Integration by Substitution is the main concept which is covered under Exercise 13A of Chapter 13. The problems are solved by teachers having a high knowledge of concepts with the aim of helping students score well in the board exam. The solved examples before each exercise give an idea to the students about the kind of questions that would appear in the exam. Students who aspire to perform well in the board exam can download RS Aggarwal Solutions Class 12 Maths Chapter 13 Methods of Integration Exercise 13A PDF using links, which are provided below.

## RS Aggarwal Solutions for Class 12 Maths Chapter 13: Methods of Integration Exercise 13A Download PDF

### Access other exercise solutions of Class 12 Maths Chapter 13: Methods of Integration

### Access RS Aggarwal Solutions for Class 12 Maths Chapter 13: Methods of Integration Exercise 13A

**Evaluate the following integrals:**

**Very-Short-Answer Questions**

**1. âˆ« (2x + 9) ^{5} dx**

**Solution:**

Take 2x + 9 = t

So we get

2 dx = dt

It can be written as

**2. Â âˆ« (7 â€“ 3x) ^{4} dx**

**Solution:**

Take 7 â€“ 3x = t

So we get

– 3 dx = dt

It can be written as

**Solution:**

Take 3x â€“ 5 = t

So we get

3 dx = dt

It can be written as

**Solution:**

Take 4x + 3 = t

So we get

4 dx = dt

It can be written as

**Solution:**

Take 3 â€“ 4x = t

So we get

– 4 dx = dt

It can be written as

**Solution:**

Take 2x â€“ 3 = t

So we get

2 dx = dt

It can be written as

**Solution:**

Take 2x â€“ 1 = t

So we get

2 dx = dt

It can be written as

**Solution:**

Take 1 â€“ 3x = t

So we get

– 3 dx = dt

It can be written as

**Solution:**

Take 2 â€“ 3x = t

So we get

– 3 dx = dt

It can be written as

**Solution:**

Take 3x = t

So we get

3 dx = dt

It can be written as

By integrating w.r.t. t

**Short-Answer Questions**

**Solution:**

Take 5 + 6x = t

So we get

6 dx = dt

It can be written as

**Solution:**

It is given that

By integrating w.r.t. t

**Solution:**

Take 2x + 5 = t

So we get

2 dx = dt

By integrating w.r.t. t

**Solution:**

Take sin x = t

So we get

cos x dx = dt

By integrating w.r.t. t

**Solution:**

Take sin x = t

So we get

cos x dx = dt

By integrating w.r.t. t

**Solution:**

Take cos x = t

So we get

– sin x dx = dt

By integrating w.r.t. t

**Solution:**

Take sin ^{-1} x = t

So we get

**Solution:**

Take tan^{-1} x = t

So we get

**Solution:**

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

âˆ«cos t dt = sin t + c

By substituting the value of t

= sin (log x) + c

**Solution:**

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

By substituting the value of t

= – cot (log x) + c

**Solution:**

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

By substituting the value of t

= log (log x) + c

**Solution:**

It is given that

**Solution:**

Take log x = t

So we get

1/x dx = dt

By integrating w.r.t. t

**Solution:**

Take âˆšx = t

So we get

1/2âˆšx dx = dt

By integrating w.r.t. t

âˆ«cos t 2 dt = 2 sin t + c

By substituting the value of t

= 2 sin âˆšx + c

**Solution:**

Take tan x = t

So we get

sec^{2} x dx = dt

By integrating w.r.t. t

âˆ«e^{t} dt = e^{t} + c

By substituting the value of t

= e ^{tan x} + c

**Solution:**

Take cos^{2} x = t

So we get

– sin 2x dx = dt

By integrating w.r.t. t

âˆ«- e^{t} dt = – e^{t} + c

By substituting the value of t

**Solution:**

Take ax + b = t

So we get

a dx = dt

It can be written as

**Solution:**

We know that

cos 3x = 4 cos^{3} x â€“ 3 cos x

It can be written as

**Solution:**

Take -1/x = t

So we get

1/x^{2} dx = dt

By integrating w.r.t. t

âˆ«e^{t} dt = e^{t} + c

By substituting the value of t

**Solution:**

Take -1/x = t

So we get

1/x^{2} dx = dt

By integrating w.r.t. t

âˆ«cos (-t) dt = âˆ«cos t dt = sin t + cÂ Â Â [Since, cos is an even function]

By substituting the value of t

= – sin 1/x + c

**Solution:**

It is given that

**Solution:**

Take e^{2x} â€“ 2 = t

So we get

2 e^{2x} dx = dt

By integrating w.r.t. t

**Solution:**

Take log (sin x) = t

So we get

cos x/ sin x dx = dt

By cross multiplication

cot x dx = dt

By integrating w.r.t. t

**Solution:**

Take log (sin x) = t

So we get

cos x/ sin x dx = dt

By cross multiplication

cot x dx = dt

By integrating w.r.t. t

âˆ«1/t dt = log t + c

By substituting the value of t

= log (log sin x) + c

**Solution:**

Take x^{2} + 1 = t

So we get

2x dx = dt

By integrating w.r.t. t

âˆ«sin t dt = – cos t + c

By substituting the value of t

= – cos (x^{2} + 1) + c