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

• Last Updated : 16 May, 2021

### Question 41. ∫cos-1⁡((1 – x2)/(1 + x2))dx

Solution:

Given that, I = ∫cos-1⁡((1 – x2)/(1 + x2))dx)

Let us considered x = tan⁡t

dx = sec²tdt

I = ∫cos-1⁡((1 – tan2t)/(1 + tan2⁡t)) sec2tdt

= ∫cos-1(cos⁡2t)sec2tdt

= ∫2tsec2⁡tdt

Using integration by parts,

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

We get

I = 2[t∫sce2tdt – ∫(1∫sec2⁡tdt)dt]

= 2[t × tan2t – ∫tan⁡tdt]

= 2[t × tan2t – log⁡sec⁡t] + c

= 2[xtan-1x – log⁡√(1 + x2)] + c

Hence, I = 2xtan-1x – log⁡|1 + x2| + c

### Question 42. ∫tan-1⁡(2x/(1 – x2))dx

Solution:

Given that, I = ∫tan-1⁡(2x/(1 – x2))dx

Let us considered x = tan⁡θ

dx = sec2θdθ

I = ∫tan-1⁡((2tan⁡θ)/(1 – tan2θ)) sec2θdθ

= ∫tan-1⁡(tan⁡2θ)sec2θdθ

= ∫2θsec2θdθ

Using integration by parts,

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

We get

I = 2[θ∫sec2θdθ – ∫(1∫ sec2⁡θdθ)dθ]

= 2[θtan⁡θ – ∫tan⁡θdθ]

= 2[θtan⁡θ – log⁡sec⁡θ] + c

= 2[xtan-1⁡x – log⁡√(1 + x2)] + c

Hence, I = 2xtan-1⁡x – log⁡|1 + x2| + c

### Question 43. ∫(x + 1)log⁡xdx

Solution:

Given that, I = ∫(x + 1)log⁡xdx

Using integration by parts,

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

We get

I = log⁡x∫ (x + 1)dx – ∫(1/x ∫(x + 1)dx)dx

= (x2/2 + x)log⁡x – ∫1/x (x2/2 + x)dx

= (x2/2 + x)log⁡x – 1/2 ∫xdx – ∫dx

= (x + x2/2)log⁡x – 1/2 × x2/2 – x + c

Hence, I = (x + x2/2)log⁡x – 1/2 × x2/2 – x + c

### Question 44. ∫ x2 tan-1xdx

Solution:

Given that, I = ∫ x2 tan-1⁡xdx

Using integration by parts,

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

We get

I = tan-1⁡x∫x2 dx – ∫(1/(1 + x2) ∫x2 dx) dx

= tan-1⁡x(x3/3) – 1/3∫x3/(1 + x2) dx

= 1/3 x3 tan-1⁡x – 1/3 ∫(x – x/(1 + x2))dx

= 1/3 x3 tan-1⁡x – 1/3 × x2/2 + 1/3 ∫x/(1 + x2) dx

Hence, I = 1/3 x3tan-1x – 1/6 x2 + 1/6 log⁡|1 + x2| + c

### Question 45. ∫(elogx + sin⁡x) cos⁡xdx

Solution:

Given that, I = ∫(elogx + sin⁡x)cos⁡xdx

= ∫(x + sin⁡x)cos⁡xdx

= ∫xcos⁡xdx + ∫sin⁡xcos⁡xdx

= [x∫cos⁡xdx – ∫(1]cos⁡xdx)dx] + 1/2 ∫sin⁡2xdx

= [xsin⁡x – ∫ sin⁡xdx] + 1/2 (-(cos⁡2x)/2) + c

I = xsin⁡x+cos⁡x – 1/4 cos⁡2x + c

= xsin⁡x + cos⁡x – 1/4 [1 – 2sin2⁡x] + c

= xsin⁡x + cos⁡x – 1/4 + 1/2 sin2x + c

= xsin⁡x + cos⁡x – 1/4 + 1/2 sin2x + c

Hence, I = xsin⁡x + cos⁡x + 1/2 sin2⁡x + d   [d = c-/4]

### Question 46. ∫((xtan-1⁡x))/(1 + x2)3/2 dx

Solution:

