# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.25 | Set 3

### Evaluate the following integrals:

### Question 41. ∫cos^{-1}((1 – x^{2})/(1 + x^{2}))dx

**Solution:**

Given that, I = ∫cos

^{-1}((1 – x^{2})/(1 + x^{2}))dx)Let us considered x = tant

dx = sec²tdt

I = ∫cos

^{-1}((1 – tan^{2}t)/(1 + tan^{2}t)) sec^{2}tdt= ∫cos

^{-1}(cos2t)sec^{2}tdt= ∫2tsec

^{2}tdtUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 2[t∫sce

^{2}tdt – ∫(1∫sec^{2}tdt)dt]= 2[t × tan

^{2}t – ∫tantdt]= 2[t × tan

^{2}t – logsect] + c= 2[xtan

^{-1}x – log√(1 + x^{2})] + cHence, I = 2xtan

^{-1}x – log|1 + x^{2}| + c

### Question 42. ∫tan^{-1}(2x/(1 – x^{2}))dx

**Solution:**

Given that, I = ∫tan

^{-1}(2x/(1 – x^{2}))dxLet us considered x = tanθ

dx = sec

^{2}θdθI = ∫tan

^{-1}((2tanθ)/(1 – tan^{2}θ)) sec^{2}θdθ= ∫tan

^{-1}(tan2θ)sec^{2}θdθ= ∫2θsec

^{2}θdθUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 2[θ∫sec

^{2}θdθ – ∫(1∫ sec^{2}θdθ)dθ]= 2[θtanθ – ∫tanθdθ]

= 2[θtanθ – logsecθ] + c

= 2[xtan

^{-1}x – log√(1 + x^{2})] + cHence, I = 2xtan

^{-1}x – log|1 + x^{2}| + c

### Question 43. ∫(x + 1)logxdx

**Solution:**

Given that, I = ∫(x + 1)logxdx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = logx∫ (x + 1)dx – ∫(1/x ∫(x + 1)dx)dx

= (x

^{2}/2 + x)logx – ∫1/x (x^{2}/2 + x)dx= (x

^{2}/2 + x)logx – 1/2 ∫xdx – ∫dx= (x + x

^{2}/2)logx – 1/2 × x^{2}/2 – x + cHence, I = (x + x

^{2}/2)logx – 1/2 × x^{2}/2 – x + c

### Question 44. ∫ x^{2} tan^{-1}xdx

**Solution:**

Given that, I = ∫ x

^{2}tan^{-1}xdxUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = tan

^{-1}x∫x^{2}dx – ∫(1/(1 + x^{2}) ∫x^{2}dx) dx= tan

^{-1}x(x^{3}/3) – 1/3∫x^{3}/(1 + x^{2}) dx= 1/3 x

^{3}tan^{-1}x – 1/3 ∫(x – x/(1 + x^{2}))dx= 1/3 x

^{3}tan^{-1}x – 1/3 × x^{2}/2 + 1/3 ∫x/(1 + x^{2}) dxHence, I = 1/3 x

^{3}tan^{-1}x – 1/6 x^{2 }+ 1/6 log|1 + x^{2}| + c

### Question 45. ∫(e^{logx }+ sinx) cosxdx

**Solution:**

Given that, I = ∫(e

^{logx }+ sinx)cosxdx= ∫(x + sinx)cosxdx

= ∫xcosxdx + ∫sinxcosxdx

= [x∫cosxdx – ∫(1]cosxdx)dx] + 1/2 ∫sin2xdx

= [xsinx – ∫ sinxdx] + 1/2 (-(cos2x)/2) + c

I = xsinx+cosx – 1/4 cos2x + c

= xsinx + cosx – 1/4 [1 – 2sin

^{2}x] + c= xsinx + cosx – 1/4 + 1/2 sin

^{2}x + c= xsinx + cosx – 1/4 + 1/2 sin

^{2}x + cHence, I = xsinx + cosx + 1/2 sin

^{2}x + d [d = c-/4]

