n-12h2-2=limh→0 h-2n+3h212+22+32………+n-12=limh→0 h-2n+3h2nn-12n-16=limn→∞ 2n-2n+2n-12n-1n=limn→∞2-2+21-1n2-1n=-4+8=4

Q22.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+2, h=2-0n=2nTherefore,I=∫02×2+2dx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+2+h2+2+……………+n-12h2+2=limh→0 h2n+h212+22+32………+n-12=limh→0 h2n+h2nn-12n-16=limn→∞ 2n2n+2n-12n-13n=limn→∞22+231-1n2-1n=4+83=203

Q23.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here, a=0, b=4, fx=x+e2x, h=4-0n=4nTherefore,I=∫04x+e2xdx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+e0+h+e2h+……………+n-1h+e2n-1h=limh→0 hh1+2+…..+n-1h+e0+e2h+e4h+………+e2n-1h=limh→0 hhnn-12+e2hn-1e2h-1=limn→∞ 16n2×nn-12+limh→0 e8-1e2h-1h=limn→∞81-1n+limh→0 e8-12(e2h-1)2h=8+e8-12=15+e82

Q24.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+x, h=2-0n=2nTherefore,I=∫02×2+xdx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+0+h2+h+……………+n-12h2+h=limh→0 hh212+22+32………+n-12+h1+2+3……..+n-1h=limh→0 hh2nn-12n-16+hnn-12=limn→∞ 2n2n-12n-13n+n-1=limn→∞2231-1n2-1n+1-1n=83+2=143

Q25.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2+2x+1, h=2-0n=2nTherefore,I=∫02×2+2x+1dx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+0+1+h2+2h+1+……………+n-12h2+2n-1h+1=limh→0 hn+h212+22+32………+n-12+2h1+2+………+n-1h=limh→0 hn+h2nn-12n-16+2hnn-12=limn→∞ 2nn+2n-12n-13n+2n-2=limn→∞23+231-1n2-1n-2n =6+83=263

Q26.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=3, fx=2×2+3x+5, h=3-0n=3nTherefore,I=∫032×2+3x+5dx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+0+5+2h2+3h+5+……………+2n-12h2+3n-1h+5=limh→0 h5n+2h212+22+32………+n-12+3h1+2+…….+n-1h=limh→0 h5n+2h2nn-12n-16+3hnn-12=limn→∞ 3n5n+3n-12n-1n+9n-12=limn→∞35+31-1n2-1n+921-1n=15+18+272=933

Q27.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=a, b=b, fx=x, h=b-anTherefore,I=∫abx dx=limh→0 hfa+fa+h+………………..+fa+n-1h=limh→0 ha+a+h+ a+2h+……….+a+n-1h=limh→0 hna+h1+2+3+……..+n-1=limh→0 hna+hnn-12=limh→0 b-anna+b-an-12=limh→0b-aa+b-ab-a-h2=b-aa+b-a22=2ab-2a2+b2+a2-2ab2=b2-a22

Q28.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=5, fx=x+1, h=5-0n=5nTherefore,I=∫05x+1dx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0+1+h+1+……………+n-1h+1=limh→0 hn+h1+2+3+……………..+n-1h=limh→0 hn+hnn-12=limn→∞ 5nn+5n-12=limn→∞51+521-1n=5+252=352

Q29.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=2, b=3, fx=x2, h=3-2n=1nTherefore,I=∫23×2 dx=limh→0 hf2+f2+h+………………..+f2+n-1h=limh→0 h22+2+h2+………..+2+n-1h2=limh→0 h4n+h212+22+32………+n-12+4h1+2+……+n-1h =limh→0 h4n+h2nn-12n-16+4hnn-12=limn→∞ 1n4n+n-12n-16n+2n-2=limn→∞6+161-1n2-1n-2n =6+13=193

Q30.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=1, b=4, fx=x2-x, h=4-1n=3nTherefore,I=∫14×2-xdx=limh→0 hf1+f1+h+………………..+f1+n-1h=limh→0 h1-1+1+h2-1+h+……………+n-1h+12-n-1h+1=limh→0 hh212+22+32………+n-12-h1+2+…….+n-1=limh→0 hh2nn-12n-16-hnn-12=limn→∞ 3n3n-12n-12n+3n-12=limn→∞3321-1n2-1n+321-1n=9+93=383

Q31.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here a=0, b=2, fx=x2-x, h=2-0n=2nTherefore,I=∫02×2-xdx=limh→0 hf0+f0+h+………………..+f0+n-1h=limh→0 h0-0+h2-h+……………+n-12h2-n-1h=limh→0 hh212+22+32………+n-12-h1+2+…..+n-1h=limh→0 hh2nn-12n-16-hnn-12=limn→∞ 2n2n-12n-13n-n+1=limn→∞2231-1n2-1n-1+1n=83-2=23

Q32.

Answer :

∫abfxdx=limh→0 hfa+fa+h+fa+2h……………+fa+n-1hwhere h=b-an

Here, a=1, b=3, fx=2×2+5x, h=3-1n=2nTherefore,I=∫132×2+5xdx=limh→0 hf1+f1+h+………………..+f1+n-1h=limh→0 h2+5+21+h2+51+h+……………+21+n-1h2+51+n-1h=limh→0 h212+1+h2+…………+1+n-1h2+51+1+h+1+2h+……..+1+n-1h=limh→0 h2n+2h212+22+32………+n-12+4h1+2+……+n-1+5n+5h1+2+……+n-1=limh→0 h7n+2h2nn-12n-16+9hnn-12=limn→∞ 2n7n+4n-12n-13n+9n-9=limn→∞216+431-1n2-1n-9n=32+163=1123

Page 20.95 (Very Short Answers)

Q1.

Answer :

∫0π2sin2x dx=∫0π21-cos2x2 dx=12∫0π21-cos2x dx=12x-sin2x20π2=12π2-0=π4

Q2.

Answer :

∫0π2cos2x dx=∫0π21+cos2x2 dx=12∫0π21+cos2x dx=12x+sin2x20π2=12π2+0=π4

Q3.

Answer :

∫-π2π2sin2x dx=∫-π2π21-cos2x2 dx=12∫-π2π21-cos2xdx=12x-sin2x2-π2π2=12π2-0+π2-0=π2

Q4.

Answer :

∫-π2π2cos2 xdx=∫-π2π21+cos2x2 dx=12∫-π2π21+cos2x dx=12x+sin2x2-π2π2=12π2+0+π2-0=π2

Q5.

Answer :

Let I=∫-π2π2sin3xdx=∫-π2π2sinx sin2x dx=∫-π2π2sinx1-cos2x dxLet cosx =t, then -sinx dx =dt,When, x→-π2 ; t→0 and x→π2 ; t→0I=∫00-1+t2 dt=0

Q6.

Answer :

We have,I=∫-π2π2x cos2x dxLet fx=x cos2x⇒ f-x=-x cos2-x=-x cos2x∴f-x=-fxi.e., fx is odd functionWe know that ∫-aafx dx=0 , if fx is odd function.∴I=∫-π2π2x cos2x dx=0

Q7.

Answer :
∫0π4tan2x dx=∫0π4sec2x-1 dx=tanx-x0π4=1-π4-0=1-π44430. →

Q8.

Answer :

∫0111+x2dx=tan-1×01=π4-0=π4ode is 4430. →

Q9.

Answer :

Let, I=∫-21xxdxWe have,x=x 0≤x≤1-x -2≤x<0∴xx=1 0≤x≤1-1 -2≤x<0Therefore,I=∫-20-1dx+∫01 1 dx =-x-20+x01 =0-2+1-0 =-1

Q10.

Answer :

∫0∞e-xdx=-e-x0∞=-0-1=0+1=1

Q11.

Answer :

∫04116-x2dx=∫04142-x2dx=sin-1×404=π2-0=π2s

Q12.

Answer :

∫031×2+9dx=∫031×2+32dx =13 tan-1×303=13tan-11- tan-10=13π4- 0=π12

Q13.

Answer :

∫0π21-cos2xdx=∫0π22sin2x dx =∫0π22 sinx dx=-2 cosx0π2=-0-2=2

Q14.

