The various methods of integrating the given set of questions are explained in brief under this chapter. Students who have difficulty in solving these problems can use the solutions PDF to clarify their doubts. It mainly improves problem-solving abilities among students in order to boost their exam preparation. The solutions are prepared by experts in a step by step manner based on CBSE exam pattern. For a better hold on the concepts covered under this chapter, RS Aggarwal Solutions for Class 12 Chapter 13 Methods of Integration PDF links are available below.

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

### Exercises of RS Aggarwal Solutions Class 12 Maths Chapter 13 – Methods of Integration

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

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Exercise 13a page: 621

**Evaluate the following integrals:**

**Very-Short-Answer Questions**

**1. ∫ (2x + 9) ^{5} dx**

**Solution:**

Take 2x + 9 = t

So we get

2 dx = dt

It can be written as

**2. ∫ (7 – 3x) ^{4} dx**

**Solution:**

Take 7 – 3x = t

So we get

– 3 dx = dt

It can be written as

**Solution:**

Take 3x – 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 – 4x = t

So we get

– 4 dx = dt

It can be written as

**Solution:**

Take 2x – 3 = t

So we get

2 dx = dt

It can be written as

**Solution:**

Take 2x – 1 = t

So we get

2 dx = dt

It can be written as

**Solution:**

Take 1 – 3x = t

So we get

– 3 dx = dt

It can be written as

**Solution:**

Take 2 – 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

∫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 √x = t

So we get

1/2√x dx = dt

By integrating w.r.t. t

∫cos t 2 dt = 2 sin t + c

By substituting the value of t

= 2 sin √x + c

**Solution:**

Take tan x = t

So we get

sec^{2} x dx = dt

By integrating w.r.t. t

∫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

∫- 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 – 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

∫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

∫cos (-t) dt = ∫cos t dt= sin t + c [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} – 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

∫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

∫sin t dt = – cos t + c

By substituting the value of t

= – cos (x^{2} + 1) + c

Exercise 13B page: 658

**Evaluate the following integrals:**

**1. (i) ∫sin ^{2} x dx**

**(ii) ∫cos ^{2} x dx**

**Solution:**

(i) ∫sin^{2} x dx

Here, we know that 1 – cos 2x = 2 sin^{2} x

It can be written as

(ii) ∫cos^{2} x dx

Here, we know that 1 + cos 2x = 2 cos^{2} x

It can be written as

**2. (i) cos ^{2} (x/2) dx**

**(ii) cot ^{2} (x/2) dx**

**Solution:**

(i) cos^{2} (x/2) dx

Here, we know that 1 + cos x = 2 cos^{2} (x/2)

It can be written as

(ii) cot^{2} (x/2) dx

Here, we know that cosec^{2} x – cot^{2} x = 1

It can be written as

**3. (i) ∫ sin ^{2} nx dx**

**(ii) ∫ sin ^{5} x dx**

**Solution:**

(i) ∫ sin^{2} nx dx

Here, we know that 1 – cos 2nx = 2 sin^{2} nx

It can be written as

(ii) ∫ sin^{5} x dx

Here, we know that 1 – cos^{2} x = sin^{2} x

It can be written as

∫sin^{5} x dx = ∫(1 – cos^{2} x)^{2} sin x dx

Take cos x = t

So we get

– sin x dx = dt

By considering cos x = t we get

∫(1 – cos^{2} x)^{2} sin x dx = – ∫(1 – t^{2})^{2} dt

Here

– ∫(1 – t^{2})^{2} dt = – ∫(1 + t^{4} – 2t^{2}) dt

We get

– ∫(1 – t^{2})^{2} dt = – ∫ dt + ∫ 2t^{2} dt – ∫ t^{4} dt

By integrating w.r.t t

**4. ∫ cos ^{3} (3x + 5) dx**

**Solution:**

It is given that

∫ cos^{3} (3x + 5) dx

By substituting 3x + 5 = u

We get 3 dx = du

Here dx = du/3

So we get

Now by re-substituting the value of t = sin u and u = 3x + 5

**5. ∫ sin ^{7} (3 – 2x) dx**

**Solution:**

We can write it as

= – ∫ sin^{7} (2x – 3) dx

By substituting the value of 2x – 3 = u

So we get

2 dx = du where dx = du/2

**Solution:**

**Solution:**

**8. ∫ sin 3x cos 4x dx**

**Solution:**

It is given that

∫ sin 3x cos 4x dx

Using the formula sin x × cos y=1/2(sin(x + y)-sin(y-x))

∫ sin 3x cos 4x dx = ½ ∫ (sin 7x – sin x) dx

On further simplification

∫ sin 3x cos 4x dx = ½ ∫ sin 7x dx – ½ ∫ sin x dx

By integrating w.r.t. x

∫ sin 3x cos 4x dx =

**9. ∫ cos 4x cos 3x dx**

**Solution:**

It is given that

∫ cos 4x cos 3x dx

Using the formula cos x × cos y=1/2(cos(x + y)+cos(x-y))

∫ cos 4x cos 3x dx = ½ ∫ (cos 7x + cos x) dx

On further simplification

∫ cos 4x cos 3x dx = ½ ∫ cos 7x dx + ½ ∫ cos x dx

By integrating w.r.t. x

∫ cos 4x cos 3x dx =

**10. ∫ sin 4x sin 8x dx**

**Solution:**

It is given that

∫ sin 4x sin 8x dx

Using the formula sin x × sin y=1/2(cos(y-x)-cos(y + x))

∫ sin 4x sin 8x dx = ½ ∫ (cos 4x – cos 12x) dx

On further simplification

∫ sin 4x sin 8x dx = ½ ∫ cos 4x dx – ½ ∫ cos 12x dx

By integrating w.r.t. x

∫ sin 4x sin 8x dx =

**11. ∫ sin 6x cos x dx**

**Solution:**

It is given that

∫ sin 6x cos x dx

Using the formula sin x × cos y=1/2(sin(y + x)-sin(y-x))

∫ sin 6x cos x dx = ½ ∫ (sin 7x – sin (-5x)) dx

On further simplification

∫ sin 6x cos x dx = ½ ∫ sin 7x dx + ½ ∫ sin 5x dx

By integrating w.r.t. x

∫ sin 6x cos x dx =

**Solution:**

We know that

1 + cos 2x = 2 cos^{2} x

So it can be written as

**13. ∫ cos ^{4} x dx**

**Solution:**

**14. ∫ cos 2x cos 4x cos 6x dx**

**Solution:**

It is given that

∫ cos 2x cos 4x cos 6x dx

By multiplying and dividing by 2

= ½ ∫ 2 cos 2x. cos 4x. cos 6x dx

We can write it as

= ½ ∫ cos 2x (2 cos 4x cos 6x) dx

On further calculation

= ½ ∫ cos 2x [cos (4x + 6x) + cos (4x – 6x)] dx

So we get

= ½ ∫ cos 2x (cos 10x + cos 2x) dx

By further simplification

= ½ ∫ cos 2x cos 10 x dx + ½ ∫ cos^{2} 2x dx

It can be written as

= ½. ½ ∫ 2 cos 2x. cos 10x dx + ½ ∫ (1 + cos 4x)/2 dx

By further multiplication

= ¼ ∫ [cos (2x + 10x) + cos (2x – 10x)] dx + ¼ ∫ (1 + cos 4x) dx

So we get

= ¼ ∫ [cos 12x + cos 8x] dx + ¼ [x + sin 4x/4]

By integrating w.r.t x

= ¼ [sin 12x/12 + sin 8x/8] + ¼x + sin 4x/16 + c

We get

= sin 12x/48 + sin 8x/32 + x/4 + sin 4x/16 + c

**15. ∫ sin ^{3} x cos x dx**

**Solution:**

It is given that

∫ sin^{3} x cos x dx

By taking sin x = t

We get

cos x = dt/dx

It can be written as

cos x dx = dt

So we get

= ∫ t^{3} dt

By integrating w.r.t t

= t^{4}/4 + c

By substituting the value of t

= sin^{4} x/4 + c

**16. ∫ sec ^{4} x dx**

**Solution:**

It is given that

∫ sec^{4} x dx

We can write it as

= ∫ sec^{2} x sec^{2} x dx

So we get

= ∫ (1 + tan^{2} x) sec^{2} x dx

Take tan x = t

By differentiation we get

sec^{2} x dx = dt

It can be written as

= ∫ (1 + t^{2}) dt

By integrating w.r.t. t

= t + t^{3}/3 + c

By substituting the value of t

= tan x + tan^{3} x/3 + c

**17. ∫ cos ^{3} x sin^{4}x dx**

**Solution:**

It is given that

∫ cos^{3} x sin^{4}x dx

We can write it as

= ∫ sin^{4} x cos^{2} x cos x dx

By taking sin^{4} x as common

= ∫ sin^{4} x (1 – sin^{2} x) cos x dx

Take sin x = t

We get cos x = dx/dt

Here cos x dx = dt

It can be written as

= ∫ t^{4} (1 – t^{2}) dt

On further simplification

= ∫ t^{4} dt – ∫ t^{6} dt

By integration w.r.t. t

= t^{5}/5 – t^{7}/7 + c

By substituting the value of t

= sin^{5} x/5 – sin^{7} x/7 + c

Exercise 13C page: 678

**Evaluate the following integrals:**

**1. ∫ x e ^{x} dx**

**Solution:**

It is given that

∫ x e^{x} dx

We can write it as

By integrating w.r.t x

= x e^{x} – ∫ 1 e^{x} dx

So we get

= x e^{x} – e^{x} + c

By taking e^{x} as common

= e^{x} (x – 1) + c

**2. ∫ x cos x dx**

**Solution:**

It is given that

∫ x cos x dx

We can write it as

By integrating w.r.t x

= x sin x – ∫ 1 sin x dx

So we get

= x sin x + cos x + c

**3. ∫ x e ^{2x} dx**

**Solution:**

It is given that

∫ x e^{2x} dx

We can write it as

**4. ∫ x sin 3x dx**

**Solution:**

It is given that

∫ x sin 3x dx

We can write it as

**5. ∫ x cos 2x dx**

**Solution:**

It is given that

∫ x cos 2x dx

We can write it as

So we get

**6. ∫ x log 2x dx**

**Solution:**

It is given that

∫ x log 2x dx

We can write it as

**7. ∫ x cosec ^{2} x dx**

**Solution:**

It is given that

∫ x cosec^{2} x dx

We can write it as

By integrating w.r.t x

= x (- cot x) – ∫ 1 (- cot x) dx

So we get

= – x cot x + ∫ cot x dx

Again by integration we get

= – x cot x + log |sin x| + c

**8. ∫ x ^{2} cos x dx**

**Solution:**

It is given that

∫ x^{2} cos x dx

We can write it as

By integrating w.r.t x

= x^{2} sin x – ∫ 2x sin x dx

So we get

= x^{2} sin x – 2 [∫ x sin x dx]

Now apply by the part method

= x^{2} sin x – 2 ∫ x sin x dx

Again by integration

We get

= x^{2} sin x – 2 [x (- cos x) – ∫ 1. (- cos x) dx]

By integrating w.r.t x

= x^{2} sin x – 2 (- x cos x + sin x) + c

On further simplification

= x^{2} sin x + 2x cos x – 2 sin x + c

**9. ∫ x sin ^{2} x dx**

**Solution:**

It is given that

∫ x sin^{2} x dx

We can write it as

sin^{2} x = (1 + cos 2x)/ 2

By integration

It can be written as

**10. ∫ x tan ^{2} x dx**

**Solution:**

It is given that

∫ x tan^{2} x dx

We can write it as

tan^{2} x = sec^{2} x – 1

So we get

∫ x tan^{2} x dx = ∫ x ( sec^{2} x – 1) dx

By further simplification

= ∫ x sec^{2} x dx – ∫ x dx

Taking first function as x and second function as sec^{2} x

∫ x sec^{2} x dx – ∫ x dx

We get

By integration w.r.t x

= {x tan x – ∫ tan x dx} – x^{2}/ 2

It can be written as

= x tan x – log |sec x| – x^{2}/ 2 + c

So we get

= x tan x – log |1/cos x| – x^{2}/ 2 + c

Here

= x tan x + log |cos x| – x^{2}/ 2 + c

**11. ∫ x ^{2} e^{x} dx**

**Solution:**

It is given that

∫ x^{2} e^{x} dx

By integrating w.r.t x

We get

= x^{2} e^{x} – 2 [x e^{x} – ∫1. e^{x} dx]

