# Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.2 | Set 3

**Evaluate the following definite integrals:**

**Question 42. **

**Solution:**

We have,

I =

Let 5 – 4 cos θ = t. So, we have

=> 4 sin θ dθ = dt

=> sin θ dθ = dt/4

Now, the lower limit is, θ = 0

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos 0

=> t = 5 – 4

=> t = 1

Also, the upper limit is, θ = π

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos π

=> t = 5 + 4

=> t = 9

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I = 9√3 – 1

Therefore, the value ofis 9√3 – 1.

**Question 43. **

**Solution:**

We have,

I =

I =

I =

I =

I =

Let tan 2θ = t. So, we have

=> 2 sec

^{2}2θ dθ = dt=> sec

^{2}2θ dθ = dt/2Now, the lower limit is, θ = 0

=> t = tan 2θ

=> t = tan 0

=> t = 0

Also, the upper limit is, θ = π/6

=> t = tan 2θ

=> t = tan π/3

=> t = √3

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 44. **

**Solution:**

We have,

I =

Let = t. So, we have

=> = dt

Now, the lower limit is, x = 0

=> t =

=> t =

=> t = 0

Also, the upper limit is, x =

=> t =

=> t =

=> t = π

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 45. **

**Solution:**

We have,

I =

Let 1 + log x = t. So, we have

=> 1/x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + log x

=> t = 1 + log 0

=> t = 1

Also, the upper limit is, x = 2

=> t = 1 + log x

=> t = 1 + log 2

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 46. **

**Solution:**

We have,

I =

I =

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 47. **

**Solution:**

We have,

I =

Let 30 – x

^{3/2}= t. So, we have=> = dt

=> = – dt

Now, the lower limit is, x = 4

=> t = 30 – x

^{3/2}=> t = 30 – 4

^{3/2}=> t = 30 – 8

=> t = 22

Also, the upper limit is, x = 9

=> t = 30 – x

^{3/2}=> t = 30 – 9

^{3/2}=> t = 30 – 27

=> t = 3

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 48. **

**Solution:**

We have,

I =

Let cos x = t. So, we have

=> – sin x dx = dt

=> sin x dx = –dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π

=> t = cos x

=> t = cos π

=> t = –1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 49. **

**Solution:**

We have,

I =

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 50. **

**Solution:**

We have,

I =

I =

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 51. **

**Solution:**

We have,

I =

On using integration by parts, we get,

I =

I =

Let cos

^{-1}x = t. So, we have=> = dt

Now, the lower limit is, x = 0

=> t = cos

^{-1}x=> t = cos

^{-1}0=> t = π/2

Also, the upper limit is, x = 1

=> t = cos

^{-1}x=> t = cos

^{-1}1=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I = π – 2

Therefore, the value ofis π – 2.

**Question 52. **

**Solution:**

We have,

I =

Let x = a tan

^{2}t. So, we have=> dx = 2a tan t sec

^{2 }t dtNow, the lower limit is, x = 0

=> a tan

^{2}t = x=> a tan

^{2}t = 0=> tan t = 0

=> t = 0

Also, the upper limit is, x = a

=> a tan

^{2}t = x=> a tan

^{2}t = a=> tan

^{2}t = 1=> tan t = 1

=> t = π/4

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 53. **

**Solution:**

We have,

I =

I =

I =

I =

I =

Let cot x/2 = t. So, we have

=> = dt

Now, the lower limit is, x = π/3

=> t = cot x/2

=> t = cot π/6

=> t = √3

Also, the upper limit is, x = π/2

=> t = cot x/2

=> t = cot π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I = 1

Therefore, the value ofis 1.

**Question 54. **

**Solution:**

We have,

I =

Let x

^{2}= a^{2}cos 2t. So, we have=> 2x dx = – 2a

^{2}sin 2t dtNow, the lower limit is, x = 0

=> a

^{2}cos 2t = x^{2}=> a

^{2}cos 2t = 0=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = a

=> a

^{2}cos 2t = x^{2}=> a

^{2}cos 2t = a^{2}=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 55. **

**Solution:**

We have,

I =

Let x = a cos 2t. So, we have

=> dx = –2a sin 2t

Now, the lower limit is, x = –a

=> a cos 2t = x

=> a cos 2t = –a

=> cos 2t = –1

=> 2t = π

=> t = π/2

Also, the upper limit is, x = a

=> a cos 2t = x

=> a cos 2t = a

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = πa

Therefore, the value ofis πa.

**Question 56. **

**Solution:**

We have,

I =

Let cos x = t. So, we have

=> – sin x dx = dt

=> sin x dx = –dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I = – log 2 + 2 log 3 + 0 – 2 log 2

I = 2 log 3 – 3 log 2

I = log 9 – log 8

I = log 9/8

Therefore, the value ofis log 9/8.

**Question 57. **

**Solution:**

We have,

I =

I =

I =

Let sin

^{2}x = t. So, we have=> 2 sin x cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin

^{2}x=> t = sin

^{2}0=> t = 0

Also, the upper limit is, x = π/2

=> t = sin

^{2}x=> t = sin

^{2}π/2=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 58. **

**Solution:**

We have,

I =

Let x = sin t. So, we have

=> dx = cos t dt

Now, the lower limit is, x = 0

=> sin t = x

=> sin t = 0

=> t = 0

Also, the upper limit is, x = 1/2

=> sin t = x

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I =

I =

I =

I =

I =

Let tan t = u. So, we have

=> sec

^{2}t dt = duNow, the lower limit is, t = 0

=> u = tan t

=> u = tan 0

=> t = 0

Also, the upper limit is, t = π/6

=> u = tan t

=> u = tan π/6

=> t = 1/√3

So, the equation becomes,

I =

I =

I =

Therefore, the value ofis.

**Question 59. **

**Solution:**

We have,

I =

Let 1/x

^{2}– 1 = t. So, we have=> –2/x

^{3}dx = dtNow, the lower limit is, x = 1/3

=> t = 1/x

^{2}– 1=> t = 9 – 1

=> t = 8

Also, the upper limit is, x = 1

=> t = 1/x

^{2}– 1=> t = 1 – 1

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I = 6

Therefore, the value ofis 6.

**Question 60. **

**Solution:**

We have,

I =

I =

Let tan x = t. So, we have

=> sec

^{2}x dx = dtNow, the lower limit is, x = 0

=> t = tan t

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/4

=> t = tan t

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

Let t

^{3 }= u. So, we have=> 3t

^{2}dt = du=> t

^{2}dt = du/3Now, the lower limit is, t = 0

=> u = t

^{3}=> u = 0

^{3}=> u = 0

Also, the upper limit is, t = 1

=> u = t

^{3}=> u = 1

^{3}=> u = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 61. **

**Solution:**

We have,

I =

I =

I =

I =

I =

Let cos x = t. So, we have

=> – sin x dx = dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 62. **

**Solution:**

We have,

I =

I =

I =

Let cos x/2 + sin x/2 = t. So, we have

=> (cos x/2 – sin x/2) dx = 2 dt

Now, the lower limit is, x = 0

=> t = cos x/2 + sin x/2

=> t = cos 0 + sin 0

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x/2 + sin x/2

=> t = cos π/2 + sin π/2

=> t = 1/√2 + 1/√2

=> t = √2

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

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