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

• Last Updated : 30 Jun, 2021

### 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 – x3/2 = t. So, we have

=>  = dt

=>  = – dt

Now, the lower limit is, x = 4

=> t = 30 – x3/2

=> t = 30 – 43/2

=> t = 30 – 8

=> t = 22

Also, the upper limit is, x = 9

=> t = 30 – x3/2

=> t = 30 – 93/2

=> 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 .

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