By further simplification

= x^{2} e^{x} – 2 [x e^{x} – e^{x}] + c

By multiplication

= x^{2} e^{x} – 2x e^{x} + 2 e^{x} + c

By taking e^{x} as common

= e^{x} (x^{2} – 2x + 2) + c

**13. ∫ x ^{2} e^{3x} dx**

**Solution:**

It is given that

∫ x^{2} e^{3x} dx

We can write it as

**14. ∫ x ^{2} sin^{2} x dx**

**Solution:**

It is given that

∫ x^{2} sin^{2} x dx

We know that

Now by integrating the second part we get

**15. ∫ x ^{3} log 2x dx**

**Solution:**

It is given that

∫ x^{3} log 2x dx

We can write it as

**16. ∫ x log (x + 1) dx**

**Solution:**

It is given that

**Solution:**

It is given that

**Solution:**

It is given that

We can write it as

= t e^{t} – ∫1. e^{t} dt

So we get

= t e^{t} – e^{t} + C

Now by replacing t with x^{2}

**19. ∫ x sin ^{3} x dx**

**Solution:**

It is given that

∫ x sin^{3} x dx

We know that

sin 3x = 3 sin x – 4 sin^{3} x

It can be written as

sin^{3} x = (3 sin x – sin 3x)/4

We get

So we get

**20. ∫ x cos ^{3} x dx**

**Solution:**

It is given that

∫ x cos^{3} x dx

We know that

cos^{3} x = (cos 3x + 3 cos x)/4

It can be written as

**21. ∫ x ^{3} cos x^{2} dx**

**Solution:**

It is given that

∫ x^{3} cos x^{2} dx

We can write it as

∫ x x^{2} cos x^{2} dx

Take x^{2} = t

So we get 2x dx = dt

x dx = dt/2

**22. ∫ sin x log (cos x) dx**

**Solution:**

It is given that

∫ sin x log (cos x) dx

Consider first function as log (cos x) and second function as sin x

So we get

= – cos x log (cos x) – ∫ sin x dx

Again by integrating the second term

= – cos x log (cos x) + cos x + c

**23. ∫ x sin x cos x dx**

**Solution:**

It is given that

∫ x sin x cos x dx

We know that

sin 2x = 2 sin x cos x

It can be written as

Consider first function as x and second function as sin 2x

**24. ∫ cos √x dx**

**Solution:**

It is given that

∫ cos √x dx

Take √x = t

So we get

1/2√x dx = dt

By cross multiplication

dx = 2 √x dt

Here dx = 2t dt

It can be written as

∫ cos √x dx = 2 ∫t cos t dt

Take first function as t and second function as cos t

By integrating w.r.t t

= 2 (t sin t – ∫ sin t dt)

