The problems of exercise 13C are based on the concept of integration by parts. The theorem which is explained under this exercise mainly provides an exact idea about the steps to be followed in solving problems. Students can clear their doubts instantly by solving problems using solutions prepared by our experts. It improves the time management skills among students, which are important from the exam point of view. RS Aggarwal Solutions Class 12 Maths Chapter 13 Methods of Integration Exercise 13C PDF links are available below.

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

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

**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