# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.9 | Set 3

• Last Updated : 30 Apr, 2021

### Question 49. Solution:

Given that I = ……..(i)

Let us considered x + tan-1⁡x = t then,

On differentiating both side we get,

d(x + tan-1⁡x) = dt

(1 + 1/(1 + x2))dx = dt

((1 + x2 + 1)/(1 + x2))dx = dt

((x2 + 2))/((x2 + 1)) dx = dt

Now on putting x + tan-1⁡x = t and ((x2 + 2)/(x2 + 1))dx = dt in equation (i), we get

I = ∫5tdt

= 5t/(log⁡5) + c /(log⁡5) + c

Hence, I = /(log⁡5) + c

### Question 50. Solution:

Given that I = ……(i)

Let us considered msin-1⁡x = t then,

On differentiating both side we get,

d(msin-1x) = dt

m 1/√(1 – x2) dx = dt

dx/√(1 – x2) = dt/m

Now on putting msin-1⁡x = t and dx/√(1 – x2) = dt/m in equation (i), we get

I = ∫etdt/m

= 1/m et+c Hence, I = ### Question 51. ∫(cos⁡√x)/√x dx

Solution:

Given that I = ∫(cos⁡√x)/√x dx

Let us considered √x = t then,

On differentiating both side we get,

1/(2√x) dx = dt

Now,

= ∫ (cos⁡√x)/√x dx

= 2∫ cos⁡tdt2

= 2sin⁡t + c

Hence, I = 2sin⁡√x + c

### Question 52. ∫sin(tan-1x)/(1 + x2) dx

Solution:

Given that I = ∫sin(tan-1x)/(1 + x2) dx   ……(i)

Let us considered tan-1 = t then,

On differentiating both side we get,

d(tan-1⁡x) = dt

1/(1 + x2) dx = dt

Now on putting tan-1x = t and dx/(1 + x2) = dt in equation (i), we get

I = ∫ sin⁡tdt

= -cos⁡t + c

= -cos⁡(tan-1x) + c

Hence, I = -cos⁡(tan-1⁡x) + c

### Question 53. ∫(sin⁡(log⁡x))/x dx

Solution:

Given that I = ∫(sin⁡(log⁡x))/x dx   ……..(i)

Let us considered log⁡x = t then,

On differentiating both side we get,

d(log⁡x) = dt

1/x dx = dt

Now on putting log⁡x = t and 1/x dx = dt in equation (i), we get

I = ∫sin⁡tdt

= -cos⁡t + c

= -cos⁡(log⁡x) + c

Hence, I = -cos⁡(log⁡x) + c

### Question 54. Solution:

Given that I = Let us considered tan-1⁡x = t, then

On differentiating the above function we have,

1/(1 + x2) dx = dt = ∫emt × dt

= = emt/m

On Substituting the value of t, we

I = + c

### Question 55. ∫x/(√(x2 + a2) + √(x2 – a2)) dx

Solution:

Given that I = ∫x/(√(x2 + a2) + √(x2 – a2)) dx

= ∫ x/(√(x2 + a²) + √(x2 – a2)) × (√(x2 + a2) – √(x2 – a2 ))/(√(x2 + a2) – √(x2 – a2)) dx

= ∫ x(√(x2 + a2) – √(x2 – a2))/(x2 + a2 – x2 + a2) dx

= ∫ x/(2a2) (√(x2 + a2) – √(x2 – a2))dx

I = 1/(2a2) ∫x(√(x2 + a2) – √(x2 – a2))dx ……(i)

Let us considered x2 = t then,

On differentiating the above function we have,

d(x2) = dt

2xdx = dt

xdx = dt/2

Now on putting x2 = t and xdx = dt/2 in equation (i), we get

I = 1/(2a2) ∫(√(t + a2) – √(t – a2)) dx/2

Hence, I = 1/(4a2) [2/3 (t + a2)3/2 – 2/3 (t – a2)3/2] + c

### Question 56. ∫x(tan-1⁡x2)/(1 + x4) dx

Solution:

Given that I = ∫x(tan-1⁡x2)/(1 + x4) dx  ……..(i)

Let us considered tan-1x2 = t then,

On differentiating the above function we have,

d(tan-1x2) = dt

(1 × 2x)/(1 + (x2)2) dx = dt

(1 × x)/(1 + x4) dx = dt/2

Now on putting tan-1⁡x2 = t and x/(1 + x4) dx = dt/2 in equation (i), we get

I = ∫ t dx/2

= 1/2 ∫tdt

= 1/2 × t2/2 + c

= t2/4 + c – 1

= (tan-1x2)2/4 + c

Hence, I = 1/4 (tan-1⁡x2)2 + c

### Question 57. ∫(sin-1x)3/√(1 – x2) dx

Solution:

Given that I = ∫(sin-1x)3/√(1 – x2) dx ……(i)

Let us considered sin-1x = t then,

On differentiating the above function we have,

d(sin-1⁡x) = dt

1/√(1 – x2) dx = dt

Now on putting sin-1⁡x = t and 1/√(1 – x2) dx = dt in equation (i), we get

I = ∫t3 dt

= t4/4 + c

Hence, I = 1/4 (sin-1x)4 + c

### Question 58.∫(sin⁡(2 + 3log⁡x))/x dx

Solution:

Given that I = ∫(sin⁡(2 + 3log⁡x))/x dx  ……..(i)

Let us considered 2 + 3log⁡x = t then,

On differentiating the above function we have,

d(2 + 3log⁡x) = dt

3 1/x dx = dt

dx/x = dt/3

Now on putting 2 + 3log⁡x = t and dx/x = dt/3 in equation (i), we get

I = ∫sin⁡t dt/3

= 1/3(-cos⁡t) + ct

= -1/3 cos⁡(2 + 3log⁡x) + c

Hence, I = -1/3 cos⁡(2 + 3log⁡x) + c

### Question 59. Solution:

Given that I = ……(i)

Let us considered x2 = t then,

On differentiating the above function we have,

d(x2) = dt

2xdx = dt

xdx = dt/2

Now on putting x2 = t and xdx = dt/2 in equation (i), we get

I = ∫etdt/2

= 1/2 et+c

= 1/2 + c

Hence, I = 1/2 + c

### Question 60. ∫e2x/(1 + ex) dx

Solution:

Given that I = ∫e2x/(1 + ex) dx  …….(i)

Let us considered 1 + ex = t then,

On differentiating the above function we have,

d(1 + ex) = dt

exdx = dt

dx = dt/ex

Now on putting 1 + ex = t and dx = dt/ex in equation (i), we get

I = ∫e2x/t × dt/ex

= ∫ex/t dt

= ∫ (t – 1)/t dt

= ∫ (t/t – 1/t)dt

= t – log⁡|t| + c

= (1 + ex) – log⁡|1 + ex| + c

Hence, I = 1 + ex – log⁡|1 + ex| + c

### Question 61. ∫(sec2⁡√x)/√x dx

Solution:

Given that I = ∫(sec2√x)/√x dx   ……(i)

Let us considered √x = t then,

On differentiating the above function we have,

d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

dx = 2tdt            [√x = t])

Now on putting √x = t and dx = 2tdt in equation (i), we get

I = ∫ (sec2t)/t × 2tdt

= 2∫ sec2⁡tdt

= 2tan⁡t + c

= 2tan⁡√x + c

Hence, I = 2tan⁡√x + c

### Question 62. ∫tan32x sec⁡2x dx

Solution:

Given that I = ∫tan32x sec⁡2x dx

= tan22xtan⁡2x sec⁡2x

= (sec22x – 1)tan⁡2x sec⁡2x

= sec22x.tan⁡2xsec⁡2x – tan⁡2xsec⁡2x

= ∫ sec2⁡2xtan⁡2xsec⁡2xdx – ∫tan⁡2xsec⁡2xdx

= ∫ sec22xtan⁡2xsec⁡2xdx – (sec⁡2x)/2 + c

Let us considered sec⁡2x = t

2sec⁡2xtan⁡2xdx = dt

I = 1/2 ∫t2 dt – (sec⁡2x)/2 + c

I = t3/6 – (sec⁡2x)/2 + c

Hence, I = (sec⁡2x)3/6 – (sec⁡2x)/2 + c

### Question 63. ∫(x + √(x + 1))/(x + 2) dx

Solution:

Given that I = ∫(x+√(x+1))/(x+2) dx  …….(i)

Let us considered x + 1 = t2 then,

On differentiating the above function we have,

d(x + 1) = d(t2)

dx = 2tdt

Now on putting x + 1 = t2 and dx = 2tdt in equation (i), we get

I = ∫ (x + √(t2))/(x + 2) 2tdt

= 2∫((t2 – 1) + t)/((t2 – 1) + 2) × tdt   [x + 1 = t2]

= 2∫(t2 + t – 1)/(t2 + 1) tdt

= 2∫ (t3 + t2 – t)/(t2 + 1) dt

= 2[∫ t3/(t2 + 1) dt + ∫ t2/(t2 + 1) dt – ∫ t/(t2 + 1) dt]

I = 2[∫t3/(t2 + 1) dt + ∫t2/(t2 + 1) dt – ∫t/(t2 + 1) dt]    ……(ii)

