Class 12 RD Sharma Solutions- Chapter 20 Definite Integrals – Exercise 20.4 Part B

Evaluate of each of the following integrals (1-46):

Question 1. 

Solution:

We have,

Let

—— 1

So,,

———— 2

Hence, by adding 1 and 2 ..

Question 2. ,

Solution:

We have,

Let, I=—– 1

So,—— 2

Adding 1 & 2 ——-

Question 3. ,

Solution:

We have ,

Let,

So,

————— 2

Adding 1 and 2 ——–

Question 4. 

Solution:

Let,—— 1

———- 2

Adding 1 & 2 —–

Question 5. 

Solution:

Let,———– (1)

So,

Question 6. 

Solution:

———- 1

—– 2

Adding 1 & 2 ——-

Question 7. 

Solution:

Let,

Let,

Now, x=0 ,, then

——————– 1

So,

————- 2

Adding (1) and (2) —————-

Question 8. 

Solution:

Let,

then

———— (1)

——————- (2)

Adding (1) & (2) ————

Question 9.  

Solution:

Let,

Question 10. 

Solution:

Let,

Equating coefficients, we get

So,

Question 11. 

Solution:

————- (1)

—————— (2)

Adding (1) & (2) —————-

Question 12. 

Solution:

Let ,————– 1

So,

Let,

As, x=0, t=1 ; x=π , t=-1

Hence,

Question 13. 

Solution:

Let,

Question 14. 

Solution:

we have,

=

Since, f(x) = f(-x) , f(x) is an even function.

————– 1

—————- 2

Adding 1 and 2 —————-

Now, let

Putting 2x=t, we get

Question 15. 

Solution:

Let,

Question 16. 

Solution:

We have , I=\int\limits_0^{π} \frac{x}{1+cos\alpha sinx}dx ———- 1

[Tex]=\int\limits_0^{π} \frac{(π-x)}{1+cos\alpha sinx}dx [/Tex]——- 2

Adding 1 and 2 —-

substituting s

when x=0 , t=0 ; x=π ,

Question 17. 

Solution:

Let,

Question 18. 

Solution:

Question 19. 

Solution:

Question 20. 

Solution:

Question 21. 

Solution:

Now,

Let cosx=t

sinx dx=-dt

[Tex]I= \frac{π }{8}[\fracπ 4+\fracπ 4] dt[/Tex]

Question 22. 

Solution:

—————— 1

———— 2

Adding 1 & 2 ————-

Let ,

Question 23. 

Solution:

Let,

Here, f(x)=-f(x)

Hence, f(x) is odd function

Question 24. 

Solution:

We have,  is an even function.

Question 25. 

Solution:

we have,

Since,

this is an odd function

Question 26. 

Solution:

we have,

sin²x is even function

Hence,

Question 27. 

Solution:

Question 28. 

Solution:

we have ,

Let,

Then,

Question 29. 

Solution:

Put cosx = t then -sinx dx = dt

Question 30. 0" title="Rendered by QuickLaTeX.com" height="60" width="261" style="vertical-align: -24px;">

Solution:

Let

is an odd function

Question 31. 

Solution:

Question 32. 

Solution:

Substitute π+x=u then dx=du

Question 33. 

Solution:

Let, 

Question 34. 

Solution:

Applying the property , 

Thus,

–

Question 35. 

Solution:

Let,

Question 36. 

Solution:

[Tex][\because \int\limits_0^{2a} f(x)dx= 2\int\limits_0^{a} f(x)dx, f(2a-x)=f(x)][/Tex]

let tanx = v

dv = sec²xdx

Question 37. 

Solution:

Putthen

x=0 ⇒ t=0 and x=π ⇒

Question 38. 

Solution:

we know,

Also here,

f(x) = f(2π -x)

So,

Question 39. 

Solution:

then,

Question 40. If f is an integrable function such that f(2a-x)=f(x), then prove that

Solution:

We have ,

Then,

Let , 2a-t =x then dx=-dt

if t=a ⇒x=a

if t=2a ⇒ x=0

[Tex]=2\int\limits_{0}^{a}f(x)dx[/Tex]

Hence Proved.

Question 41. If, prove that

Solution:

We have, 

Let 2a-t=x then dx=-dt

t=a , x=a ; t=2a , x=0

Question 42. If f is an integrable function, show that

(i)

Solution:

we have ,

clearly f(x²) is an even function .

So,

(ii)

Solution:

clearly , xf(x²) is odd function .

So,

Question 43. If f(x) is a continuous function defined on [0,2a] . Then, prove that

Solution:

We have from LHS,

substituting

we get,

Question 44. If f(a+b-x) = f(x), then prove that

Solution:

——————[ Given that f(a+b-x) = f(x) ]

Question 45. If f(x) is a continuous function defined on [-a,a], then prove that

Solution:

we have ,

Let, x=-t, then dx=-dt

x=-a ⇒ t=a

x=0 ⇒ t=0

Hence, Proved.

Question 46. Prove that:

Solution: