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

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 â€“ sin x)

=> sin x = sin x (A â€“ B) + cos x (A + B)

On comparing both sides, we get

A â€“ B = 1 and A + B = 0

On solving, we get A = 1/2 and B = â€“1/2.

Therefore, the expression becomes,

I =

I =

I =

I =

Therefore, the value of  is .

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 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 = Ï€

=> t = tan x/2

=> t = tan Ï€/2

=> t = âˆž

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 23.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 24.

Solution:

We have,

I =

Let sinâ€“1 x = t. So, we have

=>  = dt

Now, the lower limit is, x = 0

=> t = sinâ€“1 x

=> t = sinâ€“1 0

=> t = 0

Also, the upper limit is, x = 1/2

=> t = sinâ€“1 x

=> t = sinâ€“1 1/2

=> t = Ï€/6

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 25.

Solution:

We have,

I =

I =

I =

I =

I =

Let sinx â€“ cosx = t. So, we have

=> (cos x + sin x) dx = dt

Now, the lower limit is, x = 0

=> t = sinx â€“ cosx

=> t = sin 0 â€“ cos 0

=> t = 0 â€“ 1

=> t = â€“1

Also, the upper limit is, x = Ï€/4

=> t = sinx â€“ cosx

=> t = sin Ï€/4 â€“ cos Ï€/4

=> t = sin Ï€/4 â€“ sin Ï€/4

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 26.

Solution:

We have,

I =

I =

I =

I =

Let tan x = t. So, we have

=> sec2 x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = Ï€/4

=> t = tan x

=> t = tan Ï€/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

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 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 = Ï€

=> t = tan x/2

=> t = tan Ï€/2

=> t = âˆž

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 28.

Solution:

We have,

I =

I =

I =

I =

Let tan x = t. So, we have

=> sec2 x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = Ï€/2

=> t = tan x

=> t = tan Ï€/2

=> t = âˆž

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 29.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 30.

Solution:

We have,

I =

Let tanâ€“1 x = t. So, we have

=>  = dt

Now, the lower limit is, x = 0

=> t = tanâ€“1 x

=> t = tanâ€“1 0

=> t = 0

Also, the upper limit is, x = 1

=> t = tanâ€“1 x

=> t = tanâ€“1 1

=> t = Ï€/4

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

Question 31.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

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 of  is .

Question 33.

Solution:

We have,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 34.

Solution:

We have,

I =

Let 1 + x2 = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + x2

=> t = 1 + 02

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = Ï€

=> t = 1 + x2

=> t = 1 + 12

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I = 1

Therefore, the value of  is 1.

Question 35.

Solution:

We have,

I =

Let x â€“ 4 = t3. So, we have

=> dx = 3t2 dt

Now, the lower limit is, x = 4

=> t3 = x â€“ 4

=> t3 = 4 â€“ 4

=> t3 = 0

=> t = 0

Also, the upper limit is, x = 12

=> t3 = x â€“ 4

=> t3 = 12 â€“ 4

=> t3 = 8

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 36.

Solution:

We have,

I =

On using integration by parts, we get

I =

I =

I =

I =

I =

I = Ï€ + 0 â€“ 0 â€“ 0 â€“ 2

I = Ï€ â€“ 2

Therefore, the value of  is Ï€ â€“ 2.

Question 37.

Solution:

We have,

I =

Let x = cos 2t. So, we have

=> dx = â€“ 2 sin 2t dt

Now, the lower limit is, x = 0

=> cos 2t = x

=> cos 2t = 0

=> 2t = Ï€/2

=> t = Ï€/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 of  is .

Question 38.

Solution:

We have,

I =

I =

I =

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

=> (1 â€“ 1/x2)dx = dt

Now, the lower limit is, x = 0

=> t = x + 1/x

=> t = âˆž

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 of  is .

Question 39.

Solution:

We have,

I =

Let x5 + 1 = t. So, we have

=> 5x4 dx = dt

Now, the lower limit is, x = â€“1

=> t = x5 + 1

=> t = (â€“1)5 + 1

=> t = â€“1 + 1

=> t = 0

Also, the upper limit is, x = 1

=> t = x5 + 1

=> t = (1)5 + 1

=> t = 1 + 1

=> t = 2

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value of  is .

Question 40.

Solution:

We have,

I =

I =

Let tan x = t. So, we have

=> sec2 x dx = dt

Now, the lower limit is, x = 0

=> t = tan x

=> t = tan 0

=> t = 0

Also, the upper limit is, x = Ï€/2

=> t = tan x

=> t = tan Ï€/2

=> t = âˆž

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value of  is .

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 = Ï€/4

=> u = sin 2t

=> u = sin Ï€/2

=> u = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!

Previous
Next