Answer :

Let, I=∫0π2log tanxdx … (i)=∫0π2log tanπ2-x dx Using, ∫0a fx dx=∫0a fa-x dx=∫0π2log cotx dx … (ii)Adding (i) and (ii) we get2I=∫0π2log tanx dx+∫0π2log cotx dx =∫0π2logtanx×cotxdx =∫0π2log1 dx=0Hence, I=0

Q15.

Answer :

Let, I=∫0π2log3+5cosx3+5sinxdx … (i)=∫0π2log3+5cosπ2-x3+5sinπ2-x dx=∫0π2log3+5sinx3+5cosx dx … (ii)Adding (i) and (ii)2I=∫0π2log3+5cosx3+5sinx+log3+5sinx3+5cosx dx =∫0π2log3+5cosx3+5sinx×3+5sinx3+5cosx dx =∫0π2log1 dx=0Hence I=0

Q16.

Answer :

Let I=∫0π2sinnxsinnx+cosnxdx …(i)=∫0π2sinnπ2-xsinnπ2-x+cosnπ2-xdx=∫0π2cosnxcosnx+sinnxdx=∫0π2cosnxsinnx+cosnxdx …(ii)Adding (i) and (ii)2I=∫0π2sinnxsinnx+cosnx+cosnxsinnx+cosnxdx = ∫0π2sinnx+cosnxsinnx+cosnx dx=∫0π2dx=x0π2=π2Hence I=π4

Q17.

Answer :

Let I=∫0πcos5x dx =∫0πcosxcos2x2 dx =∫0πcosx1-sin2x2 dx Let sinx =t, then cosx dx = dtWhen, x→0 ; t→0 and x→π ; t→0 Therefore,I=∫001-t22 dt =0

Q18.

Answer :

We have,I=∫02xdxWe know that,x=0, 0<x<11, 1<x<2∴I=∫02xdx=∫01xdx+∫12xdx=∫010dx+∫121dx=0+x12=2-1=1

Q19.

Answer :

We have,I=∫01.5x dx=∫01x dx+∫11.5x dx=∫010 dx+∫11.51dx ∵x=0 0≤x<11 1≤x<1.5=0+x11.5=1.5-1=0.5=12

Q20.

Answer :

We have,I=∫01x dxWe know x=x, 0<x<1∴I=∫01x dx=x2201=12-02=12
s 4430. →

Q21.

Answer :

We have,I=∫01exdxWe know that,x=x, when 0<x<1∴I=∫01exdx=ex01=e1-e0=e-1

Q22.

Answer :

We have,I=∫02xx dxWe know that,xx=x×0, 0<x<1x×1, 1<x<2i.e.,xx=0, 0<x<1x, 1<x<2∴I=∫02xx dx=∫01xx dx+∫12xx dx=∫010 dx+∫12x dx=0+x2212=222-122=42-12=32

Q23.

Answer :

We have,I=∫012x-x dx=∫012x-0 dx ∵x=0 where, 0<x<1=∫012x dx=2xloge201=21loge2-20loge2=2loge2-1loge2=1loge2

Q24.

Answer :

Let, I=∫-π2π2loga-sinθa+sinθdθHere, fθ=loga-sinθa+sinθConsider, f-θ=loga-sin-θa+sin-θ=-loga-sinθa+sinθ=-fθi.e., fθ is odd function.Therefore, I=0

Q25.

Answer :

x = -x , -1<x<0 x , 0<x<1∴xx = -x2 , -1<x<0 x2 , 0<x<1Now, ∫-11xxdx=∫-10- x2 dx+∫01 x2 dx=-∫-10 x2 dx+∫01 x2 dx=-x33-10+x3301=-0+13+13-0=0-13+13-0=0

Q26.

Answer :

We have,I=∫12loge x dxWe know that,x=1, when 1<x<2∴I=∫12loge 1 dxI=∫120 dx=0

Q27.

Answer :

Let I=∫abfxfx+fa+b-xdx … (i) =∫abfa+b-xfa+b-x+fa+b-a-b+xdx =∫abfa+b-xfa+b-x+fxdx∴ I=∫abfa+b-xfx+fa+b-xdx …(ii)Adding (i) and (ii) we get2I=∫abfxfx+fa+b-x+fa+b-xfx+fa+b-xdx =∫abfx+fa+b-xfx+fa+b-x dx =xab =b-aHence, I=b-a2

Q28.

Answer :

∫0111+x2dx=tan-1×01=tan-11-tan-10=π4-0=π4

Q29.

Answer :

∫0111+x2dx=tan-1×01=tan-11-tan-10=π4-0=π4

Q30.

Answer :

∫231xdx=logex23=loge3-loge2=loge32→

Q31.

Answer :

∫024-x2dx=∫0222-x2dx=x24-x2+12×22sin-1×202=x24-x202+2sin-1×202=0+2π2-0=π

Q32.

Answer :

We have,I=∫012×1+x2dxPutting 1+x2=t⇒2x dx=dtWhen x→0; t→1And x→1; t→2∴I=∫12dtt=loge t12=loge2-loge1=loge2-0=loge2 4430. →

Q33.

Answer :

We have,I=∫02×2 dx=∫01×2 dx+∫12×2 dx=∫010dx+∫121dx ∵x2=0 0< x<11 1<x<2=0+x12=2-1

Q34.

Answer :

We have,∫013×2+2x+kdx=0⇒x3+x2+kx01=0⇒1+1+k-0=0⇒k=-2. →

Q35.

Answer :

We have,I=∫0π/4sin x dxWe know that,x=x, when 0<x<π4 As π=3.14 ⇒ π4=0.785<1∴I=∫0π/4sin x dx=-cos x0π4=-cos π4-cos 0=cos 0-cos π4=1-12=2-12

Page 20.96 (Very Short Answers)

Q36.

Answer :

∫-33ax2+bx+cdx=ax33+bx22+cx-33=9a+92b+3c+9a-92b+3c=18a+6c

Hence, the given integral is independent of b

Q37.

Answer :

We have,∫0a3x2dx=8⇒3x330a=8⇒x30a=8⇒a3-0=8⇒a=83 =20. →

Page 20.96 (Multiple Choice Questions)

Q1.

Answer :

(d) π/8

Let, I=∫01×1-xdx =∫01x-x2dx =∫0114-x2-x+14dx =∫01122-x-122 dx =x-122x-x2+12×14sin-12x-101 =18 sin-11-sin-1-101=18π2+π2 =π8

Q2.

Answer :

(c) 2

∫0π11+sinxdx=∫0π11+sinx×1-sinx1-sinxdx=∫0π1-sinx1-sin2xdx=∫0π1-sinxcos2xdx =∫0πsec2x-secx tanx dx=tanx-secx0π=0+1-0+1=2

Q3.

Answer :

a π24π24

Let I=∫0πxtanxsecx+cosxdx (i)=∫0ππ-xtanπ-xsecπ-x+cosπ-xdx=∫0ππ-xtanxsecx+cosx dx (ii)Adding (i) and (ii)2I=∫0πxtanxsecx+cosx+π-xtanxsecx+cosxdx =∫0ππ tanxsecx+cosxdx =π∫0πsinx1+cos2x dx =-πtan-1cosx0π =-π-π4-π4=π22Hence I=π24We have, I=∫0πx tanxsecx+cosxdx …..1=∫0ππ-xtanπ-xsecπ-x+cosπ-xdx=∫0ππ-xtanxsecx+cosx dx …..2Adding 1 and 2, we get2I=∫0πxtanxsecx+cosx+π-xtanxsecx+cosxdx⇒I=12∫0ππ tanxsecx+cosxdx =π2∫0πsinx1+cos2x dxPutting cos x=t⇒-sinx dx=dt⇒sinx dx=-dtWhen x→0; t→1and x→π; t→-1⇒I=π2∫1-1-dt1+t2=π2∫-11dt1+t2=π2tan-1t-11=π2tan-11-tan-1-1=π2π4–π4=π2×π2=π24Hence I=π24

Q4.

Answer :

(c) 8

∫02π1+sinx2dx=∫02πsin2x4+cos2x4+2sinx4cosx4 dx=∫02πsinx4+cosx4dx=-cosx414+sinx41402π=4sinx4-cosx402π=4sin2π4-cos2π4-sin 0+cos 0=4sinπ2-cosπ2-0+1=41-0-0+1=4×2=8

Q5.

Answer :

(c) π/4

Let I=∫0π2cosxcosx+sinxdx … (i) =∫0π2cosπ2-xcosπ2-x+sinπ2-xdx = ∫0π2sinxsinx+cosxdx = ∫0π2sinxcosx+sinxdx … (ii)Adding (i) and (ii)2I=∫0π2cosxcosx+sinx+sinxcosx+sinxdx =∫0π2dx =x0π2=π2Hence I=π4

Q6.

Answer :

(b) log 2

We have,I=∫0∞11+exdxPutting ex=t⇒ exdx= dt⇒dx= dttWhen x→0; t→1and x→∞; t→∞∴I=∫1∞1t1+tdt=∫1∞1t+t2dt=∫1∞1t+122-122dt

=12×12logt+12-12t+12+121∞=logtt+11∞=logtttt+1t1∞=log11+1t1∞=log11+0-log11+1=log1-log12=0–log2=log2

Page 20.97 (Multiple Choice Questions)

Q7.

Answer :

(a) 2

∫0π24sinxxdxLet x=t, then12xdx=dtWhen x=0, t=0, x=π24, t=π2Therefore the integral becomes∫0π22 sint dt=-2cost0π2=2

Q8.

Answer :

(d) log43

Let, I= ∫0π2cosx2+sinx1+sinxdxLet sinx , then cosx dx =dtWhen x=0, t=0, x=π2, t=1Therefore the integral becomesI=∫01dt2+t1+t=∫01-12+t+11+t dt=-log2+t+log1+t01=log1+t-log2+t01=log2-log3-log1+log2=log43

Q9.

Answer :

b 23tan-113

We have,I=∫0π212+cosxdx=∫0π212+1-tan2x21+tan2x2dx=∫0π21+tan2x22+2 tan2x2+1-tan2x2dx=∫0π2sec2x23+tan2x2dx

Putting tan x2=t⇒12sec2x2dx=dt⇒sec2x2dx=2dtWhen, x→0; t→0and x→π2; t→1∴I=∫0123+t2dt=2∫01132+t2dt=23tan-1t301=23tan-113-tan-103=23tan-113

Q10.

Answer :

Disclaimer: None of the given option is correct.

We have,I=∫0π1-x1+xdx=∫0π1-x1+x×1-x1-xdx=∫0π1-x1-x2dx=∫0π11-x2dx-∫0πx1-x2dxPutting 1-x2=t⇒-2x dx=dt⇒x dx=-dt2When x→0; t→1and x→π; t→1-π2∴I=∫0π11-x2dx-∫11-π2 -dt2t=sin-1×0π+22 t11-π2=0-0+ 1-π2-1=1-π2-1

Q11.

Answer :

a πa2-b2We have,I=∫0π1a+bcosxdx=∫0π1a+b1-tan2x21+tan2x2dx

=∫0π1+tan2x2a1+tan2x2+b1-tan2x2dx=∫0π1+tan2x2a+b+a-btan2x2dx=∫0πsec2x2a+b+a-btan2x2dx

Putting tanx2=t⇒12sec2x2dx=dt⇒sec2x2dx=2 dtWhen x→0; t→0and x→π; t→∞∴I=∫0∞2dta+b+a-bt2=2a-b∫0∞1a+ba-b+t2dt

=2a-b∫0∞1a+ba-b2+t2dt=2a-b×a-ba+btan-1ta+ba-b0∞=2a2-b2π2-0=2a2-b2π2=πa2-b2

Q12.

Answer :

(c) π12Let, I=∫π6π311+cotxdx … (i)=∫π6π311+cotπ3+π6-xdx Using ∫abfxdx=∫abfa+b-xdx=∫π6π311+tanxdx … (ii)Adding (i) and (ii) we get 2I= ∫π6π3 11+cotx+11+tanxdx =∫π6π32+cotx+tanx1+cotx1+tanxdx =∫π6π32+cotx+tanx2+cotx+tanxdx =∫π6π3dx =xπ6π3 =π3-π6 =π6Hence, I=π12

Q13.

Answer :
(b) π60∫0∞1×2+4×2+9dx=15∫0∞1×2+4-1×2+9dx=1512tan-1×2-13tan-1×30∞=1512×π2-13×π2=15×π12=π60

Q14.

Answer :

(a) 1

∫1elogx dx=∫1elogx x0 dx=x logx1e-∫1e1xx dx=x logx1e-x1e=e-0-e-1=e-e+1=1

Q15.

Answer :

(a) π12

∫1311+x2dx=tan-1×13=π3-π4=π12

Page 20.98 (Multiple Choice Questions)

Q16.

Answer :

a π12+log22

We have,I=∫033x+1×2+9dxI=∫033xx2+9dx+∫031×2+9dxI1=∫033xx2+9dx and I2=∫031×2+9dxPutting x2+9=t in I1⇒2x dx=dt⇒x dx=dt2When x→0; t→9and x→3; t→18∴I=∫9183 dt2 t+∫031×2+9dx=32∫918dtt+∫031×2+32dx=32logt918+13tan-1×303=32log18-log9+13π4-0=32log189+π12=32log 2+π12=log8+π12=log22+π12=π12+log22

Q17.

Answer :

b π4

We have,I=∫0∞x1+x1+x2 dxPutting x=tan θ⇒dx=sec2θ dθWhen x→0 ; θ→0and x→∞ ; θ→π2Now, integral becomes

I=∫0π2tan θ1+tan θ sec2θ sec2θ dθ=∫0π2tan θ1+tan θ dθ=∫0π2sin θcos θ1+sin θcos θdθ⇒I=∫0π2sin θsin θ+cos θdθ …..1⇒I=∫0π2sinπ2-θsinπ2-θ+cosπ2-θdθ ∵∫0afxdx=∫0afa-xdx⇒I=∫0π2cos θcos θ+sin θdθ⇒I=∫0π2cosθsinθ+cosθdθ …..2

Adding 1 and 2, we get2I=∫0π2sinθ+cosθsinθ+cosθ dθ⇒2I=∫0π2dθ⇒2I=π2⇒I=π4∴∫0∞x1+x1+x2 dx=π4

Q18.

Answer :

(b) 2

∫-π2π2sinxdx=-∫-π20sinx dx+∫0π2sinx dx=–cosx-π20+-cosx0π2=1-0-0+1=2

Q19.

Answer :

(a) π4
Let, I=∫0π211+tanxdx … (i) =∫0π211+tanπ2-xdx = ∫0π211+cot xdx … (ii)Adding (i) and (ii) we get2I=∫0π211+tanx+11+cotx dx =∫0π21+cotx+1+tan x1+tan x1+cotx dx =∫0π22+ tan x+cot x1+tan x+cotx +tan x cot x dx =∫0π22+ tan x+cot x2+ tan x+cot x dx =∫0π2dx =x0π2=π2Hence, I=π4

Q20.

Answer :

(b) e − 1
Let, I=∫0π2cosx esinx dxLet sinx =t, then cosx dx =dtWhen x=0, t=0 and x=π2, t=1Therefore the integral becomesI=∫01 et dt=et01=e-1

Q21.

Answer :

(b) 12
∫0α11+4x2dx=π8⇒∫0α11+2x2dx=π8⇒12 tan-12×0α=π8⇒12tan-12α=π8⇒2α=tanπ4⇒2α=1∴ α=124430. →

Q22.

Answer :

(b) 0

∫01a-x2 fxdx=a2∫01fxdx+∫01×2 fx dx-2a∫01x fxdx=a2×1+a2-2aa As per given values=2a2-2a2=0

Q23.

Answer :

(c) 0

∫-ππsin3x cos2 xdx=∫-ππsinx1-cos2x cos2x dxLet cos x =t, then -sin x dx =dt,When, x=-π, t=-1, x=π,t=-1Therefore the integral becomes∫-1-1-1-t2t2 dt=0

Q24.

Answer :

(b) loge3

∫π6π31sin2xdx=∫π6π3cosec2x dx=12∫π6π32cosec2x dx=-12logcosec2x+cot2xπ6π3=-12-2log3=log3

Q25.

Answer :

(b) 2

∫-111-xdx=∫-101-x dx+∫011-x dx=x-x22-10+x-x2201=0+1+12+1-12-0=2

Q26.

Answer :

(c) (ln x)−1 x (x − 1)

Using Newton Leibnitz formula

f'(x)=1logex3(3×2)−1logex2(2x)=3x23lnx−2x2lnx=x2lnx−xlnx=1lnxx(x−1)=(lnx)−1x(x−1).

Page 20.99 (Multiple Choice Questions)

Q27.

Answer :

(b) 10π29We have,I10=∫0π/2×10 sin x dx=x10 -cos x0π2-∫0π/210 x9 ∫sin x dxdx=-x10cos x0π2-10∫0π/2 x9 -cos x dx=-x10 cos x0π2+10∫0π/2 x9 cos x dx=-x10 cos x0π2+10×9 sin x0π2-10∫0π/2 9×8 sin x dx=-π210 ×0-010 cos 0+10π29 ×1-09 ×0-90∫0π/2 x8 sin x dx=10π29 ×1-90 I8=10π29-90 I8∴I10+90 I8=10π29

Q28.

Answer :

Disclaimer: The question given is not correct because the function provided does not converge in the given domain.

Q29.

Answer :

(c) ln(3/2)

limn→∞12n+1+12n+2+……….+12n+n=limn→∞∑r=1n12n+r=limn→∞1n∑r=1n12+rnlet rn=x=∫0∞12+xdx=log2+x0∞=log3-log2=log32=ln32

Q30.

Answer :

(a) 4

We have,I=∫-221-x2dx1-x2=-1-x2, -2<x<-11-x2, -1<x<1-1-x2, 1<x<2∴I=∫-2-11-x2dx+∫-111-x2dx+∫121-x2dx=∫-2-1-1-x2dx+∫-111-x2dx+∫12-1-x2dx=-∫-2-11-x2dx+∫-111-x2dx-∫121-x2dx=-x-x33-2-1+x-x33-11-x-x3312=–1+13+2-83+1-13+1-13-2-83-1+13=-1-73+2-23-1-73=-1+73+2-23-1+73=4

Q31.

Answer :

(d) π/4

We have, I=∫0π211+cot3xdx …..1=∫0π211+cot3π2-xdx ∴I=∫0π211+tan3xdx …..2Adding 1 and 2 we get2I=∫0π211+cot3x+11+tan3xdx

=∫0π21+tan3x+1+cot3x1+cot3x1+tan3x dx=∫0π22+tan3x+cot3x1+tan3x+cot3x+cot3x tan3xdx=∫0π22+tan3x+cot3x1+tan3x+cot3x+1dx=∫0π22+tan3x+cot3x2+tan3x+cot3x dx=∫0π2[1]dx=x0π2=π2Hence I=π4

Q32.

Answer :

(d) π/4

We have, I=∫0π2sinxsinx+cosxdx …..1⇒I=∫0π2sinπ2-xsinπ2-x+cosπ2-xdx⇒I=∫0π2cosxcosx+sinx dx ∴I=∫0π2cosxsinx+cosx dx …..2Adding 1 and 2, we get2I=∫0π2sinxsinx+cosx+cosxcosx+sinxdx=∫0π2sinx+cosxsinx+cosxdx =∫0π2dx=x0π2=π2Hence I=π4

Q33.

Answer :

(c) π/2

We have,I=∫01ddxsin-12×1+x2dxWe know since ∫f'(x) = f(x)f(x) =sin-12×1+x2 and f'(x)=ddxsin-12×1+x2 Therefore, I=sin-12×1+x201=sin-11-sin-10=π2

Q34.

Answer :

(d) 1

We have, I=∫0π2x sinx dx =-x cosx0π2-∫0π21-cosx dx=-x cosx0π2+∫0π2cosx dx=-x cosx0π2+sinx0π2=-0-0+ 1-0=1

Q35.

Answer :

(c) 0

I=∫0π2sin2x log tanx dx …..1I=∫0π2sinπ-2x log tanπ2-x dxI=∫0π2sin2x log cotx dx …..2Adding 1 and 2, we get,2I=∫0π2sin2x log tanx +log cotx dx2I=∫0π2sin2x log tanx cotx dx2I=∫0π2sin2x log1 dxI=0

Q36.

Answer :

(a) π/4
∫0π15+3 cosxdx=∫0π15+3 1-tan2x21+tan2x2dx=∫0π1+tan2x25+5tan2x2+3-3tan2x2dx=∫0πsec2x28+2tan2x2dxLet tanx2=t, then sec2x2 dx=2dtWhen x=0, t=0, x=π, t=∞Therefore the integral becomes12∫0∞dt4+t2=12tan-1t20∞=12π2-0=π4

Q37.

Answer :

(a) π ln 2

∫0∞log x+1x 11+x2dx
Substitute x = tan θ
⇒ dx = sec2 θ dθ.
when,
x = 0 ⇒ θ = 0
x=∞⇒θ=π2∫0π2 tan θ+1tan θ11+tan2θ×sec2θ dθ∫0π2log tan2θ+1tanθ 11+tan2θ×sec2θdθ⇒∫0π2log sec2θtan θ1sec2θ×sec2θdθ ∵1+tan2θ=sec2θ⇒∫0π2log sec2θtan θdθ⇒∫0π2log 1sin θ.cos θdθ⇒-∫0π2log sin θ.cos θdθ⇒-∫0π2 log sin θ+log cos θdθ⇒-∫0π2log sin θdθ-∫0π2log cos θ dθ
Let us consider,
∫0π2log sin θdθ=I …..(i)⇒I=∫0π2log sin π2-θdθ=∫0π2log cos θdθ …..iiAdding i and ii2I=∫0π2log sin θdθ+∫0π2log cos θdθ =∫0π2log sin θ.cos θdθ =∫0π2log sin 2θdθ-∫0π2log 2dθLet us consider 2θ=t2dθ=dt2I=12∫0πlog sin tdt-π2log 22I=22∫0π2log sin tdt-π2log 2 ∵sin θ is positive in both 1st and 2nd quadrants2I=I-π2log 22I-I=-π2log 2I=-π2log 2, where I=∫0π2log sin θdθNow,-∫0π2logsin θdθ-∫0π2log cos θdθ-2∫0π2log sin θdθ=-2×I=-2×-π2log 2 ∵where I=-π2log2=π log 2

Page 20.100 (Multiple Choice Questions)

Q38.

Answer :

c ∫0afx dx+∫0af2a-x dx

According to the additivity property of integrals,∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx, where a<c<busing this property, ∫02af(x)dx=∫0af(x)dx +∫02af(x)dx ……(1)Now, consider the integral, ∫02af(x)dxLet x=2a-t. Then dx=d(2a-t)⇒dx=-dtAlso, x=a⇒t=a and x=2a⇒t=0Therefore, ∫a2af(x)dx=-∫a0f(2a-t)dt⇒∫a2af(x)dx=∫0af(2a-t)dt⇒∫a2af(x)dx=∫0af(2a-x)dxSubstituting this in equation (1) we get,∫02af(x)dx=∫0af(x)dx +∫0af(2a-x)dx →

Q39.

Answer :

(d) a+b2 ∫ab fx dx
Let, I=∫abx fxdx …(i) =∫aba+b-x fa+b-xdx =∫aba+b-x fx dx …(ii) Adding (i) and (ii)2I=∫abx+a+b-x fxdx =a+b∫ab fxdx Hence I=a+b2∫ab fxdx

Q40.

Answer :

(b) 0

Let, I=∫01tan-12x-11+x-x2dx … (i)=∫01tan-121-x-11+1-x-1-x2dx=∫01tan-11-2×2-x-1-x2+2x dx=∫01tan-11-2×1+x-x2 dx=-∫01tan-12x-11+x-x2 dx … (ii)Adding (i) and (ii)2I=∫01tan-12x-11+x-x2 dx-∫01tan-12x-11+x-x2 dx =0Hence, I=0

Q41.

Answer :

(c) 0

Let I=∫0π2log4+3sinx4+3cosxdx …(i) =∫0π2log4+3sinπ2-x4+3cosπ2-x dx =∫0π2log4+3 cosx4+3sinxdx …(ii)Adding (i) and (ii)2I=∫0π2log4+3sinx4+3cosx+log4+3 cosx4+3sinx dx = ∫0π2log4+3sinx4+3cosx×4+3 cosx4+3sinx dx = ∫0π2log1 dx=0Hence I=0

Q42.

Answer :

(c) π

∫-π2π2×3+xcosx+tan5x+1dx=x44-π2π2+x sinx-π2π2-∫-π2π2sinx dx+∫-π2π2tan3x sec2x-1dx+x -π2π2=π464-π464+π2-π2–cosx-π2π2+∫-π2π2tan3x sec2x dx-∫-π2π2tan3x dx+π2+π2=π+0+tan4x4-π2π2-∫-π2π2tanx sec2x dx-∫-π2π2tanx dx=π-tan2x2-π2π2–logcosx-π2π2=π

Page 20.100 (Revision Exercise)

Q1.

Answer :

Let, I=∫04×4-x dx =∫044-x4-4+x dx =∫044-xx dx =∫044x-x32 dx =8×32304-2×52504 =643-645 =12815 →

Q2.

Answer :

∫12x3x-2dxLet, 3x-2=t, then 3dx=dtwhen, x=1 ; t=1 and x=2 ; t=4Therefore the integral becomes∫14t+23t dt3=19∫14t32+2t dt=192t525+4t32314=19645+323-25-43=46135

Q3.

Answer :

Let I=∫15x2x-1dxLet, 2x-1=t, then 2dx=dt,When, x→1 ; t→1 and x→5 ; t→9x=t+12I=12∫19t+1t×dt2=142t323+2t19=1418+6-23-2=163

Q4.

Answer :

∫01cos-1x dx=∫01cos-1x ×1 dx= cos-1x x01-∫01-x1-x2dx=x cos-1×01-221-x201=0+1=1

Q5.

Answer :

∫01tan-1x dx=∫01tan-1x×1 dx=tan-1x x01-∫01×1+x2dx=xtan-1×01-12log1+x201=π4-0-12log2+0=π4-12log2

Q6.

Answer :

∫01cos-11-x21+x2dxLet, x= tanθ, dx= sec2θ dθWhen, x→0 ; θ→0and x→1 ; θ→π4Therefore, the integral becomes∫0π4cos-11-tan2θ1+tan2θ sec2θ dθ= ∫0π4cos-1cos2θ sec2θ dθ=2∫0π4θ sec2θ dθ=2θ tanθ0π4-2∫0π4tanθ dθ=2θ tanθ0π4+2logcosθ0π4=2π4-0+2log12-0=π2-log2

Q7.

Answer :

∫01tan-12×1-x2dxLet, x= tanθ, then dx= sec2θ dθWhen, x→0 ; θ→0And x→1 ; θ→π4Therefore the integral becomes∫0π4tan-12tanθ1-tan2θ sec2θ dθ=∫0π4tan-1tan2θ sec2θ dθ=2∫0π4θ sec2θ dθ=2θ tanθ0π4-2∫0π4tanθ dθ=2θ tanθ0π4-2-logcosθ0π4=2π4-0+2log12-0=π2-log2

Q8.

Answer :

∫013tan-13x-x31-3x2dxLet x = tanθ, then dx= sec2θ dθWhen, x→0 ; θ→0And x→13 ; θ→π6Therefore the integral becomes∫0π6tan-13tanθ-tan3θ1-3tan2θsec2θ dθ=∫0π6tan-1tan3θsec2θ dθ=3∫0π6θ sec2θ dθ=3θ tanθ0π6-3∫0π6tanθ dθ=3θ tanθ0π6-3-logcosθ0π6=3π6×13-0+3log32=π23+3log32=π23-32log43

Q9.

Answer :

∫011-x1+x dx=∫011-x-1+11+xdx=∫012-x+11+xdx=∫0121+x-∫011+x1+xdx=∫0121+x-∫01dx=2log1+x01-x01=2log2-1

Page 20.101 (Revision Exercise)

Q10.

Answer :

∫0π3cosx3+4sinxdxLet, sin x = t ⇒ cosx dx = dtWhen, sinx →0 ; t→0And sinx →π3 ; t→32=∫032dt3+4t=14log3+4t032=14loglog3+23-log3+0=14loglog23+3-log3=14log23+33

Q11.

Answer :

∫0π2sin2x1+cosx2dx=∫0π21-cos2x1+cosx2dx=∫0π21+cosx1-cosx1+cosx2dx=∫0π21-cosx1+cosxdx=∫0π21-cosx-1+11+cosxdx=∫0π22-1+cosx1+cosxdx=∫0π221+cosxdx-∫0π2dx=∫0π221-cosx1+cosx1-cosxdx-∫0π2dx=2∫0π21-cosxsin2xdx-x0π2=2∫0π2cosec2x-cosecx cotx dx-x0π2=2-cotx+cosecx0π2-x0π2=2-π2

Q12.

Answer :

∫0π2sinx1+cosxdxLet 1+cosx =t, then -sinx dx = dtWhen, x→0, t→2 and x→π2, t→1Therefore, the integral becomes∫21-1tdt=∫121tdt=2t12=22-1

Q13.

Answer :

∫0π2cosx1+sin2xdxLet sinx =t, then cosx dx = dtWhen x→0 ; t→0And x→π2; t→1Therefore the integral becomes∫01dt1+t2=tan-1×01=π4

Q14.

Answer :

We have,I=∫0πsin3x1+2cosx1+cosx2dx=∫0πsin2x1+2cosx1+cosx2sinx dx=∫0π1-cos2x1+2cosx1+cosx2sinx dxPutting cosx=t⇒-sinx dx=dtWhen x→0; t→1and x→π; t→-1∴I=-∫1-11-t21+2t1+t2dt=∫-111-t21+2t1+t2dt=∫-111+2t-t2-2t31+2t+t2dt=∫-111+2t+t2+2t+4t2+2t3-t2-2t3-t4-2t3-4t4-2t5dt=∫-111+4t+4t2-2t3-5t4-2t5dt=t+2t2+4t33-t42-t5-t63-11=1+2+43-12-1-13–1-2-12-4-133+-142+-15+-163=1+2+43-12-1-13+1-2+43+12-1+13=83

Q15.

Answer :

I = ∫0∞x1+x1+x2 dx

using partial fraction,
x(1+x)(1+x2)A1+x + Bx+C1+x2x=A(1+x2) + (Bx+C)(1+x)x=A+Ax2+Bx+Bx2+C+CxB+C=1A+C=0A+B=0so, A=-12, B=12, C=12

putting the values of A,B and C we get

-121+x+12x+121+x2=-1211+x + 12x+11+x2Therefore, I=∫0∞-1211+x + 12x+11+x2I=-12log1+x0∞ + 12∫0∞x1+x2+11+x2I=-12log1+x0∞ + 12×2∫0∞2×1+x2 +12∫0∞11+x2
I=-12log1+x0∞ + 14log1+x20∞ + 12tan-1×0∞I=-12log1+x0∞ + 12×12log1+x20∞ + 12tan-1×0∞I=12logx2+1x+10∞ + 12tan-1×0∞I=12log1+1×21+1×0∞ + 12tan-1×0∞I=120 + 12tan-1∞ – tan-10
I = π4

Q16.

Answer :

Let, I=∫0π4sin2x sin3x dx …i⇒I=-sin2xcos3x30π4+∫0π42cos2xcos3x3dx⇒I=-sin2xcos3x30π4+23cos2xsin3x30π4+49∫0π4sin2x sin3x dx⇒I=-sin2xcos3x30π4+23cos2xsin3x30π4+49I From i⇒59I=-sin2xcos3x30π4+23cos2xsin3x30π4⇒59I=132+0⇒59I=132∴ I=352

Q17.

Answer :

∫011-x1+xdx=∫011-x1+x×1-x1-xdx=∫011-x1-x2dx=∫0111-x2dx-∫01×1-x2dx=sin-1×01+1-x201=π2-1

Q18.

Answer :

∫121x2e-1xdxLet -1x=t, then 1×2 dx=dtWhen, x→1 ; t→-1And x→2 ; t→-12Therefore the integral becomes∫-1-12etdt=et-1-12=e-12-e-1=e-1e

Q19.

Answer :

∫0π4cos4x sin3x dx=∫0π4cos4x sinx 1-cos2x dx=∫0π4cos4x sinx dx-∫0π4cos6x sinx dx=-cos5x50π4+cos7x70π4=-1202+15+1562-17=-240+235+2112=235-92560

Q20.

Answer :

∫π3π21+cosx1-cosx52dx=∫π3π21+cosx1-cosx52×1-cosx1-cosxdx=∫π3π2sinx1-cosx3dx=-121-cosx-2π3π2=-121-4=32

Q21.

Answer :

∫0π2×2 cos2x dx=x2sin2x20π2-∫0π22x sin2x2dx=x2sin2x20π2-∫0π2x sin 2x dx=x2sin2x20π2–xcos2x20π2+-∫0π2cos2x2dx=x2sin2x20π2+xcos2x20π2-∫0π2cos2x2dx=x2sin2x20π2+xcos2x20π2-12sin2x20π2=0-π4-0=-π4

Q22.

Answer :

∫01log1+xdx=∫01log1+x×1 dx=log1+x x01-∫01×1+xdx=log1+x x01-∫011-11+xdx=xlog1+x01-x-log1+x01=log2-1+log2=2log2-1=log4-loge=log4e

Q23.

Answer :

∫24×2+x2x+1dx=x2+x2x+124-∫242x+12x+1 dx=x2+x2x+124-∫242x+132 dx=x2+x2x+124-2x+152524=60-65-2435-2555=575-5

Q24.

Answer :

We have,I=∫01xtan-1x2dxPutting tan-1x=u⇒x=tan u⇒dx=sec2u duWhen x→0; u→0and x→1; u→π4∴I=∫0π4tan uu2sec2u du=∫0π4u2 tan u sec2u du=u2 tan2 u20π4-∫0π42u tan2 u2 du=u2 tan2 u20π4-∫0π4u sec2 u-1 du=u2 tan2 u20π4-∫0π4u sec2 u du+∫0π4u du=u2 tan2 u20π4-u tan u0π4+∫0π4 tan u du+u220π4=u2 tan2 u20π4-u tan u0π4+log sec u0π4+u220π4=π216 ×12-π4+log2+π232=π216-π4+log2=π216-π4+12log 2

Q25.

Answer :
I=∫01(cos-1x)2dxlet cos-1x = θ⇒x = cosθ⇒dx = -sinθ dθwhen x=0, θ=π2 and when x=1, θ=0therefore, I = ∫π20θ2(-sinθ)dθ I = -∫π20θ2(sinθ)dθI = ∫0π2θ2(sinθ)dθI = [θ2(-cosθ)]0π2 -∫0π22θ∫0π2sinθdθ]I = [θ2(-cosθ)]0π2 – ∫0π2[2θ(-cosθ)dθ]
=[-θ2cosθ]0π2 + ∫0π22θ(cosθ)dθ=[-θ2cosθ]0π2 + 2[θsinθ – ∫0π2sinθdθ]=[-θ2cosθ]0π2+ 2[θsinθ + cosθ]0π2

I = 2[(π2 + 0) – 1] I = π – 2

Q26.

Answer :

∫12x+3xx+2dx=∫12x+2+1xx+2dx=∫121xdx+∫121xx+2dx=∫121xdx+12∫12x+2-xxx+2dx=∫121xdx+12∫121xdx-12∫121x+2dx=32∫121xdx-12∫121x+2dx=32logx12-12logx+212=32log2-12log4+12log3=32log2- log2+12log3=12log2+12log3=12log6

Q27.

Answer :

Let, I=∫0π4ex sinxdx …i =-excosx0π4+∫0π4ex cosx dx =-excosx0π4+exsinx0π4-∫0π4ex sinx dx⇒ I =-excosx0π4+exsinx0π4- I Using i ⇒ 2I=-excosx0π4+exsinx0π4 =-12eπ4+1+12eπ4-0 =1Hence I = 12

Q28.

Answer :

∫0π4tan4x dx=∫0π4tan2xsec2x-1 dx=∫0π4tan2x sec2x dx-∫0π4tan2x dx=tan3x30π4-tanx-x0π4=13-1+π4=π4-23

Q29.

Answer :

We have,2x-1=-2x-1, 0≤x≤12 2x-1, 12≤x≤1∴∫012x-1dx=∫012-2x-1 dx+∫1212x-1 dx=-x2+x012+x2-x121=-14+12+1-1-14+12=12

Q30.

Answer :
We have,x2-2x=-x2-2x, 1≤x≤2 x2-2x, 2≤x≤3∴∫13×2-2xdx=∫12-x2-2x dx+∫23×2-2x dx=-x33+x212+x33-x223=-83+4+13-1+9-9-83+4=2

Q31.

Answer :

∫0π2sinx-cosxdx=2∫0π2sinx12-cosx12dx=2∫0π2sinx cosπ4-cosx sinπ4dx=2∫0π2sinx-π4dxWe have,sinx-π4=-sinx-π4, 0≤x≤π4 sinx-π4, π4≤x≤π2∴∫0π2sinx-cosxdx=2∫0π4-sinx-π4dx+2∫π4π2sinx-π4dx =2cosx-π40π4-2cosx-π4π4π2 =2cos 0-cos-π4-2cosπ4-cos 0 =21-12-12+1 =22-22 = 22-2 = 22-1

Q32.

Answer :
We have,sin2πx=sin2πx, 0≤x≤12-sin2πx, 12≤x≤1∴∫01sin2πxdx =∫012sin2πx dx+∫121-sin2πx dx =-cos2πx2π012+cos2πx2π121 =12π+12π+12π+12π =2π

Q33.

Answer :

∫13×2-4dx=∫12-x2-4 dx+∫23×2-4 dx=-x33+4×12+x33-4×23=-83+8+13-4+9-12-83+8=4

Q34.

Answer :

∫-π2π2sin9xdxLet fx=sin9xConsider, f-x=sin9-x=-sin9x=-fxThus fx is an odd functionTherefore,∫-π2π2sin9xdx=0

Q35.

Answer :

∫-1212cosx log1+x1-xdxLet fx=cosx log1+x1-xConsider f-x=cos-x log1-x1+x =cosx-log1+x1-x=-cosx log1+x1-x=-fxThus fx is an odd functionTherefore,∫-1212cosx log1+x1-xdx=0

Q36.

Answer :

∫-aax ex21+x2dxLet fx=x ex21+x2Consider f-x=-x ex21+x2=-fxThus fx is an odd functionTherefore,∫-aax ex21+x2dx=0ode is 4430. →

Q37.

Answer :

Let, I=∫0π211+cot7xdx … (i) =∫0π211+cot7π2-xdx =∫0π211+tan7xdx …(ii)Adding (i) and (ii)2I=∫0π211+cot7x+11+tan7xdx =∫0π22+cot7x+tan7x1+cot7x1+tan7xdx =∫0π22+cot7x+tan7x2+cot7x+tan7xdx =∫0π2dx =x0π2 =π2Hence, I=π4

Q38.

Answer :

Let, I=∫02πcos7xdx …i =∫02πcos72π-xdx =∫02π-cos7xdx⇒ I=-∫02πcos7xdx …iiAdding i and ii we get, 2I=∫02πcos7xdx-∫02πcos7xdx⇒2I=0∴I = 0

Q39.

