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 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 e^x = t. So, we have

=> e^x dx = dt

Now, the lower limit is, x = 0

=> t = e^x

=> t = e⁰

=> t = 1

Also, the upper limit is, x = a

=> t = e^x

=> 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 .

Summary

This exercise typically covers:

• Basic evaluation of definite integrals
• Application of fundamental theorem of calculus
• Integrals involving algebraic and trigonometric functions
• Use of substitution method in definite integrals
• Properties of definite integrals