### Question 46. ∫((xtan^{-1}x))/(1 + x^{2})^{3/2} dx

**Solution:**

Given that, I = ∫((xtan

^{-1}x))/(1 + x^{2})^{3/2}dxLet us considered tan

^{-1}x = t1/(1 + x

^{2}) dx = dtI = ∫(t tant)/√(1 + tan

^{2}t) dt= ∫(t × tant)/(sect) dt

= ∫t (sint)/(cost) costdt

= ∫tsintdt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = [t]sintdt – ∫(1)sintdt)dt]

= [-tcost + ∫costdt]

= [-tcost + sint] + c

= -(tan

^{-1}x)/√(1 + x^{2}) + x/√(1 + x^{2}) + cHence, I = -(tan

^{-1}x)/√(1 + x^{2}) + x/√(1 + x^{2}) + c

### Question 47. ∫ tan^{-1}(√x)dx

**Solution:**

Given that, I = ∫ tan

^{-1}(√x)dxLet us considered x = t

^{2}dx = 2tdt

I = ∫2ttan

^{-1}tdtUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 2[tan

^{-1})t∫tdt – ∫(1/(1 + t^{2}) ∫tdt)dt]= 2[t

^{2}/2 tan^{-1}t – ∫t^{2}/2(1 + t^{2})dt]= t

^{2}tan^{-1}t – ∫(t^{2}+ 1 – 1)/(1 + t^{2})dt= t

^{2}tan^{-1}t – ∫(1 – 1/(1 + t^{2}))dt= t

^{2}tan^{-1}t – t + tan^{-1}t + c= (t

^{2}+ 1) tan^{-1}t – t + cHence, I = (x + 1)tan

^{-1}√x – √x + c

### Question 48. ∫x^{3} tan^{-1}xdx

**Solution:**

Given that, I = ∫x

^{3}tan^{-1}xdx= tan

^{-1}x∫x^{3}dx – (∫(dtan^{-1}x)/dx (∫x^{3}dx)dx)= tan

^{-1}x x^{4}/4 – (∫1/(1 + x^{2}) (x^{4}/4)dx)= tan

^{-1}x x^{4}/4 – (∫1/(1 + x^{2}) (x^{4}/4)dx)= tan

^{-1}x x^{4}/4 – (∫1/(1 + x^{2}) (x^{4}/4)dx)∫ 1/(1 + x

^{2}) (x^{4}/4)dx = 1/4 [∫1/(1 + x^{2}) dx + (x^{2}– 1)dx]∫ 1/(1 + x

^{2}) (x^{4}/4)dx = 1/4 [tan^{-1}x + x^{3}/3 – x]Hence, I = x

^{4}/4 tan^{-1}x – 1/4 [tan^{-1}x + x^{3}/3 – x] + c

### Question 49. ∫xsinxcos2xdx

**Solution:**

Given that, I = ∫xsinxcos2xdx

= 1/2 ∫x(2sinxcos2x)dx

= 1/2 ∫x(sin(x + 2x) – sin(2x – x))dx

= 1/2 ∫x(sin3x – sinx)dx

= 1/2[x](sin3x – sinx)dx – ∫ (1)(sin3x – sinx)dx)dx]

= 1/2 [x((-cos3x)/3 + cosx) – ∫(-(cos3x)/3 + cosx)dx]

Hence, I = 1/2 [-x (cos3x)/3 + xcosx + 1/9 sin3x – sinx] + c

### Question 50. ∫(tan^{-1}x^{2})xdx

**Solution:**

Given that, I = ∫(tan

^{-1}x^{2})xdxLet us considered x

^{2}= t2xdx = dt

I = 1/2∫tan

^{-1}tdt= 1/2∫1tan

^{-1}tdt= 1/2 [tan

^{-1}t∫dt – (∫1/(1 + t^{2})∫dt)dt]= 1/2 [t × tan

^{-1}t – ∫t/(1 + t^{2}) dt]= 1/2 t × tan

^{-1}t – 1/4∫2t/(1 + t^{2}) dt= 1/2 t × tan

^{-1}t – 1/4 log|1 + t^{2}| + cHence, I = 1/2 x

^{2}tan^{-1}x^{2}– 1/4 log|1 + x^{4}| + c

### Question 51. ∫xdx/√(1 – x^{2})

**Solution:**

Given that, I = ∫xdx/√(1 – x

^{2})Let first function be sin

^{-1}x and second function be x/√(1 – x^{2}).Now, first we find the integral of the second function,

