# Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.2 | Set 2

• Last Updated : 30 Jun, 2021

### Question 21.

Solution:

We have,

I =

Let sin x = A (sin x + cos x) + B

=> sin x = A (sin x + cos x) + B (cos x – sin x)

=> sin x = sin x (A – B) + cos x (A + B)

On comparing both sides, we get

A – B = 1 and A + B = 0

On solving, we get A = 1/2 and B = –1/2.

Therefore, the expression becomes,

I =

I =

I =

I =

Therefore, the value of  is .

### Question 22.

Solution:

We have,

I =

On putting cos x =  and sin x = , we get,

I =

I =

I =

I =

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

=> 1/2 sec2 x/2 dx = dt

=> sec2 x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π

=> t = tan x/2

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 23.

Solution:

We have,

I =

I =

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I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 24.

Solution:

We have,

I =

Let sin–1 x = t. So, we have

=>  = dt

Now, the lower limit is, x = 0

=> t = sin–1 x

=> t = sin–1 0

=> t = 0

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

=> t = sin–1 x

=> t = sin–1 1/2

=> t = π/6

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 25.

Solution:

We have,

I =

I =

I =

I =

I =

Let sinx – cosx = t. So, we have

=> (cos x + sin x) dx = dt

Now, the lower limit is, x = 0

=> t = sinx – cosx

=> t = sin 0 – cos 0

=> t = 0 – 1

=> t = –1

Also, the upper limit is, x = π/4

=> t = sinx – cosx

=> t = sin π/4 – cos π/4

=> t = sin π/4 – sin π/4

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 26.

Solution:

We have,

I =

I =

I =

I =

Let tan x = t. So, we have

=> sec2 x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/4

=> t = tan x

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

### Question 27.

Solution:

We have,

I =

On putting cos x = , we get

I =

I =

I =

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

=> 1/2 sec2 x/2 dx = dt

=> sec2 x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π

=> t = tan x/2

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 28.

Solution:

We have,

I =

I =

I =

I =

Let tan x = t. So, we have

=> sec2 x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 29.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 30.

Solution:

We have,

I =

Let tan–1 x = t. So, we have

=>  = dt

Now, the lower limit is, x = 0

=> t = tan–1 x

=> t = tan–1 0

=> t = 0

Also, the upper limit is, x = 1

=> t = tan–1 x

=> t = tan–1 1

=> t = π/4

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

### Question 31.

Solution:

We have,

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I =

I =

I =

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I =

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Therefore, the value of  is .

### Question 32.

Solution:

We have,

I =

On using integration by parts, we get

I =

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I =

Therefore, the value of  is .

### Question 33.

Solution:

We have,

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Therefore, the value of  is .

### Question 34.

Solution:

We have,

I =

Let 1 + x2 = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + x2

=> t = 1 + 02

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π

=> t = 1 + x2

=> t = 1 + 12

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

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I =

I =

I = 1

Therefore, the value of  is 1.

### Question 35.

Solution:

We have,

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Let x – 4 = t3. So, we have

=> dx = 3t2 dt

Now, the lower limit is, x = 4

=> t3 = x – 4

=> t3 = 4 – 4

=> t3 = 0

=> t = 0

Also, the upper limit is, x = 12

=> t3 = x – 4

=> t3 = 12 – 4

=> t3 = 8

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

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I =

Therefore, the value of  is .

### Question 36.

Solution:

We have,

I =

On using integration by parts, we get

I =

I =

I =

I =

I =

I = π + 0 – 0 – 0 – 2

I = π – 2

Therefore, the value of  is π – 2.

### Question 37.

Solution:

We have,

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Let x = cos 2t. So, we have

=> dx = – 2 sin 2t dt

Now, the lower limit is, x = 0

=> cos 2t = x

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = 1

=> cos 2t = x

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

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Therefore, the value of  is .

### Question 38.

Solution:

We have,

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

=> (1 – 1/x2)dx = dt

Now, the lower limit is, x = 0

=> t = x + 1/x

=> t = ∞

Also, the upper limit is, x = 1

=> t = x + 1/x

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

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Therefore, the value of  is .

### Question 39.

Solution:

We have,

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Let x5 + 1 = t. So, we have

=> 5x4 dx = dt

Now, the lower limit is, x = –1

=> t = x5 + 1

=> t = (–1)5 + 1

=> t = –1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x5 + 1

=> t = (1)5 + 1

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

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Therefore, the value of  is .

### Question 40.

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 x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x

=> t = tan π/2

=> t = ∞

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 41.

Solution:

We have,

I =

Let sin 2t = u. So, we have

=> 2 cos 2t dt = du

=> cos 2t dt = du/2

Now, the lower limit is, x = 0

=> u = sin 2t

=> u = sin 0

=> u = 0

Also, the upper limit is, x = π/4

=> u = sin 2t

=> u = sin π/2

=> u = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

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