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

**Evaluate the following definite integrals:**

**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 ofis.

**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 sec

^{2}x/2 dx = dt=> sec

^{2}x/2 dx = 2 dtNow, 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 ofis.

**Question 23. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**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 ofis.

**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 ofis.

**Question 26. **

**Solution:**

We have,

I =

I =

I =

I =

Let tan x = t. So, we have

=> sec

^{2}x dx = dtNow, 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 ofis.

**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 sec

^{2}x/2 dx = dt=> sec

^{2}x/2 dx = 2 dtNow, 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 ofis.

**Question 28. **

**Solution:**

We have,

I =

I =

I =

I =

Let tan x = t. So, we have

=> sec

^{2 }x dx = dtNow, 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 ofis.

**Question 29. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**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 ofis.

**Question 31. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 32. **

**Solution:**

We have,

I =

On using integration by parts, we get

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 33. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 34. **

**Solution:**

We have,

I =

Let 1 + x

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

Now, the lower limit is, x = 0

=> t = 1 + x

^{2}=> t = 1 + 0

^{2}=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π

=> t = 1 + x

^{2}=> t = 1 + 1

^{2}=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I = 1

Therefore, the value ofis 1.

**Question 35. **

**Solution:**

We have,

I =

Let x – 4 = t

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

^{2}dtNow, the lower limit is, x = 4

=> t

^{3}= x – 4=> t

^{3}= 4 – 4=> t

^{3}= 0=> t = 0

Also, the upper limit is, x = 12

=> t

^{3}= x – 4=> t

^{3}= 12 – 4=> t

^{3}= 8=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**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 ofis π – 2.

**Question 37. **

**Solution:**

We have,

I =

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,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 38. **

**Solution:**

We have,

I =

I =

I =

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

=> (1 – 1/x

^{2})dx = dtNow, 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 =

I =

I =

Therefore, the value ofis.

**Question 39. **

**Solution:**

We have,

I =

Let x

^{5}+ 1 = t. So, we have=> 5x

^{4}dx = dtNow, the lower limit is, x = –1

=> t = x

^{5}+ 1=> t = (–1)

^{5}+ 1=> t = –1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x

^{5}+ 1=> t = (1)

^{5}+ 1=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 40. **

**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 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 ofis.

**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 ofis.

Attention reader! Don’t stop learning now. Join the **First-Step-to-DSA Course for Class 9 to 12 students ****, **specifically designed to introduce data structures and algorithms to the class 9 to 12 students