Let I1 = ∫t3/(t2 + 1) dt

I2 = ∫t2/(t2 + 1) dt

and I3 = ∫t/(t2 + 1) dt

Now, I1 = ∫t3/(t2 + 1) dt

= ∫(t – t/(t2 + 1))dt

= t2/2 – 1/2 log⁡(t2 + 1)

I1 = t2/2 – 1/2 log⁡(t2 + 1) + c1    ……..(iii)

Since, I2 = ∫t2/(t2 + 1) dt

= ∫ (t2 + 1 – 1)/(t2 + 1) dt

= ∫(t2 + 1)/(t2 + 1) dt – ∫1/(t2 + 1) dt

= ∫dt – ∫1/(t2 + 1) dt

I2 = t – tan-1⁡(t2) + c2 ………….(iv)

and,

I3 = ∫t/(t2 + 1) dt

= 1/2 log⁡(1 + t2) + c3   ……..(v)

Using equations (ii), (iii), (iv) and (v), we get

I = 2[t2/2 – 1/2 log⁡(t2 + 1) + c1 + t-tan-1⁡(t2) + c2 – 1/2 log⁡(1 + t2) + c3]

= 2[t2/2 + t-tan-1⁡(t2) – log⁡(1 + t2) + c1 + c2 + c3]

= 2[t2/2 + t – tan-1⁡(t2) – log⁡(1 + t2) + c4  [Putting c1 + c2 + c3 = c4]

= t2 + 2t – 2tan-1(t2) – 2log⁡(1 + t2) + 2c4

= (x + 1) + 2√(x + 1) – 2tan-1(√(x + 1)) – 2log⁡(1 + x + 1) + 2c4

= (x + 1) + 2√(x + 1) – 2tan-1⁡(√(x + 1)) – 2log⁡(x + 2) + c [Putting 2c4 = c]

Hence, I = (x + 1) + 2√(x + 1) – 2tan-1⁡(√(x + 1)) – 2log⁡(x + 2) + c

### Question 64. Solution:

Given that I = …….(1)

Let us considered = t then

On differentiating the above function we have,

d( ) = dt × × 5x × (log⁡5)3 dx = dt dx = dt/((log⁡5)^3 ))

Now on putting = t and dx = dt/((log⁡5)^3 )) in equation (i), we get

I = ∫dt/((log⁡5)3)

= 1/((log⁡5)3) ∫dt

= t/((log⁡5)^3) + c

Hence, I = /((log⁡5)3) + c)

### Question 65. ∫1/(x√(x4 – 1)) dx

Solution:

Given that I = ∫1/(x√(x4 – 1)) dx   ……….(i)

Let us considered x2 = t then,

On differentiating the above function we have,

d(x2) = dt

2xdx = dt

dx = dt/2x

Now on putting x2 = t and dx = dt/2x in equation (i), we get

I = ∫ 1/(x√(t2 – 1)) × dt/2x

= 1/2 ∫ 1/(x2 √(t2 – 1)) dt

= 1/2 ∫ 1/(t√(t2 – 1)) dt

= 1/2 sec-1t + c

= 1/2 sec-1x2 + c

Hence, I = 1/2 sec-1x2 + c

### Question 66. ∫√(ex – 1) dx

Solution:

Given that I = ∫√(ex – 1) dx   ……..(i)

Let us considered ex – 1 = t2 then,

On differentiating the above function we have,

d(ex – 1) = dt(t2)

ex dx = 2tdt

dx = 2t/ex dt

dx = 2t/(t2 + 1) dt               [ex – 1 = t2

Now on putting ex – 1 = t² and dx = 2tdt/(t2 + 1) in equation (i), we get

I = ∫√(t2) × 2tdt/(t2 + 1)

= 2∫(t × t)/(t2 + 1) dt

= 2∫t2/(t2 + 1) dt

= 2∫(t2 + 1 – 1)/(t2 + 1) dt

= 2∫[(t2 + 1)/(t2 + 1) – 1/(t2 + 1)]dt

= 2∫dt – 2∫1/(t2 + 1) dt

= 2t – 2tan-1⁡(t) + c

= 2√(ex – 1) – 2tan-1(√(ex – 1)) + c

Hence, I = 2√(ex – 1) – 2tan-1⁡√(ex – 1) + c

### Question 67. ∫ 1/(x + 1)(x2 + 2x + 2) dx

Solution:

Given that I = ∫1/(x + 1)(x2 + 2x + 2) dx

= ∫1/(x + 1)((x + 1)2 + 1) dx

Let us considered x + 1 = tan u then,                               [tanu = Perpendicular/Base = (x + 1)/1]

On differentiating the above function we have,

dx = sec2u du                                                                   [Hypotenuses = √(x2 + 2x + 2)]

