# RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals

Here we provide RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 Definite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 20.2 Solutions Chapter 20 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 = a+ x2

=> t2 = a+ 02

=> t2 = a2

=> t = a

Also, the upper limit is, x = a

=> t2 = a2 + x2

=> t= 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 ex = t. So, we have

=> ex dx = dt

Now, the lower limit is, x = 0

=> t = ex

=> t = e0

=> t = 1

Also, the upper limit is, x = a

=> t = ex

=> 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

=> t= 2

=> t = √2

Also, the upper limit is, θ = π/2

=> t2 = 1 + cos θ

=> t= 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

=> t= 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

=> secx/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 .

### 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 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 = π

=> t = tan x/2

=> t = tan π/2

=> t = ∞

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 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 of  is .

### 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 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 = π/4

=> t = tan x

=> t = tan π/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 = 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 of  is .

### Question 28.

Solution:

We have,

I =

I =

I =

I =

Let tan x = t. So, we have

=> secx dx = dt

Now, 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 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 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 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 = π

=> 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 – 4 = t3. So, we have

=> dx = 3t2 dt

Now, the lower limit is, x = 4

=> t3 = x – 4

=> t3 = 4 – 4

=> t3 = 0

=> t = 0

Also, the upper limit is, x = 12

=> t3 = x – 4

=> t3 = 12 – 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 = π + 0 – 0 – 0 – 2

I = π – 2

Therefore, the value of  is 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 of  is .

### Question 38.

Solution:

We have,

I =

I =

I =

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

=> (1 – 1/x2)dx = dt

Now, 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 of  is .

### Question 39.

Solution:

We have,

I =

Let x5 + 1 = t. So, we have

=> 5x4 dx = dt

Now, the lower limit is, x = –1

=> t = x5 + 1

=> t = (–1)5 + 1

=> t = –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

=> secx dx = dt

Now, 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 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 = π/4

=> u = sin 2t

=> u = sin π/2

=> u = 1

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

### Question 42.

Solution:

We have,

I =

Let 5 – 4 cos θ = t. So, we have

=> 4 sin θ dθ = dt

=> sin θ dθ = dt/4

Now, the lower limit is, θ = 0

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos 0

=> t = 5 – 4

=> t = 1

Also, the upper limit is, θ = π

=> t = 5 – 4 cos θ

=> t = 5 – 4 cos π

=> t = 5 + 4

=> t = 9

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I = 9√3 – 1

Therefore, the value of  is 9√3 – 1.

### Question 43.

Solution:

We have,

I =

I =

I =

I =

I =

Let tan 2θ = t. So, we have

=> 2 sec2 2θ dθ = dt

=> sec2 2θ dθ = dt/2

Now, the lower limit is, θ = 0

=> t = tan 2θ

=> t = tan 0

=> t = 0

Also, the upper limit is, θ = π/6

=> t = tan 2θ

=> t = tan π/3

=> t = √3

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

### Question 44.

Solution:

We have,

I =

Let  = t. So, we have

=>  = dt

Now, the lower limit is, x = 0

=> t =

=> t =

=> t = 0

Also, the upper limit is, x =

=> t =

=> t =

=> t = π

So, the equation becomes,

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 45.

Solution:

We have,

I =

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

=> 1/x dx = dt

Now, the lower limit is, x = 0

=> t = 1 + log x

=> t = 1 + log 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 =

Therefore, the value of  is .

### Question 46.

Solution:

We have,

I =

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 47.

Solution:

We have,

I =

Let 30 – x3/2 = t. So, we have

=>  = dt

=>  = – dt

Now, the lower limit is, x = 4

=> t = 30 – x3/2

=> t = 30 – 43/2

=> t = 30 – 8

=> t = 22

Also, the upper limit is, x = 9

=> t = 30 – x3/2

=> t = 30 – 93/2

=> t = 30 – 27

=> t = 3

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 48.

Solution:

We have,

I =

Let cos x = t. So, we have

=> – sin x dx = dt

=> sin x dx = –dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π

=> t = cos x

=> t = cos π

=> t = –1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 49.

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 =

Therefore, the value of  is .

### Question 50.

Solution:

We have,

I =

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 =

Therefore, the value of  is .

### Question 51.

Solution:

We have,

I =

On using integration by parts, we get,

I =

I =

Let cos-1 x = t. So, we have

=>  = dt

Now, the lower limit is, x = 0

=> t = cos-1 x

=> t = cos-1 0

=> t = π/2

Also, the upper limit is, x = 1

=> t = cos-1 x

=> t = cos-1 1

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I = π – 2

Therefore, the value of  is π – 2.