∫xdx/√(1 – x

^{2})Now, put t = 1 – x

^{2}Then dt = -2xdx

Therefore,

∫ xdx/√(1 – x

^{2}) = -1/2 ∫dt/√t = -√t = -√(1 – x^{2})Hence,

∫(xsin

^{-1}x)/√(1 – x^{2}) dx= (sin

^{-1}x)(-√(1 – x^{2}) – ∫1/√(1 – x^{2}) * (-√(1 – x^{2}))dx= -√(1 – x

^{2}) sin^{-1}x + x + c= x – √(1 – x

^{2}) sin^{-1}x + c

### Question 52. ∫sin^{3}√x dx

**Solution:**

Given that, I = ∫sin

^{3}√x dxLet us considered √x = t

x = t

^{2}dx = 2tdt

I = 2∫ tsin

^{3}tdt= 2∫t((3sint – sin3t)/4)dt

= 1/2 ∫t(3sint – sin3t)dt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 1/2 [t(-3cost + 1/3 cos3t) – ∫(-3cost + (cos3t)/3)dt]

= 1/2 [(-9tcost + tcos3t)/3 – {-3sint + (sin3t)/9}] + c

= 1/2 [(-9tcost + tcos3t)/3 + (27sint – 3sin3t)/9] + c

= 1/18[-27tcost + 3tcos3t + 27sint – 3sin3t] + c

Hence, I = 1/18[3√x cos3√x + 27sin√x – 27√x cos√x – 3sin3√x] + c

### Question 53. ∫ xsin^{3}xdx

**Solution:**

Given that, I = ∫ xsin

^{3}xdx= ∫x((3sinx – sin3x)/4)dx

= 1/4 ∫x(3sinx – sin3x)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

= 1/4 [x∫ (3sinx – sin3x)dx – ∫(1)(3sinx – sin3x)dx)dx]

= 1/4 [x(-3cosx + (cos3x)/3) – ∫(-3cosx + (cos3x)/3)dx]

= 1/4 [-3xcosx + (xcos3x)/3 + 3sinx – (sin3x)/9] + c

Hence, I = 1/36[3xcos3x – 27xcosx + 27sinx – sin3x] + c

### Question 54. ∫cos^{3}√x dx

**Solution:**

Given that, I = ∫cos

^{3}√x dxLet us considered x = t²

dx = 2tdt

= 2∫tcos

^{3}tdt= 2∫t((3cost + cos3t)/4)dt

= 1/2 ∫t(3cost + cos3t)dt

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 1/2 [t(3sint + 1/3 sin3t) + ∫(1 × 3sint + (sin3t)/3)dt]

= 1/2 [t((9sint + sin3t)/3) + 3cost(cos3t)/9] + c

= 1/18[27tsint + 3tsin3t + 9cost + cos3t] + c

Hence, I = 1/18[27√x sin√x + 3√x sin3√x + 9cos√x + cos3√x] + c

### Question 55. ∫xcos^{3}xdx

**Solution:**

Given that, I = ∫xcos

^{3}xdx= ∫x((3cosx + cos3x)/4)dx

= 1/4 ∫x(3cosx + cos3x)dx

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 1/4 [x∫(3cosx + cos3x)dx – ∫(1)(3cosx + cos3x)dx)dx]

= 1/4 [x(3sinx + (sin3x)/3) – ∫ (3sinx + (sin3x)/3)dx]

= 1/4 [3xsinx + (xsin3x)/3 + 3cosx + (cos3x)/9] + c

Hence, I = (3xsinx)/4 + (xsin3x)/12 + (3cosx)/4 + (cos3x)/36 + c

### Question 56. ∫tan^{-1}√((1 – x)/(1 + x))

**Solution:**

Given that, I = ∫tan

^{-1}√((1 – x)/(1 + x))Let us considered x = cosθ

dx = -sinθdθ

I = ∫ tan

^{-1}(tanθ/2)(-sinθ)dθ=-1/2 ∫θsinθdθ

Let θ = u and sinθdθ = v

So that sinθ = ∫vdθ

Using integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = -1/2 (-θcosθ – ∫-cosθdθ)

