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 of  is 9√3 – 1.  

Question 43. 

Solution:

We have,

I = 

I = 

I = 

I = 

I = 

Let tan 2θ = t. So, we have

=> 2 sec² 2θ dθ = dt

=> sec² 2θ dθ = dt/2

Now, 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 of  is .

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 of  is .

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 of  is .

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 of  is .

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 of  is .

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 of  is .

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 of  is .

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 of  is .

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 of  is π – 2.

Question 52. 

Solution:

We have,

I = 

Let x = a tan² t. So, we have

=> dx = 2a tan t sec² t dt

Now, the lower limit is, x = 0

=> a tan² t = x

=> a tan² t = 0

=> tan t = 0

=> t = 0

Also, the upper limit is, x = a

=> a tan² t = x

=> a tan² t = a

=> tan² 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 of  is .

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 of  is 1.

Question 54. 

Solution:

We have,

I = 

Let x² = a² cos 2t. So, we have

=> 2x dx = – 2a² sin 2t dt

Now, the lower limit is, x = 0

=> a² cos 2t = x²

=> a² cos 2t = 0

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = a

=> a² cos 2t = x²

=> a² cos 2t = a²

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

Therefore, the value of  is .

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 of  is 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 of  is log 9/8.

Question 57. 

Solution:

We have,

I = 

I = 

I = 

Let sin² x = t. So, we have

=> 2 sin x cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin² x

=> t = sin² 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin² x

=> t = sin² π/2

=> t = 1

So, the equation becomes,

I = 

I = 

I  = 

I = 

I = 

I = 

I = 

I = 

Therefore, the value of  is .

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

=> sec² t dt = du

Now, 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 of  is .

Question 59. 

Solution:

We have,

I = 

Let 1/x² – 1 = t. So, we have

=> –2/x³ dx = dt

Now, the lower limit is, x = 1/3

=> t = 1/x² – 1

=> t = 9 – 1 

=> t = 8

Also, the upper limit is, x = 1

=> t = 1/x² – 1

=> t = 1 – 1

=> t = 0

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

I = 6

Therefore, the value of  is 6.

Question 60. 

Solution:

We have,

I = 

I = 

Let tan x = t. So, we have

=> sec² x dx = dt 

Now, 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 t³ = u. So, we have

=> 3t² dt = du

=> t² dt = du/3

Now, the lower limit is, t = 0

=> u = t³

=> u = 0³

=> u = 0

Also, the upper limit is, t = 1

=> u = t³

=> u = 1³

=> u = 1

So, the equation becomes,

I = 

I = 

I = 

I = 

I = 

I = 

I = 

Therefore, the value of  is .

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 of  is .

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 of  is .