# Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.2 | Set 3

### Question 42.

Solution:

We have,

I =

Let 5 â€“ 4 cos Î¸ = t. So, we have

=> 4 sin Î¸ dÎ¸ = dt

=> sin Î¸ dÎ¸ = dt/4

Now, the lower limit is, Î¸ = 0

=> t = 5 â€“ 4 cos Î¸

=> t = 5 â€“ 4 cos 0

=> t = 5 â€“ 4

=> t = 1

Also, the upper limit is, Î¸ = Ï€

=> t = 5 â€“ 4 cos Î¸

=> t = 5 â€“ 4 cos Ï€

=> t = 5 + 4

=> t = 9

So, the equation becomes,

I =

I =

I =

I =

I =

I =

I =

I = 9âˆš3 â€“ 1

Therefore, the value of  is 9âˆš3 â€“ 1.

### Question 43.

Solution:

We have,

I =

I =

I =

I =

I =

Let tan 2Î¸ = t. So, we have

=> 2 sec2 2Î¸ dÎ¸ = dt

=> sec2 2Î¸ dÎ¸ = dt/2

Now, the lower limit is, Î¸ = 0

=> t = tan 2Î¸

=> t = tan 0

=> t = 0

Also, the upper limit is, Î¸ = Ï€/6

=> t = tan 2Î¸

=> t = tan Ï€/3

=> t = âˆš3

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 = Ï€

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 = Ï€/2

=> t = sin x

=> t = sin Ï€/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 â€“ x3/2 = t. So, we have

=>  = dt

=>  = â€“ dt

Now, the lower limit is, x = 4

=> t = 30 â€“ x3/2

=> t = 30 â€“ 43/2

=> t = 30 â€“ 8

=> t = 22

Also, the upper limit is, x = 9

=> t = 30 â€“ x3/2

=> t = 30 â€“ 93/2

=> t = 30 â€“ 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

=> â€“ sin x dx = dt

=> sin x dx = â€“dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = Ï€

=> t = cos x

=> t = cos Ï€

=> t = â€“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 = Ï€/2

=> t = sin x

=> t = sin Ï€/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 = Ï€/2

=> t = sin x

=> t = sin Ï€/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 = Ï€/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 = Ï€ â€“ 2

Therefore, the value of  is Ï€ â€“ 2.

### Question 52.

Solution:

We have,

I =

Let x = a tan2 t. So, we have

=> dx = 2a tan t sec2 t 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 = Ï€/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 = Ï€/3

=> t = cot x/2

=> t = cot Ï€/6

=> t = âˆš3

Also, the upper limit is, x = Ï€/2

=> t = cot x/2

=> t = cot Ï€/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 = â€“ 2a2 sin 2t dt

Now, the lower limit is, x = 0

=> a2 cos 2t = x2

=> a2 cos 2t = 0

=> cos 2t = 0

=> 2t = Ï€/2

=> t = Ï€/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 = â€“2a sin 2t

Now, the lower limit is, x = â€“a

=> a cos 2t = x

=> a cos 2t = â€“a

=> cos 2t = â€“1

=> 2t = Ï€

=> t = Ï€/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 = Ï€a

Therefore, the value of  is Ï€a.

### Question 56.

Solution:

We have,

I =

Let cos x = t. So, we have

=> â€“ sin x dx = dt

=> sin x dx = â€“dt

Now, the lower limit is, x = 0

=> t = cos x

=> t = cos 0

=> t = 1

Also, the upper limit is, x = Ï€/2

=> t = cos x

=> t = cos Ï€/2

=> t = 0

So, the equation becomes,

I =

I =

I =

I =

I = â€“ log 2 + 2 log 3 + 0 â€“ 2 log 2

I = 2 log 3 â€“ 3 log 2

I = log 9 â€“ 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 = Ï€/2

=> t = sin2 x

=> t = sin2 Ï€/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 = Ï€/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 = Ï€/6

=> u = tan t

=> u = tan Ï€/6

=> t = 1/âˆš3

So, the equation becomes,

I =

I =

I =

Therefore, the value of  is .

### Question 59.

Solution:

We have,

I =

Let 1/x2 â€“ 1 = t. So, we have

=> â€“2/x3 dx = dt

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

=> t = 1/x2 â€“ 1

=> t = 9 â€“ 1

=> t = 8

Also, the upper limit is, x = 1

=> t = 1/x2 â€“ 1

=> t = 1 â€“ 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 = Ï€/4

=> t = tan t

=> t = tan Ï€/4

=> t = 1

So, the equation becomes,

I =

Let t3 = 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

=> â€“ 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 = Ï€/2

=> t = cos x

=> t = cos Ï€/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 â€“ 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 = Ï€/2

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

=> t = cos Ï€/2 + sin Ï€/2

=> t = 1/âˆš2 + 1/âˆš2

=> t = âˆš2

So, the equation becomes,

I =

I =

I =

I =

Therefore, the value of  is .

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