### Question 52.

Solution:

We have,

I =

Let x = a tan2 t. So, we have

=> dx = 2a tan t sect dt

Now, the lower limit is, x = 0

=> a tan2 t = x

=> a tan2 t = 0

=> tan t = 0

=> t = 0

Also, the upper limit is, x = a

=> a tan2 t = x

=> a tan2 t = a

=> tan2 t = 1

=> tan t = 1

=> t = π/4

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 53.

Solution:

We have,

I =

I =

I =

I =

I =

Let cot x/2 = t. So, we have

=>  = dt

Now, the lower limit is, x = π/3

=> t = cot x/2

=> t = cot π/6

=> t = √3

Also, the upper limit is, x = π/2

=> t = cot x/2

=> t = cot π/4

=> t = 1

So, the equation becomes,

I =

I =

I =

I =

I = 1

Therefore, the value of  is 1.

### Question 54.

Solution:

We have,

I =

Let x2 = a2 cos 2t. So, we have

=> 2x dx = – 2a2 sin 2t dt

Now, the lower limit is, x = 0

=> a2 cos 2t = x2

=> a2 cos 2t = 0

=> cos 2t = 0

=> 2t = π/2

=> t = π/4

Also, the upper limit is, x = a

=> a2 cos 2t = x2

=> a2 cos 2t = a2

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 55.

Solution:

We have,

I =

Let x = a cos 2t. So, we have

=> dx = –2a sin 2t

Now, the lower limit is, x = –a

=> a cos 2t = x

=> a cos 2t = –a

=> cos 2t = –1

=> 2t = π

=> t = π/2

Also, the upper limit is, x = a

=> a cos 2t = x

=> a cos 2t = a

=> cos 2t = 1

=> 2t = 0

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I =

I =

I =

I = πa

Therefore, the value of  is πa.

### Question 56.

Solution:

We have,

I =

Let cos x = t. So, we have

=> – sin x dx = dt

=> sin x dx = –dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I = – log 2 + 2 log 3 + 0 – 2 log 2

I = 2 log 3 – 3 log 2

I = log 9 – log 8

I = log 9/8

Therefore, the value of  is log 9/8.

### Question 57.

Solution:

We have,

I =

I =

I =

Let sin2 x = t. So, we have

=> 2 sin x cos x dx = dt

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 =

I =

Therefore, the value of  is .

### Question 58.

Solution:

We have,

I =

Let x = sin t. So, we have

=> dx = cos t dt

Now, the lower limit is, x = 0

=> sin t = x

=> sin t = 0

=> t = 0

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

=> sin t = x

=> sin t = 1/2

=> t = π/6

So, the equation becomes,

I =

I =

I =

I =

I =

Let tan t = u. So, we have

=> sec2 t dt = du

Now, the lower limit is, t = 0

=> u = tan t

=> u = tan 0

=> t = 0

Also, the upper limit is, t = π/6

=> u = tan t

=> u = tan π/6

=> t = 1/√3

So, the equation becomes,

I =

I =

I =

Therefore, the value of  is .

### Question 59.

Solution:

We have,

I =

Let 1/x2 – 1 = t. So, we have

=> –2/x3 dx = dt

Now, the lower limit is, x = 1/3

=> t = 1/x2 – 1

=> t = 9 – 1

=> t = 8

Also, the upper limit is, x = 1

=> t = 1/x2 – 1

=> t = 1 – 1

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I = 6

Therefore, the value of  is 6.

### Question 60.

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 t

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/4

=> t = tan t

=> t = tan π/4

=> t = 1

So, the equation becomes,

I =

Let t= u. So, we have

=> 3t2 dt = du

=> t2 dt = du/3

Now, the lower limit is, t = 0

=> u = t3

=> u = 03

=> u = 0

Also, the upper limit is, t = 1

=> u = t3

=> u = 13

=> u = 1

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 61.

Solution:

We have,

I =

I =

I =

I =

I =

Let cos x = t. So, we have

=> – sin x dx = dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x

=> t = cos π/2

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I =

I =

Therefore, the value of  is .

### Question 62.

Solution:

We have,

I =

I =

I =

Let cos x/2 + sin x/2 = t. So, we have

=> (cos x/2 – sin x/2) dx = 2 dt

Now, the lower limit is, x = 0

=> t = cos x/2 + sin x/2

=> t = cos 0 + sin 0

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = π/2

=> t = cos x/2 + sin x/2

=> t = cos π/2 + sin π/2

=> t = 1/√2 + 1/√2

=> t = √2

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

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