Given that, I = ∫((xtan-1⁡x))/(1 + x2)3/2dx

Let us considered tan-1⁡x = t

1/(1 + x2) dx = dt

I = ∫(t tan⁡t)/√(1 + tan2⁡t) dt

= ∫(t × tan⁡t)/(sec⁡t) dt

= ∫t (sin⁡t)/(cos⁡t) cos⁡tdt

= ∫tsin⁡tdt

Using integration by parts,

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

We get

I = [t]sin⁡tdt – ∫(1)sin⁡tdt)dt]

= [-tcos⁡t + ∫cos⁡tdt]

= [-tcos⁡t + sin⁡t] + c

= -(tan-1⁡x)/√(1 + x2) + x/√(1 + x2) + c

Hence, I = -(tan-1⁡x)/√(1 + x2) + x/√(1 + x2) + c

### Question 47. ∫ tan-1(√x)dx

Solution:

Given that, I = ∫ tan-1(√x)dx

Let us considered x = t2

dx = 2tdt

I = ∫2ttan-1tdt

Using 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 + t2) ∫tdt)dt]

= 2[t2/2 tan-1⁡t – ∫t2/2(1 + t2)dt]

= t2 tan-1⁡t – ∫(t2 + 1 – 1)/(1 + t2)dt

= t2 tan-1⁡t – ∫(1 – 1/(1 + t2))dt

= t2 tan-1t – t + tan-1⁡t + c

= (t2 + 1) tan-1⁡t – t + c

Hence, I = (x + 1)tan-1⁡√x – √x + c

### Question 48. ∫x3 tan-1xdx

Solution:

Given that, I = ∫x3 tan-1xdx

= tan-1⁡x∫x3dx – (∫(dtan-1⁡x)/dx (∫x3 dx)dx)

= tan-1⁡x x4/4 – (∫1/(1 + x2) (x4/4)dx)

= tan-1⁡x x4/4 – (∫1/(1 + x2) (x4/4)dx)

= tan-1⁡x x4/4 – (∫1/(1 + x2) (x4/4)dx)

∫ 1/(1 + x2) (x4/4)dx = 1/4 [∫1/(1 + x2) dx + (x2 – 1)dx]

∫ 1/(1 + x2) (x4/4)dx = 1/4 [tan-1⁡x + x3/3 – x]

Hence, I = x4/4 tan-1⁡x – 1/4 [tan-1⁡x + x3/3 – x] + c

### Question 49. ∫xsin⁡xcos⁡2xdx

Solution:

Given that, I = ∫xsin⁡xcos⁡2xdx

= 1/2 ∫x(2sin⁡xcos⁡2x)dx

= 1/2 ∫x(sin⁡(x + 2x) – sin⁡(2x – x))dx

= 1/2 ∫x(sin⁡3x – sin⁡x)dx

= 1/2[x](sin⁡3x – sin⁡x)dx – ∫ (1)(sin⁡3x – sin⁡x)dx)dx]

= 1/2 [x((-cos⁡3x)/3 + cos⁡x) – ∫(-(cos⁡3x)/3 + cos⁡x)dx]

Hence, I = 1/2 [-x (cos⁡3x)/3 + xcos⁡x + 1/9 sin⁡3x – sin⁡x] + c

### Question 50. ∫(tan-1x2)xdx

Solution:

Given that, I = ∫(tan-1⁡x2)xdx

Let us considered x2 = t

2xdx = dt

I = 1/2∫tan-1tdt

= 1/2∫1tan-1tdt

= 1/2 [tan-1⁡t∫dt – (∫1/(1 + t2)∫dt)dt]

= 1/2 [t × tan-1⁡t – ∫t/(1 + t2) dt]

= 1/2 t × tan-1⁡t – 1/4∫2t/(1 + t2) dt

= 1/2 t × tan-1⁡t – 1/4 log⁡|1 + t2| + c

Hence, I = 1/2 x2 tan-1⁡x2 – 1/4 log⁡|1 + x4| + c

### Question 51. ∫xdx/√(1 – x2)

Solution:

Given that, I = ∫xdx/√(1 – x2)

Let first function be sin-1⁡x and second function be x/√(1 – x2).

