# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.9 | Set 2

• Last Updated : 30 Apr, 2021

### Question 25. ∫(1 + cos⁡x)/((x + sin⁡x)3) dx

Solution:

Given that I = ∫(1 + cos⁡x)/((x + sin⁡x)3) dx    …..(i)

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Let us considered x + sin⁡x = t then,

On differentiating both side we get,

d(x + sin⁡x) = dt

(1 + cos⁡x)dx = dt

Now on putting x + sin⁡x = t and (1 + cos⁡x)dx = dt in equation (i), we get

I = ∫ dt/t3

= ∫ t-3 dt

= t-2/-2 + c

= -1/(2t2) + c

= (-1)/(2(x + sin⁡x)2) + c

Hence, I = (-1)/(2(x + sin⁡x)2) + c

### Question 26. ∫(cos⁡x – sin⁡x)/(1 + sin⁡2x) dx

Solution:

Given that I = (cos⁡x – sin⁡x)/(1 + sin⁡2x)

= (cos⁡x – sin⁡x)/((sin2⁡x + cos2⁡x) + 2sin⁡xcos⁡x)  [Because sin2⁡x + cos2⁡x = 1 and sin⁡2x = 2sin⁡xcos⁡x]

Let us considered sin⁡x + cos⁡x = t

On differentiating both side we get,

(cos⁡x – sin⁡x)dx = dt

Now,

= ∫(cos⁡x – sin⁡x)/(1 + sin⁡2x) dx

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

= ∫dt/t2

= ∫t-2 dt

= -t-1 + c

= -1/t + c

Hence, I = (-1)/(sin⁡x + cos⁡x) + c

### Question 27. ∫(sin⁡2x)/(a + bcos⁡2x)2 dx

Solution:

Given that I = ∫(sin⁡2x)/((a + bcos⁡2x)2) dx   ……(i)

Let us considered a + bcos⁡2x = t then,

On differentiating both side we get,

(a + bcos⁡2x) = dt

b(-2sin⁡2x)dx = dt

sin⁡2x dx = -dt/2b

Now on putting a + bcos⁡2x = t and sin⁡2xdx = -dt/2b in equation (i), we get

I = ∫1/t2 × (-dt)/2b

= (-1)/2b ∫ t-2 dt

= -1/2b (-1t-1) + c

= 1/2bt + c

= 1/(2b(a + bcos⁡2x)) + c

Hence, I = 1/(2b(a + bcos⁡2x)) + c

### Question 28. ∫(log⁡x2)/x dx

Solution:

Given that I = ∫(log⁡x2)/x dx  ……..(i)

Let us considered log⁡x = t then,

On differentiating both side we get,

d(log⁡x) = dt

1/x dx = dt

dx/x = dt

Now, I = ∫(log⁡x2)/x dx

= ∫(2log⁡x)/x dx

= 2∫(log⁡x)/x dx …….(ii)

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

I = 2∫tdt

= (2t2)/2 + c

= t2 + c

I = (log⁡x)2 + c

### Question 29. ∫(sin⁡x)/(1 + cos⁡x)2 dx

Solution:

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

Let us considered 1 + cos⁡x = t then,

On differentiating both side we get,

d(1 + cos⁡x) = dt

-sin⁡xdx = dt

sin⁡xdx = -dt

Now on putting 1 + cos⁡x = t and sin⁡dx = -dt in equation (i), we get

I = ∫(-dt)/t2

= -∫t-2dt

= -(-1t-1) + c

= 1/t + c

= 1/(1 + cos⁡x) + c

Hence, I = 1/(1 + cos⁡x) + c

### Question 30. ∫cot⁡x log⁡ sin⁡x dx

Solution:

Given that I = ∫cot⁡x log⁡ sin⁡x dx

Let us considered log⁡ sin⁡x = t

1/(sin⁡x).cos⁡xdx = dt

cot⁡x dx = dt

∫cot⁡x log⁡ sin⁡x dx = ∫tdt

= t2/2 + c

= 1/2(log⁡sin⁡x)2 + c

### Question 31. ∫sec⁡x.log⁡(sec⁡x + tan⁡x)dx

Solution:

Given that I = ∫sec⁡x.log⁡(sec⁡x + tan⁡x)dx  ……..(i)

Let us considered log⁡(sec⁡x + tan⁡x) = t then,

On differentiating both side we get,

d[log⁡(sec⁡x + tan⁡x)] = dt

sec⁡x dx = dt    [Since, d/dx(log⁡(sec⁡x + tan⁡x)) = sec⁡x]

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

I = ∫tdt

= t2/2 + c

= 1/2[log⁡(sec⁡x + tan⁡x)]2 + c

Hence, I = 1/2[log⁡(sec⁡x + tan⁡x)]2 + c

### Question 32. ∫cosec⁡x log⁡(cosec⁡x – cot⁡x)dx

Solution:

Given that I = ∫cosec⁡x log⁡(cosec⁡x – cot⁡x)dx    ……(i)