Answer :

Let, I=∫0axx+a-xdx …(i) =∫0aa-xa-x+a-a+xdx =∫0aa-xa-x+xdx⇒I =∫0aa-xx+a-xdx …(ii)Adding (i) and (ii)2I=∫0axx+a-x+a-xx+a-xdx =∫0adx =x0a =aHence, I=a2

Q40.

Answer :

Let, I=∫0π211+tan3xdx … (i) =∫0π211+tan3π2-xdx =∫0π211+cot3xdx ….(ii)Adding (i) and (ii)2I=∫0π211+tan3x+11+cot3xdx =∫0π22+tan3x+cot3x1+tan3x1+cot3xdx =∫0π22+tan3x+cot3x2+tan3x+cot3xdx =∫0π2dx =x0π2 =π2Hence, I=π4

Q41.

Answer :

Let, I=∫0πx sinx1+cos2xdx …(i) =∫0ππ-x sinπ-x1+cos2π-xdx =∫0ππ-x sinx1+cos2xdx …(ii)Adding (i) and (ii)2I=∫0πx sinx1+cos2x+π-x sinx1+cos2x dx = ∫0ππ sinx1+cos2xdx =π-tan-1cosx0π =-πtan-1-1-tan-11 =-π-π4-π4 =π22Hence, I=π24

Q42.

Answer :

Let, I=∫0πx sinx cos4x dx …(i) =∫0ππ-x sinπ-x cos4π-x dx =∫0ππ-x sinx cos4x dx …(ii) Adding (i) and (ii)2I=∫0πx sinx cos4x +π-x sinx cos4x dx =∫0πx+π-x sinx cos4x dx = π∫0πsinx cos4x dx =π-cos5x50π =π15+15 =2π5Hence, I=π5

Q43.

Answer :

We have,I =∫0πxa2cos2x+b2sin2xdx …..1=∫0ππ-xa2cos2π-x+b2sin2π-xdx=∫0ππ-xa2cos2x+b2sin2xdx …..2Adding 1 and 22I=∫0πx+π-xa2cos2x+b2sin2xdx=π∫0π1a2cos2x+b2sin2xdx=π∫0πsec2xa2+b2tan2xdx Dividing numerator and denominator by cos2x=2π∫0π2sec2xa2+b2tan2xdx Using ∫02afxdx=∫0afxdx+∫0af2a-xdxPutting tan x=t⇒sec2x dx=dtWhen x→0; t→0and x→π2; t→∞∴2I=2π∫0π2dta2+b2t2⇒I=πb2∫0π2dta2b2+t2=πb2×batan-1bta0∞=πabπ2-0=πab×π2=π22ab Hence I=π22ab

Q44.

Answer :
∫-π4π4tanxdx=∫-π40-tanx dx+∫0π4 tanx dx=log cosx-π40+-log cosx0π4=-log12-log12=2log2=log2

Q45.

Answer :

We have,I=∫01.5×2 dx=∫01×2 dx+∫12×2 dx+∫21.5×2 dx=∫010 dx+∫121 dx+∫21.52 dx ∵x2=0 where, 0<x<11 where, 1<x<22 where, 2<x<1.5=0+x12+2×21.5=x12+2×21.5=2-1+21.5-2=2-1+3-22=2-2

Q46.

Answer :

We have,I=∫0πx1+cos α sin x dx …..1⇒I=∫0ππ-x1+cos α sin π-x dx ∵∫0afxdx=∫0afa-xdx⇒I=∫0ππ-x1+cos α sin x dx …..2Adding 1 and 2, we get

2I=∫0ππ1+cos α sin x dx ⇒I=π2∫0π11+cos α sin x dx =π2∫0π11+cos α 2tan x21+tan2x2 dx =π2∫0π1+tan2x21+tan2x2+cos α 2tan x2 dxPutting tanx2=t⇒12sec2x2dx=dtWhen x→0 ; t→0and x→π ; t→∞Now, integral becomes

I=π∫0∞dt1+t2+2t cos α =π∫0∞dtt+cos α2+1-cos2α=π∫0∞dtt+cos α2+sin2α=π1sin αtan-1t+cos αsin α0∞=πsin αtan-1t+cos αsin α0∞=πsin απ2-tan-1cot α=πsin απ2-tan-1tanπ2-α=πsin απ2-π2-α=παsin α

Q47.

Answer :

Let, I=∫0π2xsinx cosxsin4x+cos4xdx …(i)=∫0π2π2-xsinπ2-x cosπ2-xsin4π2-x+cos4π2-xdx=∫0π2π2-xcosx sinxcos4x+sin4xdx …(ii)Adding (i) and (ii)2I=∫0π2x+π2-xsinx cosxsin4x+cos4xdx =π2∫0π2sinx cosxsin2x+cos2x2-2sin2x cos2xdx = π2∫0π2sinx cosx1-2sin2x cos2x dx = π2∫0π2sinx cosx1-2sin2x 1-sin2xdx =π2∫0π2sinx cosx1-2sin2x+2sin4xdxLet, sin2x =t, then 2sinxcosx dx = dt When, x→0 ; t→0 and x→π2 ; t→1 2I=π4∫0111-2t+2t2dt =π8∫011t-122+14 =π82 tan-12t-101 =π4tan-11-tan-1-1 =π4π4+π4 =π28Hence, I=π216

Q48.

Answer :

We have,I=∫0π2cos2xsinx+cosxdx …..1=∫0π2cos2π2-xsinπ2-x+cosπ2-xdx=∫0π2sin2xcosx+sinx dx …..2

Adding 1 and 22I=∫0π2cos2xsinx+cosx+sin2xcosx+sinxdx= ∫0π21sinx+cosxdx=∫0π212tanx21+tan2x2+1-tan2x21+tan2x2dx

=-∫0π21+tan2x2tan2x2-2tanx2-1 dx=-∫0π2sec2x2tan2x2-2tanx2-1 dxPutting tanx2=t⇒12sec2x2dx=dt⇒sec2x2dx=2dtWhen x→0; t→0and x→π2; t→1

∴2I=-2∫01dtt2-2t-1⇒I=-∫01dtt-12-22=-122logt-1-2t-1+2 01=-122log-1-log-1-2-1+2 =-122log 1-log2+12-1

=-122-log2+12-1 =122log2+12+12-12+1=122log2+122-1=122log2+12=122×2 log2+1=12log2+1

Q49.

Answer :

∫0πcos2x logsinx dx=logsinx sin2x20π-∫0πcosxsinxsin2x2 dx=logsinx sin2x20π-∫0πcos2x dx=logsinx sin2x20π-∫0π1+cos2x2dx=logsinx sin2x20π-12x+sin2x20π=0-12π+0=-π2

Q50.

Answer :

Let I=∫0πxa2-cos2xdx … (i) =∫0ππ-xa2-cos2π-xdx =∫0ππ-xa2-cos2xdx …(ii)Adding (i) and (ii)2I=∫0ππa2-cos2xdx =π2a∫0π1a-cosx+1a+cosx dx =π2a∫0πsec2x2a-1+a+1tan2x2+sec2x2a+1+a-1tan2x2dxLet, tanx2=t, then 12sec2x2 dx=dt2I=πa∫0∞1a-1+a+1t2+1a+1+a-1t2 dt =πaa2-1tan-1a+1a-1t+tan-1a-1a+1t0∞ =πaa2-1π2+π2=π2aa2-1∴I=π22aa2-1

Q51.

Answer :

Let I=∫0πx tanxsecx +tanxdx …(i) =∫0ππ-x tanπ-xsecπ-x +tanπ-xdx =∫0ππ-x tanxsecx +tanxdx …(ii)Adding (i) and (ii) we get2I=∫0ππ tanxsecx +tanxdx =π∫0πsinx1+sinxdx =π∫0π1+sinx-11+sinxdx =π∫0π1-11+sinxdx =πx0π-π∫0π11+2tanx21+tan2x2dx =π2-π∫0πsec2x21+tan2x2+2tanx2dx =π2-π∫0πsec2x21+tanx22dx =π2+π21+tanx20π =π2+π0-2 =π2-2π =ππ-2Hence I=π2π-2

Q52.