I = ∫sec2u/tanu(tan2u + 1) du

= ∫ cosu/sinu du

= log| sinu |+c

= log| sin(x + 1)| + c                                         [As we know, sin(x + 1) = P/H = (x + 1)/√(x2 + 2x + 2)]

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

### Question 68. ∫x5/√(1 + x3) dx

Solution:

Given that I = ∫ x5/(√(1 + x3)) dx   …….(i)

Let us considered 1 + x3 = t2, then

On differentiating the above function we have,

d(1 + x3) = d(t2)

3x2 dx = 2t * dt

dx = 2t dt/3x2

Now on putting 1 + x3 = t2 and dx = 2tdt/3x2 in equation (i), we get

I = ∫ x5/√t2 * 2t/3x2 dt

= ∫x5/t * 2t/3x2 dt

= 2/3∫x3 dt

= 2/3 ∫( t2 – 1) dt

= 2/3[t3/3 – 2t/3] + c

Hence, I = 2/9(1 + x3)3/2 – 2 √(1 + x2)/3 + c

### Question 69. ∫4x3 √(5 – x2) dx

Solution:

Given that I = ∫4x3 √(5 – x2) dx  ……(i)

Let us considered 5 – x2 = t2 then,

On differentiating the above function we have,

d(5 – x2) = t2

-2xdx = 2tdt

dx = (-t)/x dt

Now on putting 5 – x2 = t2 and dx = (-t)/x dt in equation (i), we get

I = ∫4x3 √(t2) × (-t)/x dt

= -4∫ x2 t × tdt

= -4∫(5 – t2) t2 dt         [5 – x2 = t2]

= -4∫(5t2 – t5)dt

= -20×t3/3 + 4 t5/5 + c

= (-20)/3 × t3 + 4/5 × t5 + c

= (-20)/3 × (5 – x2)3/2 + 4/5 × (5 – x2)5/2 + c

I = (-20)/3 × (5 -x2)3/2 + 4/5 × (5 – x2)5/2 + c

### Question 70. ∫1/(√x + x) dx

Solution:

Given that I = ∫1/(√x + x) dx  ……..(i)

Let us considered √x = t then,

On differentiating the above function we have,

d(√x) = dt

1/(2√x) dx = dt

dx = 2√x dt

Now on putting √x = t and 2√x dt = dx in equation (i), we get

I = ∫1/(t + t2) 2t × dt  [Since √x = t and x = t2]

= ∫2t/(t(1 + t)) dt

= 2∫t/(1 + t) dt

= 2log⁡|1 + t| + c

= 2log⁡|1 + √x| + c

Hence, I = 2log⁡|1 + √x|+c

### Question 71. ∫1/(x2 (x4 + 1)3/4) dx

Solution:

Given that I = 1/(x2 (x4+1)3/4)

Multiplying and dividing by x-3, we obtain

(x-3/(x2x-3 (x4+ 1)3/4) = (x-3 (x4 + 1)-3/4/(x2x-3))

= (x4 + 1)-3/4/(x5(x4)-3/4

= 1/x5 ((x4 + 1)/x4)-3/4

= 1/x5 (1 + 1/x4)-3/4

Let us considered 1/x4 = t

-4/x5 dx = dt

1/x5 dx = -dt/4

I = ∫ 1/x5(1 + 1/x4)-3/4 dx

= -1/4 ∫(1 + t)-3/4 dt

= -1/4 [(1 + t)1/4)/(1/4)] + c

= -1/4(1 + 1/x4)1/4/(1/4) + c

Hence, I = -(1 + 1/x4)1/4 + c

### Question 72. ∫(sin⁡5x)/(cos4x) dx

Solution:

Given that I = ∫(sin⁡5x)/(cos4⁡x) dx  ……(i)

Let us considered cos⁡x = t then,

On differentiating the above function we have,

d(cos⁡x) = dt

-sin⁡xdx = dt

dx = -dt/(sin⁡x)

Now on putting  cos⁡x = t and dx = -dt/(sin⁡x) in equation (i), we get

I = ∫(sin5⁡x)/t4 × -dt/(sin⁡x)

= -∫(sin4⁡x)/t4 dt

= -∫(1 – cos2x)2/t4 dt

= -∫(1 – t2)2/t4 dt

= -∫(1 + t4 – 2t2)/t4dt

= -∫(1/t4 + t4/t4 – (2t2)/t4)dt

= -∫(t-4 + 1 – 2t-2)dt

= -[t-3/(-3) + t – 2 t-1/(-1)] + c

= 1/3 * 1/t3– t – 1/t + c

Hence, I = 1/3 * 1/cos3x – cosx -2/cosx + c

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