# Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.2 | Set 1

• Last Updated : 30 Jun, 2021

### Question 1.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 2.

Solution:

We have,

I =

Let 1 + log x = t, so we have,

=> (1/x) dx = 2t dt

Now, the lower limit is, x = 1

=> t = 1 + log x

=> t = 1 + log 1

=> t = 1 + 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 =

I =

I =

I =

Therefore, the value of  is .

### Question 3.

Solution:

We have,

I =

Let 9x2 – 1 = t, so we have,

=> 18x dx = dt

=> 3x dx = dt/6

Now, the lower limit is, x = 1

=> t = 9x2 – 1

=> t = 9 (1)2 – 1

=> t = 9 – 1

=> t = 8

Also, the upper limit is, x = 2

=> t = 9x2 – 1

=> t = 9 (2)2 – 1

=> t = 36 – 1

=> t = 35

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

### Question 4.

Solution:

We have,

I =

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

I =

I =

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

=>  = dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 5.

Solution:

We have,

I =

Let a2 + x2 = t2. So, we have

=> 2x dx = 2t dt

=> x dx = t dt

Now, the lower limit is, x = 0

=> t2 = a2 + x2

=> t2 = a2 + 02

=> t2 = a2

=> t = a

Also, the upper limit is, x = a

=> t2 = a2 + x2

=> t2 = a2 + a2

=> t2 = 2a2

=> t = √2 a

So, the equation becomes,

I =

I =

I =

I = √2a – a

I = a (√2 – 1)

Therefore, the value of  is a (√2 – 1).

### Question 6.

Solution:

We have,

I =

Let ex = t. So, we have

=> ex dx = dt

Now, the lower limit is, x = 0

=> t = ex

=> t = e0

=> t = 1

Also, the upper limit is, x = a

=> t = ex

=> t = e1

=> t = e

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

### Question 7.

Solution:

We have,

I =

Let x2 = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = x2

=> t = 02

=> t = 0

Also, the upper limit is, x = 1

=> t = x2

=> t = 12

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 8.

Solution:

We have,

I =

Let log x = t. So, we have

=> (1/x) dx = dt

Now, the lower limit is, x = 1

=> t = log x

=> t = log 1

=> t = 0

Also, the upper limit is, x = 3

=> t = log x

=> t = log 3

So, the equation becomes,

I =

I =

I = sin (log 3) – sin 0

I = sin (log 3) – 0

I = sin (log 3)

Therefore, the value of  is sin (log 3).

### Question 9.

Solution:

We have,

I =

Let x2 = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = x2

=> t = 02

=> t = 0

Also, the upper limit is, x = 1

=> t = x2

=> t = 12

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 10.

Solution:

We have,

I =

Let x = a sin t. So, we have

=> dx = a cos t dt

Now, the lower limit is, x = 0

=> a sin t = x

=> a sin t = 0

=> sin t = 0

=> t = 0

Also, the upper limit is, x = a

=> a sin t = a

=> a sin t = a

=> sin t = 1

=> t = π/2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 11.

Solution:

We have,

I =

I =

Let sin Ø = t. So, we have

=> cos Ø dØ = dt

Now, the lower limit is, Ø = 0

=> t = sin Ø

=> t = sin 0

=> t = 0

Also, the upper limit is, Ø = π/2

=> t = sin Ø

=> t = sin π/2

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 12.

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 =

I =

Therefore, the value of  is .

### Question 13.

Solution:

We have,

I =

Let 1 + cos θ = t2. So, we have

=> – sin θ dθ = 2t dt

=> sin θ dθ = –2t dt

Now, the lower limit is, θ = 0

=> t2 = 1 + cos θ

=> t2 = 1 + cos 0

=> t2 = 1 + 1

=> t2 = 2

=> t = √2

Also, the upper limit is, θ = π/2

=> t2 = 1 + cos θ

=> t2 = 1 + cos π/2

=> t2 = 1 + 0

=> t2 = 1

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 14.

Solution:

We have,

I =

Let 3 + 4 sin x = t. So, we have

=> 0 + 4 cos x dx = dt

=> 4 cos x dx = dt

=> cos x dx = dt/4

Now, the lower limit is, x = 0

=> t = 3 + 4 sin x

=> t = 3 + 4 sin 0

=> t = 3 + 0

=> t = 3

Also, the upper limit is, x = π/3

=> t = 3 + 4 sin x

=> t = 3 + 4 sin π/3

=> t = 3 + 4 (√3/2)

=> t = 3 + 2√3

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

### Question 15.

Solution:

We have,

I =

Let tan–1 x = t. So, we have

=> (1/1+x2) dx = 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 t

=> t = tan–1 1

=> t = π/4

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 16.

Solution:

We have,

I =

Let x + 2 = t2. So, we have

=> dx = 2t dt

Now, the lower limit is, x = 0

=> t2 = x + 2

=> t2 = 0 + 2

=> t2 = 2

=> t = √2

Also, the upper limit is, x = 2

=> t2 = x + 2

=> t2 = 2 + 2

=> t2 = 4

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 17.

Solution:

We have,

I =

Let x = tan t. So, we have

=> dx = sec2 t dt

Now, the lower limit is, x = 0

=> tan t = x

=> tan x = 0

=> x = 0

Also, the upper limit is, x = 1

=> tan t = x

=> tan x = 1

=> x = π/4

So, the equation becomes,

I =

I =

I =

I =

On applying integration by parts method, we get

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 18.

Solution:

We have,

I =

Let sin2 x = t. So, we have

=> 2 sin x cos x = dt

=> sin x cos x = dt/2

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 =

Therefore, the value of  is .

### Question 19.

Solution:

We have,

I =

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

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/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 20.

Solution:

We have,

I =

On putting sin x = , we get

I =

I =

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/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

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