Now, first we find the integral of the second function,

∫xdx/√(1 – x2)

Now, put t = 1 – x2

Then dt = -2xdx

Therefore,

∫ xdx/√(1 – x2) = -1/2 ∫dt/√t = -√t = -√(1 – x2)

Hence,

∫(xsin-1x)/√(1 – x2) dx

= (sin-1⁡x)(-√(1 – x2) – ∫1/√(1 – x2) * (-√(1 – x2))dx

= -√(1 – x2) sin-1⁡x + x + c

= x – √(1 – x2) sin-1⁡x + c

### Question 52. ∫sin3√x dx

Solution:

Given that, I = ∫sin3√x dx

Let us considered √x = t

x = t2

dx = 2tdt

I = 2∫ tsin3⁡tdt

= 2∫t((3sin⁡t – sin⁡3t)/4)dt

= 1/2 ∫t(3sin⁡t – sin⁡3t)dt

Using integration by parts,

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

We get

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

= 1/2 [(-9tcos⁡t + tcos⁡3t)/3 – {-3sin⁡t + (sin⁡3t)/9}] + c

= 1/2 [(-9tcos⁡t + tcos⁡3t)/3 + (27sin⁡t – 3sin⁡3t)/9] + c

= 1/18[-27tcos⁡t + 3tcos⁡3t + 27sin⁡t – 3sin⁡3t] + c

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

### Question 53. ∫ xsin3xdx

Solution:

Given that, I = ∫ xsin3⁡xdx

= ∫x((3sin⁡x – sin⁡3x)/4)dx

= 1/4 ∫x(3sin⁡x – sin⁡3x)dx

Using integration by parts,

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

We get

= 1/4 [x∫ (3sin⁡x – sin⁡3x)dx – ∫(1)(3sin⁡x – sin⁡3x)dx)dx]

= 1/4 [x(-3cos⁡x + (cos⁡3x)/3) – ∫(-3cos⁡x + (cos⁡3x)/3)dx]

= 1/4 [-3xcos⁡x + (xcos⁡3x)/3 + 3sin⁡x – (sin⁡3x)/9] + c

Hence, I = 1/36[3xcos⁡3x – 27xcos⁡x + 27sin⁡x – sin⁡3x] + c

### Question 54. ∫cos3√x dx

Solution:

Given that, I = ∫cos3√x dx

Let us considered x = t²

dx = 2tdt

= 2∫tcos3⁡tdt

= 2∫t((3cos⁡t + cos⁡3t)/4)dt

= 1/2 ∫t(3cos⁡t + cos⁡3t)dt

Using integration by parts,

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

We get

I = 1/2 [t(3sin⁡t + 1/3 sin⁡3t) + ∫(1 × 3sin⁡t + (sin⁡3t)/3)dt]

= 1/2 [t((9sin⁡t + sin⁡3t)/3) + 3cos⁡t(cos⁡3t)/9] + c

= 1/18[27tsin⁡t + 3tsin⁡3t + 9cos⁡t + cos⁡3t] + c

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

### Question 55. ∫xcos3xdx

Solution:

Given that, I = ∫xcos3⁡xdx

= ∫x((3cos⁡x + cos⁡3x)/4)dx

= 1/4 ∫x(3cos⁡x + cos⁡3x)dx

Using integration by parts,

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

We get

I = 1/4 [x∫(3cos⁡x + cos⁡3x)dx – ∫(1)(3cos⁡x + cos⁡3x)dx)dx]

= 1/4 [x(3sin⁡x + (sin⁡3x)/3) – ∫ (3sin⁡x + (sin⁡3x)/3)dx]

= 1/4 [3xsin⁡x + (xsin⁡3x)/3 + 3cos⁡x + (cos⁡3x)/9] + c

Hence, I = (3xsin⁡x)/4 + (xsin⁡3x)/12 + (3cos⁡x)/4 + (cos⁡3x)/36 + c

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

Solution:

Given that, I = ∫tan-1√((1 – x)/(1 + x))

Let us considered x = cos⁡θ

dx = -sin⁡θdθ

I = ∫ tan-1⁡(tan⁡θ/2)(-sin⁡θ)dθ

=-1/2 ∫θsin⁡θdθ

Let θ = u and sin⁡θdθ = v

So that sin⁡θ = ∫vdθ

Using integration by parts,

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

We get

I = -1/2 (-θcos⁡θ – ∫-cos⁡θdθ)

= -1/2(-θcos⁡θ + sin⁡θ)+c

= -1/2 (-θcos⁡θ + √(1 – cos2⁡θ)) + c

= -1/2 (-xcos-1⁡x + √(1 – x2)) + c

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

Solution:

Given that, I = ∫sin-1⁡√(x/(a + x)) dx

Let us considered x = atan2θ

dx = 2atan⁡θsec2⁡θdθ

I = ∫(sin-1⁡√((atan2⁡θ)/(a + atan2⁡θ))(2atan⁡θsec2θ)dθ

= ∫ (sin-1√((tan2θ)/(sec2θ)))(2atan⁡θsec2θ)dθ

= ∫ sin-1(sin⁡θ)(2atan⁡θsec2θ)dθ

= ∫ 2θatan⁡θsec2θdθ

= 2a∣θ(tan⁡θsec2⁡θ)dθ)

= ∫2θatan⁡θsec2θdθ

= 2a∫θ(tan⁡θsec2⁡θ)dθ

= 2a[θ]tan⁡θsec2θdθ – ∫(∫tan⁡θsec2⁡θdθ)dθ]

= 2a[θ (tan2⁡θ)/2 – ∫(tan2θ)/2 dθ]

= aθtan2θ – 2a/2∫(sec2θ – 1)dθ

= aθtan2θ – atan⁡θ + aθ + c

= a(tan-1⁡√(x/a)) x/a – a√(x/a) + atan-1⁡√(x/a) + c

Hence, I = xtan-1⁡√(x/a) – √ax + atan-1⁡√(x/a) + c

### Question 58. ∫(x3 sin-1⁡x²)/√(1 – x4) dx

Solution:

Given that, I = ∫(x3 sin-1x²)/√(1 – x4) dx

Let us considered sin-1⁡x² = t

(1/√(1 – x4)(2x)dx = dt

I = ∫(x² sin-1⁡x²)/√(1 – x4) xdx

= ∫(sin⁡t)t dt/2

= 1/2∫tsin⁡tdt

= 1/2 [t∫sin⁡tdt – ∫(1∫sin⁡tdt)dt]

= 1/2 [t(-cost)dt – ∫(1∫(-cost))dt]

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

Hence, I = 1/2 [x2 – √(1 – x4) sin(-1)⁡x2] + c

### Question 59. ∫(x2 sin-1⁡x)/(1 – x2)3/2 dx

Solution:

Given that, I = ∫(x2 sin-1x)/(1 – x2)3/2dx

Let us considered sin-1⁡x = t

(1/√(1 – x2) dx = dt

I = ∫(sin2t × t)/((1 – sin2t)) dt

= ∫(tsin2t)/(cos2t) dt

= ∫t × tan2tdt

= ∫t(sec2⁡t – 1)dt

= ∫tsec2⁡tdt – t2/2 + c

= t∫sec2tdt – ∫(1∫sec2tdt)dt – t2/2 + c

= t × tan⁡t – ∫tan⁡tdt – t2/2 + c

= t × tan⁡t – log⁡sec⁡t – t2/2 + c

Hence, I = x/√(1 – x2) sin-1x + log⁡|1 – x2| – 1/2 (sin-1x)2 + c

### Question 60. ∫cos-1(1 – x2/ 1 + x2) dx

Solution:

Given that, I = ∫cos-1(1 – x2/ 1 + x2) dx

Let us considered, x = tant

dx = sec2tdt

I = ∫cos-1(1 – tan2t/ 1 + tan2t) sec2tdt

= ∫ 2t sec2tdt

Using integration by parts,

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

We get

I = 2[t∫sec2tdt – ∫(1 ∫sec2tdt)dt]

= 2[t tan2t – ∫tant dt]

= 2[t tan2t – log sect] + c

= 2[x tan2x – log √1 + x2] + c

Hence, I = 2[xtan2x – log √1 + x2] + c

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