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

**Evaluate the following definite integrals:**

**Question 1. **

**Solution:**

We have,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**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 ofis.

**Question 3. **

**Solution:**

We have,

I =

Let 9x

^{2}– 1 = t, so we have,=> 18x dx = dt

=> 3x dx = dt/6

Now, the lower limit is, x = 1

=> t = 9x

^{2}– 1=> t = 9 (1)

^{2}– 1=> t = 9 – 1

=> t = 8

Also, the upper limit is, x = 2

=> t = 9x

^{2}– 1=> t = 9 (2)

^{2}– 1=> t = 36 – 1

=> t = 35

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**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 ofis.

**Question 5. **

**Solution:**

We have,

I =

Let a

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

=> x dx = t dt

Now, the lower limit is, x = 0

=> t

^{2}= a^{2 }+ x^{2}=> t

^{2}= a^{2 }+ 0^{2}=> t

^{2}= a^{2}=> t = a

Also, the upper limit is, x = a

=> t

^{2}= a^{2}+ x^{2}=> t

^{2 }= a^{2}+ a^{2}=> t

^{2}= 2a^{2}=> t = √2 a

So, the equation becomes,

I =

I =

I =

I = √2a – a

I = a (√2 – 1)

Therefore, the value ofis a (√2 – 1).

**Question 6. **

**Solution:**

We have,

I =

Let e

^{x}= t. So, we have=> e

^{x}dx = dtNow, the lower limit is, x = 0

=> t = e

^{x}=> t = e

^{0}=> t = 1

Also, the upper limit is, x = a

=> t = e

^{x}=> t = e

^{1}=> t = e

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value ofis.

**Question 7. **

**Solution:**

We have,

I =

Let x

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

Now, the lower limit is, x = 0

=> t = x

^{2}=> t = 0

^{2}=> t = 0

Also, the upper limit is, x = 1

=> t = x

^{2}=> t = 1

^{2}=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**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 ofis sin (log 3).

**Question 9. **

**Solution:**

We have,

I =

Let x

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

Now, the lower limit is, x = 0

=> t = x

^{2}=> t = 0

^{2}=> t = 0

Also, the upper limit is, x = 1

=> t = x

^{2}=> t = 1

^{2}=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**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 ofis.

**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 ofis.

**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 ofis.

**Question 13. **

**Solution:**

We have,

I =

Let 1 + cos θ = t

^{2}. So, we have=> – sin θ dθ = 2t dt

=> sin θ dθ = –2t dt

Now, the lower limit is, θ = 0

=> t

^{2}= 1 + cos θ=> t

^{2}= 1 + cos 0=> t

^{2}= 1 + 1=> t

^{2 }= 2=> t = √2

Also, the upper limit is, θ = π/2

=> t

^{2}= 1 + cos θ=> t

^{2 }= 1 + cos π/2=> t

^{2}= 1 + 0=> t

^{2}= 1=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value ofis.

**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 ofis.

**Question 15. **

**Solution:**

We have,

I =

Let tan

^{–1}x = t. So, we have=> (1/1+x

^{2}) dx = dtNow, 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 ofis.

**Question 16. **

**Solution:**

We have,

I =

Let x + 2 = t

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

Now, the lower limit is, x = 0

=> t

^{2}= x + 2=> t

^{2}= 0 + 2=> t

^{2}= 2=> t = √2

Also, the upper limit is, x = 2

=> t

^{2}= x + 2=> t

^{2}= 2 + 2=> t

^{2 }= 4=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**Question 17. **

**Solution:**

We have,

I =

Let x = tan t. So, we have

=> dx = sec

^{2}t dtNow, 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 ofis.

**Question 18. **

**Solution:**

We have,

I =

Let sin

^{2}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 = sin

^{2}x=> t = sin

^{2}0=> t = 0

Also, the upper limit is, x = π/2

=> t = sin

^{2}x=> t = sin

^{2}π/2=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value ofis.

**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 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/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 ofis.

**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 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/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 ofis.

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