We get

= 2t sin t + 2 cos t + c

Now by substituting t as √x

= 2√x sin √x + 2 cos √x + c

Taking 2 as common

= 2 (cos √x + √x sin √x) + c

**25. ∫ cosec ^{3} x dx**

**Solution:**

It is given that

∫ cosec^{3} x dx

We can write it as

∫ cosec^{3} x dx = ∫ cosec x cosec^{2} x dx

So we get

By integration we get

= cosec x (- cot x) – ∫ (- cosec x cot x) (- cot x) dx

It can be written as

= – cosec x cot x – ∫ cosec x cot^{2} x dx

Here cot^{2} x = cosec^{2} x – 1

We get

= – cosec x cot x – ∫ cosec x (cosec^{2} x – 1) dx

By further simplification

= – cosec x cot x – ∫ cosec^{3} x dx + ∫ cosec x dx

We know that ∫ cosec^{3} x dx = 1

∫cosec^{3} x dx = – cosec x cot x – ∫ cosec^{3} x dx + ∫ cosec x dx

On further calculation

2 ∫cosec^{3} x dx = – cosec x cot x + ∫ cosec x dx

By integration we get

2 ∫cosec^{3} x dx = – cosec x cot x + log |tan x/2| + c

Here

∫cosec^{3} x dx = – ½ cosec x cot x + ½ log |tan x/2| + c

**27. ∫ sin x log (cos x) dx**

**Solution:**

It is given that

∫ sin x log (cos x) dx

Take cos x = t

So we get

– sin x dx = dt

It can be written as

∫ sin x log (cos x) dx = – ∫log t dt = – ∫1. log t dt

Consider first function as log t and second function as 1

By integrating w.r.t t

= – log t. t + ∫1/t. t dt

Again by integrating the second term

= – t log t + t + c

Now replace t as cos x

= t (- log t + 1) + c

We get

= cos x (1 – log (cos x)) + c

**Solution:**

Take log x = t

So we get

1/x dx = dt

Here

Again by integrating the second term we get

= t log t – t + c

Replace t with log x

= log x. log (log x) – log x + c

By taking log x as common

= log x (log (log x) – 1) + c

**29. ∫ log (2 + x ^{2}) dx**

**Solution:**

It is given that ∫ log (2 + x^{2}) dx

We can write it as

= ∫1. log (2 + x^{2}) dx

Consider first function as log (2 + x^{2}) and second function as 1

So we get

**Solution:**

Consider log x = t

Where x = e^{t}

So we get

dx = e^{t} dt

We can write it as

**32. ∫ e ^{–x} cos 2x cos 4x dx**

**Solution:**

Using the formula

cos A cos B = ½ [cos (A + B) + cos (A – B)]

We get

cos 4x cos 2x = ½ [cos (4x + 2x) + cos (4x – 2x)] = ½ [cos 6x + cos 2x]

By applying in the original equation

∫ e ^{–x} cos 2x cos 4x dx = ∫ e ^{–x} (½ [cos 6x + cos 2x])

= ½ [∫e ^{–x} cos 6x dx + ∫e ^{–x} cos 2x dx]

Taking first function as cos 6x and cos 2x and second function as e ^{–x}

Now by solving both parts separately

**33. ∫e ^{√x} dx**

**Solution:**

Consider √x = t

We get

It can be written as

dx = 2 √x dt

where dx = 2t dt

We can replace it in the equation

∫e ^{√x} dx = ∫e^{t} 2t dt

So we get

= 2 ∫t e^{t} dt

Consider the first function as t and second function as e^{t}

By integration w.r.t. t

= 2 (t e^{t} – ∫1. e^{t} dt)

We get

= 2 (t e^{t} – e^{t}) + c

Taking e^{t} as common

= 2 e^{t} (t – 1) + c

Substituting the value of t we get

= 2 e ^{√x} (√x – 1) + c

**34. ∫ e ^{ sin x} sin 2x dx**

**Solution:**

We know that

sin 2x = 2 sin x cos x

So we get

∫ e^{ sin x} sin 2x dx = 2 ∫ e^{ sin x} sin x cos x dx

Take sin x = t

So we get cos x dx = dt

It can be written as

2 ∫ e^{ sin x} sin x cos x dx = 2 ∫ e^{ t} t dt

Consider first function as t and second function as e^{t}

By integrating w.r.t. t

= 2 (t e^{t} – ∫1. e^{t} dt)

We get

= 2 (t e^{t} – e^{t}) + c

Here

= 2 e^{t} (t – 1) + c

By substituting the value of t

= 2 e^{sin x} (sin x – 1) + C

**Solution:**

Take sin ^{-1} x = t

Here x = sin t

So we get

By integration we get

= t (- cos t) – ∫1. (- cos t) dt

We get

= – t cos t + sin t + c

It can be written as

cos t = √1 – sin^{2} t

Here we get

= – t (√1 – sin^{2} t) + sin t + c

By replacing t as sin^{-1} x

= – sin^{-1} x (√1 – x^{2}) + x + c