Let us considered log⁡(cosec⁡x – cot⁡x) = t then,

On differentiating both side we get,

dx[log⁡(cosec⁡x – cot⁡x)] = dt

cosec⁡x dx = dt     [ Since, d/dx(log⁡(cosec⁡x – cot⁡x)) = cosec⁡x]

Now on putting log⁡(cosec⁡x – cot⁡x) = t and cosec⁡xdx = dt in equation (i), we get

I = ∫tdt

= t2/2 + c

Hence, I = 1/2[log⁡(cosec⁡x – cot⁡x)]2 + c

### Question 33. ∫x3cos⁡x4 dx

Solution:

Given that I = ∫x3cos⁡x4 dx   …….(i)

Let us considered x4 = t then,

On differentiating both side we get,

dx(x4) = dt

4x3dx = dt

x3 = dt/4

Now on putting x4 = t and x3dx = dt/4 in equation (i), we get

I = ∫ cos⁡t dt/4

= 1/4sint + c

Hence, I = 1/4sinx4 + c

### Question 34. ∫x3 sin⁡x4 dx

Solution:

Given that I = ∫x3sin⁡x4 dx   …..(i)

Let us considered x4 = t then,

On differentiating both side we get,

d(x4) = dt

4x3dx = dt

x3 = dt/4

Now on putting x4 = t and x3dx = dt/4 in equation (i), we get

I = ∫sin⁡t dt/4

= 1/4 ∫sin⁡t dt

= -1/4 cos⁡t + c

Hence, I = -1/4 cos⁡x4 + c

### Question 35. ∫(xsin-1x2)/√(1 – x4) dx

Solution:

Given that I = ∫(xsin-1⁡x2)/√(1 – x4) dx  …….(i)

Let us considered sin-1x2 = t then,

On differentiating both side we get,

d(sin-1x2) = dt

2x × 1/√(1 – x4) dx = dt

x/√(1 – x4) dx = dt/2

Now on putting sin-1x2 = t and x/√(1 – x4) dx = dt/2 in equation (i), we get

I = ∫t dt/2

= 1/2 × t2/2 + c

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

Hence, I = 1/4 (sin-1x2)2 + c

### Question 36. ∫x3sin⁡(x4 + 1)dx

Solution:

Given that I = ∫x3 sin⁡(x4 + 1)dx ……..(i)

Let us considered x4 + 1 = t then,

On differentiating both side we get,

d(x4 + 1) = dt

x3 dx = dt/4

Now on putting x4 + 1 = t and x3dx = dt/4 in equation (i), we get

I = ∫ sin⁡t dt/4

= -1/4 cos⁡t + c

= -1/4 cos⁡(x4 + 1) + c

Hence, I = -1/4 cos⁡(x4 + 1) + c

### Question 37. ∫(x + 1)ex/(cos2⁡(xex) dx

Solution:

Given that I = ∫((x + 1)ex)/(cos2⁡(xex)) dx   ……(i)

Let us considered xex = t then,

On differentiating both side we get,

d(xex) = dt

(ex + xex)dx = dt

(x + 1)exdx = dt

Now on putting xex = t and (x + 1)exdx = dt in equation (i), we get

I = ∫dt/(cos2⁡t)

= ∫ sec2tdt

= tan⁡t + c

= tan⁡(xex) + c

Hence, I = tan⁡(xex) + c

### Question 38. Solution:

Given that I = ……..(i)

Let us considered = t then,

On differentiating both side we get,

d( ) = dt

3x2 dx = dt

x2 dx = dt/3

Now on putting = t and x2 dx = dt/3 in equation (i), we get

I = ∫cos⁡t dt/3

= (sin⁡t)/3 + c

Hence, I = sin⁡( )/3 + c

### Question 39. ∫2xsec3(x2 + 3)tan⁡(x2 + 3)dx

Solution:

Given that I = ∫2xsec3(x2 + 3)tan⁡(x2 + 3)dx    ………(i)

Let us considered sec⁡(x2 + 3) = t then,

On differentiating both side we get,

d[sec⁡(x2 + 3)] = dt

2xsec⁡(x2 + 3)tan⁡(x2 + 3)dx = dt

Now on putting sec⁡(x2 + 3) = t and 2xsec⁡(x2 + 3)tan⁡(x2 + 3)dx = dt in equation (i), we get

I = ∫t2 dt

= t3/3 + c

= 1/3 [sec⁡(x2 + 3)]3 + c

Hence, I = 1/3 [sec⁡(x2 + 3)]3 + c

### Question 40. ∫(1 + 1/x)(x + log⁡x)2 dx

Solution:

Given that I = ((x + 1)(x + log⁡x)2)/x

= ((x + 1)/x)(x + log⁡x)2

= (1 + 1/x)(x + log⁡x)2

Let us considered (x + log⁡x) = t

On differentiating both side we get,

(1 + 1/x)dx = dt

Now,

I = ∫(1 + 1/x)(x + log⁡x)2 dx

= ∫t2 dt

= t3/3 + c

Hence, I = 1/3(x + log⁡x)3 + c

### Question 41. ∫tan⁡x sec2⁡x√(1 – tan2x) dx

Solution:

Given that I = ∫tan⁡x sec2⁡x√(1 – tan2x) dx      ………(i)

Let us considered 1 – tan2⁡x = t then,

On differentiating both side we get,

d(1 – tan2x) = dt

-2tan⁡x sec2⁡x dx = dt

tan⁡x sec2⁡x dx = (-dt)/2

Now on putting 1 – tan2x = t and tan⁡x sec2⁡x dx = -dt/2 in equation (i), we get

I = ∫√t × (-dt)/2

=-1/2 ∫t1/2 dt

=-1/2×t3/2/(3/2) + c

=-1/3 t3/2 + c

Hence, I = -1/3 [1 – tan2⁡x]3/2 + c

### Question 42.∫log⁡x (sin⁡(1 + (log⁡x)2)/x dx

Solution:

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

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

I = ∫sin⁡t × dt/2

= 1/2 ∫ sin⁡tdt

= -1/2 cos⁡t + c

= -1/2 cos⁡[1 + (log⁡x)2] + c

Hence, I = -1/2 cos⁡[1 + (log⁡x)2] + c

### Question 43.∫ 1/x2 × (cos2(1/x))dx

Solution:

Given that I = ∫ 1/x2 × (cos2⁡(1/x))dx ……(i)

Let us considered 1/x = t then,

On differentiating both side we get,

d(1/x) = dt

(-1)/x2dx = dt

1/x2 dx = -dt

Now on putting 1/x = t and 1/x2dx = -dt in equation (i), we get

I = ∫cos2t(-dt)

= -∫cos2tdt

= -∫(cos2t + 1)/2 dt

= -1/2 ∫cos⁡2t dt – 1/2 ∫dt

= -1/2 × (sin⁡2t)/2 – 1/2 t + c

= -1/4 sin⁡2t – 1/2 t + c

= -1/4 sin⁡2 × 1/x – 1/2 × 1/x + c

Hence, I = -1/4 sin⁡(2/x) – 1/2 (1/x) + c

### Question 44. ∫sec4x tan⁡x dx

Solution:

Given that I = ∫sec4x tan⁡x dx  ……(i)

Let us considered tan⁡x = t then,

On differentiating both side we get,

d (tan⁡x) = dt

sec2xdx = dt

dx = dt/sec2⁡x

Now on putting tan⁡x = t and dx = dt/(sec2⁡x) in equation (i), we get

I = ∫sec4x tan⁡x dt/(sec2⁡x)

= ∫ sec2x tdt

= ∫ (1 + tan2⁡x)tdt

= ∫(1 + t2)tdt

= ∫(t + t3)dt

= t2/2 + t4/4 + c

= (tan2⁡x)/2 + (tan4⁡x)/4 + c

Hence, I = 1/2 tan2⁡x + 1/4 tan4⁡x + c

### Question 45. ∫(e√x cos⁡(e√x ))/√x dx

Solution:

Given that I = ∫(e√x cos⁡(e√x))/√x dx  …….(i)

Let us considered e√x = t then,

On differentiating both side we get,

d(e√x) = dt

e√x(1/(2√x))dx = dt

e√x/√x dx = 2dt

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

I = ∫ cos⁡t × 2dt

= 2∫ cos⁡tdt

= 2sin⁡t + c

= 2sin⁡(e√x) + c

I = 2sin⁡(e√x) + c

### Question 46. ∫(cos5x)/(sin⁡x) dx

Solution:

Given that I = ∫(cos5⁡x)/(sin⁡x) dx    …..(i)

Let us considered sin⁡x = t then,

On differentiating both side we get,

d(sin⁡x) = dt

cos⁡x dx = dt

dx = dt/(cos⁡x)

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

I = ∫(cos5⁡x)/t × dt/(cos⁡x)

= ∫(cos4⁡x)/t dt

= ∫(1 – sin2⁡x)2/t dt

= ∫(1 – t2)2/t dt

= ∫(1 + t4 – 2t2)/t dt

= ∫1/t dt + ∫t4/t dt – 2∫t2/t dt

= log⁡|t| + t4/4 – (2t2)/2 + c

= log⁡|sin⁡x| + (sin4⁡x)/4 – sin2⁡x + c

Hence, I = 1/4 sin4x – sin2x + log⁡|sin⁡x| + c

### Question 47. ∫(sin⁡√x)/√x dx

Solution:

Given that I = ∫(sin⁡√x)/√x dx

Let us considered √x = t then,

On differentiating both side we get,

1/(2√x) dx = dt

1/√x dx = 2dt

Now,

I = ∫(sin⁡√x)/√x dx

= 2 ∫sint dt

= -2 cos⁡t + c

Hence, I = -2cos⁡√x + c

### Question 48. ∫((x + 1)ex)/(sin2(xex)) dx

Solution:

Given that I = ∫((x + 1)ex)/(sin2⁡(xex)) dx   …….(i)

Let us considered xex = t then,

d(xex) = dt

(xex + ex)dx = dt

(x + 1)exdx = dt

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

I = ∫dt/(sin2t)

= ∫cos⁡ec2t dt

= -cot + c

Hence, I = -cot⁡(xex) + c

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