Answer :

Let, I=∫23×5-x+xdx …(i) =∫235-x5-5+x+5-xdx =∫235-xx+5-xdx …(ii)Adding (i) and (ii) 2I=∫23×5-x+x+5-xx+5-xdx =∫235-x+x5-x+x dx =∫23dx =x23 =3-1=1Hence, I=12

Q53.

Answer :

We have,I=∫0π2sin2xsinx+cosxdx …..1=∫0π2sin2π2-xsinπ2-x+cosπ2-xdx=∫0π2cos2xcosx+sinx dx …..2Adding 1 and 22I=∫0π2sin2xsinx+cosx+cos2xcosx+sinx dx=∫0π21sinx+cosx dx=∫0π21+tan2x22tanx2+1-tan2x2 dx=∫0π2sec2x22tanx2+1-tan2x2 dxPutting tanx2=t⇒ 12sec2x2dx=dt⇒ sec2x2dx=2 dtWhen x→0; t→0and x→π2; t→1∴2I=∫012dt2t+1-t2 dx=2∫01dt22-t-12=222log2+t-12-t+1 01=12log22 – log2-12+1 =120-log2-12+1 =-12log2-12+1=12log2+12-1=12log2+12+12-12+12I=12log2+122-12I=22log2+1Hence I=12log2+1

Q54.

Answer :

Let, I=∫0π2xsin2x+cos2xdx =∫0π2x1dx =∫0π2x dx =x220π2 ∴ I=π28

Q55.

Answer :

∫-ππx10 sin7x dxLet fx=x10 sin7xConsider f-x=-x10 sin7-x=-x10 sin7x=-fxHence fx is an odd functionTherefore ∫-ππx10 sin7x dx=0

Q56.

Answer :

∫01cot-11-x+x2dx=∫01cot-1xx-1+1dx=∫01cot-1xx-1+1x-x-1dx=∫01cot-1x-cot-1x-1 dx=xcot-1×01+∫01×1+x2dx-x-1cot-1x-101-∫01x-11+x-12dx=xcot-1×01+12log1+x201-x-1cot-1x-101-12log1+1-x201=π4-12log2+π4-12log2=π2-log2

Page 20.102 (Revision Exercise)

Q57.

Answer :

∫0π16-cosxdx=∫0π1+tan2x26+6tan2x2-1+tan2x2dx=∫0πsec2x25+7tan2x2dxLet, tanx2=t, then 12sec2x2 dx=dtTherefore the integralbecomes∫0∞2dt5+7t2 =27∫0∞dt57+t2 =235tan-17t50∞=π35

Q58.

Answer :

We have,I=∫0π212cosx+4sinxdx=∫0π21+tan2x22-2tan2x2+8tanx2dxPutting tanx2=t⇒ 12sec2x2dx=dtWhen x→0; t→0and x→π2; t→1∴I=2∫01dt2-2t2+8t=-22∫01dt t2-4 t-1=-∫01dtt-22-5=∫01dt52-t-22=125log5+t-25-t+2 01= 125log5-15 +1 -log5 -25 +2 = 125log5-15 +1 ×5 +25 -2=125log5+25-5-25-25+5-2 =125log5+3-5+3
I = 125log 3 + 53 – 5×3 + 53 + 5 I = 125log 3 + 522I =225log 3 + 52 I = 15log 3 + 52

Q59.

Answer :

∫π6π2cosecx cotx1+cosec2xdx=∫π6π2cosx1+sin2xdx=tan-1sinxπ6π2=tan-11-tan-112=tan-11-121+1×12=tan-113

Q60.

Answer :

∫0π214cosx+2sinxdx=∫0π21+tan2x24-4tan2x2+4tanx2dxLet tanx2=t, then 12sec2x2 dx=dtWhen x=0, t=0, x=π2, t=1=-14∫01dtt-122-54=-14×-45log2t-1-52t-1+501=15log5+15-1

Q61.

Answer :

∫04xdx=x2204=8-0=8

Q62.

Answer :

∫022×2+3dx=2×33+3×02=163+6=343

Q63.

Answer :

Here a =1,b=4, fx=x2+x, h=4-1n=3nTherefore,∫14×2+xdx =limh→0 hfa+fa+h+fa+2h+…………+fa+n-1h =limh→0 hf1+f1+h+……….+f1+n-1h =limh→0 h1+1+1+h2+1+h+1+2h2+1+2h+………+1+n-1h2+1+n-1h =limh→0 h2n+h212+22+…………..n-12+2h1+2+……+n-1+h1+2+……+n-1 =limh→0 h2n+h2nn-12n-16+3hnn-12 =limn→0∞6+921-1n2-1n+921-1n =6+9+92=272

Q64.

Answer :

Here a =-1,b=1, fx=e2x, h=1+1n=2nTherefore,∫-11e2xdx =limh→0 hfa+fa+h+fa+2h+…………+fa+n-1h =limh→0 hf-1+f-1+h+……….+f-1+n-1h =limh→0 he-2+e2-1+h+e2-1+2h+…….+e2-1+n-1h =limh→0 he-2e2hn-1e2h-1 =limh→0 e-2 e4-1e2h-12h×12 Since, nh=2 =12e2-e-2

Q65.

Answer :

Here a =2,b=3, fx=e-x, h=3-2n=1nTherefore,∫23e-xdx =limh→0 hfa+fa+h+fa+2h+…………+fa+n-1h =limh→0 hf2+f2+h+……….+f2+n-1h =limh→0 he-2+e-2+h+e-2+2h+…….+e-2+n-1h =limh→0 he-2e-hn-1e-h-1 =limh→0 e-2 e-1-1e-h-1-h×-1 Since nh=1 =e-2-e-3

Q66.

Answer :

Here, a =1,b=3, fx=2×2+5x, h=3-1n=2nTherefore,∫132×2+5xdx =limh→0 hfa+fa+h+fa+2h+…………+fa+n-1h =limh→0 hf1+f1+h+……….+f1+n-1h =limh→0 h2+5+21+h2+51+h+21+2h2+51+2h+………+2n-1h2+5n-1h =limh→0 h2n+2h212+22+…………..n-12+4h1+2+…………n-1+5n+5h1+2+…………n-1 =limh→0 h7n+2h2nn-12n-16+9hnn-12 =limn→0∞14+831-1n2-1n+181-1n =14+163+18 =1123

Q67.

Answer :

Here a =1,b=3, fx=x2+3x, h=3-1n=2nTherefore,∫13×2+3xdx =limh→0 hfa+fa+h+fa+2h+…………+fa+n-1h =limh→0 hf1+f1+h+……….+f1+n-1h =limh→0 h1+3+1+h2+31+h+1+2h2+31+2h+………+n-1h2+3n-1h =limh→0 hn+h212+22+…………..n-12+2h1+2+…………n-1+3n+3h1+2+…………n-1 =limh→0 h4n+h2nn-12n-16+5hnn-12 =limn→0∞8+431-1n2-1n+101-1n =8+83+10 =623

Q68.

Answer :

Here a =0,b=2, fx=x2+2, h=2-0n=2nTherefore,∫02×2+2dx =limh→0 hfa+fa+h+fa+2h+…………+fa+n-1h =limh→0 hf0+f0+h+……….+f0+n-1h =limh→0 h0+2+0+h2+2+0+2h2+2+………+n-1h2+2 =limh→0 h2n+h212+22+…………..n-12 =limh→0 h2n+h2nn-12n-16 =limn→0∞4+431-1n2-1n =4+83 =203

Q69.

Answer :

Here, a =0, b=3, fx=x2+1, h=3-0n=3nTherefore,∫03×2+1dx =limh→0 hfa+fa+h+fa+2h+…………+fa+n-1h =limh→0 hf0+f0+h+……….+f0+n-1h =limh→0 h0+1+h2+1+2h2+1+………+n-1h2+1 =limh→0 hn+h212+22+…………..n-12 =limh→0 hn+h2nn-12n-16 =limh→03+921-1n2-1n =3+9=12