= -1/2(-θcosθ + sinθ)+c

= -1/2 (-θcosθ + √(1 – cos

^{2}θ)) + c= -1/2 (-xcos

^{-1}x + √(1 – x^{2})) + c

### Question 57. ∫sin^{-1}√(x/(a + x)) dx

**Solution**:

Given that, I = ∫sin

^{-1}√(x/(a + x)) dxLet us considered x = atan

^{2}θdx = 2atanθsec

^{2}θdθI = ∫(sin

^{-1}√((atan^{2}θ)/(a + atan^{2}θ))(2atanθsec^{2}θ)dθ= ∫ (sin

^{-1}√((tan^{2}θ)/(sec^{2}θ)))(2atanθsec^{2}θ)dθ= ∫ sin

^{-1}(sinθ)(2atanθsec^{2}θ)dθ= ∫ 2θatanθsec

^{2}θdθ= 2a∣θ(tanθsec

^{2}θ)dθ)= ∫2θatanθsec

^{2}θdθ= 2a∫θ(tanθsec

^{2}θ)dθ= 2a[θ]tanθsec

^{2}θdθ – ∫(∫tanθsec^{2}θdθ)dθ]= 2a[θ (tan

^{2}θ)/2 – ∫(tan^{2}θ)/2 dθ]= aθtan

^{2}θ – 2a/2∫(sec^{2}θ – 1)dθ= aθtan

^{2}θ – atanθ + aθ + c= a(tan

^{-1}√(x/a)) x/a – a√(x/a) + atan^{-1}√(x/a) + cHence, I = xtan

^{-1}√(x/a) – √ax + atan^{-1}√(x/a) + c

### Question 58. ∫(x^{3} sin^{-1}x²)/√(1 – x^{4}) dx

**Solution:**

Given that, I = ∫(x

^{3}sin^{-1}x²)/√(1 – x^{4}) dxLet us considered sin

^{-1}x² = t(1/√(1 – x

^{4})(2x)dx = dtI = ∫(x² sin

^{-1}x²)/√(1 – x^{4}) xdx= ∫(sint)t dt/2

= 1/2∫tsintdt

= 1/2 [t∫sintdt – ∫(1∫sintdt)dt]

= 1/2 [t(-cost)dt – ∫(1∫(-cost))dt]

= 1/2[-tcost + sint] + c

Hence, I = 1/2 [x

^{2}– √(1 – x^{4}) sin^{(-1)}x^{2}] + c

### Question 59. ∫(x^{2} sin^{-1}x)/(1 – x^{2})^{3/2} dx

**Solution:**

Given that, I = ∫(x

^{2}sin^{-1}x)/(1 – x^{2})^{3/2}dxLet us considered sin

^{-1}x = t(1/√(1 – x

^{2}) dx = dtI = ∫(sin

^{2}t × t)/((1 – sin^{2}t)) dt= ∫(tsin

^{2}t)/(cos^{2}t) dt= ∫t × tan

^{2}tdt= ∫t(sec

^{2}t – 1)dt= ∫tsec

^{2}tdt – t^{2}/2 + c= t∫sec

^{2}tdt – ∫(1∫sec^{2}tdt)dt – t^{2}/2 + c= t × tant – ∫tantdt – t

^{2}/2 + c= t × tant – logsect – t

^{2}/2 + cHence, I = x/√(1 – x

^{2}) sin^{-1}x + log|1 – x^{2}| – 1/2 (sin^{-1}x)^{2}+ c

### Question 60. ∫cos^{-1}(1 – x^{2}/ 1 + x^{2}) dx

**Solution:**

Given that, I = ∫cos

^{-1}(1 – x^{2}/ 1 + x^{2}) dxLet us considered, x = tant

dx = sec

^{2}tdtI = ∫cos

^{-1}(1 – tan^{2}t/ 1 + tan^{2}t) sec^{2}tdt= ∫ 2t sec

^{2}tdtUsing integration by parts,

∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c

We get

I = 2[t∫sec

^{2}tdt – ∫(1 ∫sec^{2}tdt)dt]= 2[t tan

^{2}t – ∫tant dt]= 2[t tan

^{2}t – log sect] + c= 2[x tan

^{2}x – log √1 + x^{2}] + cHence, I = 2[xtan

^{2}x – log √1 + x^{2}] + c

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