Exercise 13C page: 678

RS Aggarwal Solutions for Class 12 Maths Chapter 13

Methods of Integration

Exercise 13C page: 678 1.

Solution:

It is given that

∫ x ex dx

We can write it as

By integrating w.r.t x

= x ex – ∫ 1 ex dx

So we get

= x ex – ex + c

By taking ex as common

= ex (x – 1) + c

2.

Solution:

It is given that

∫ x cos x dx

We can write it as


= x sin x – ∫ 1 sin x dx

So we get

= x sin x + cos x + c

3.

Solution:

It is given that

∫ x e2x dx

We can write it as

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4.

Solution:

It is given that

∫ x sin 3x dx

We can write it as

5.

Solution:

It is given that

∫ x cos 2x dx

We can write it as





So we get

6.

Solution:

It is given that

∫ x log 2x dx

We can write it as

7.





Solution:

It is given that

∫ x cosec2 x dx

We can write it as


= 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.

Solution:

It is given that

∫ x2 cos x dx

We can write it as


= x2 sin x – ∫ 2x sin x dx

So we get

= x2 sin x – 2 [∫ x sin x dx]

Now apply by the part method

= x2 sin x – 2 ∫ x sin x dx

Again by integration

We get

= x2 sin x – 2 [x (- cos x) – ∫ 1. (- cos x) dx]






= x2 sin x – 2 (- x cos x + sin x) + c

On further simplification

= x2 sin x + 2x cos x – 2 sin x + c

9.

Solution:

It is given that

∫ x sin2 x dx

We can write it as

sin2 x = (1 + cos 2x)/ 2

By integration

It can be written as





10.

Solution:

It is given that

∫ x tan2 x dx

We can write it as

tan2 x = sec2 x – 1

So we get

∫ x tan2 x dx = ∫ x ( sec2 x – 1) dx

By further simplification

= ∫ x sec2 x dx – ∫ x dx

Taking first function as x and second function as sec2 x

∫ x sec2 x dx – ∫ x dx





We get

By integration w.r.t x

= {x tan x – ∫ tan x dx} – x2/ 2


= x tan x – log |sec x| – x2/ 2 + c

So we get

= x tan x – log |1/cos x| – x2/ 2 + c

Here

= x tan x + log |cos x| – x2/ 2 + c

11.

Solution:

It is given that

∫ x2 ex dx


We get

= x2 ex – 2 [x ex – ∫1. ex dx]


= x2 ex – 2 [x ex – ex] + c

By multiplication

= x2 ex – 2x ex + 2 ex + c

By taking ex as common





= ex (x2 – 2x + 2) + c

13.

Solution:

It is given that

∫ x2 e3x dx

We can write it as

14.

Solution:





It is given that

∫ x2 sin2 x dx

We know that

Now by integrating the second part we get





15.

Solution:

It is given that

∫ x3 log 2x dx

We can write it as

16.

Solution:





It is given that

17.

Solution:

It is given that





18.

Solution:

It is given that

We can write it as

= t et – ∫1. et dt

So we get





= t et – et + C

Now by replacing t with x2

19.

Solution:

It is given that

∫ x sin3 x dx

We know that

sin 3x = 3 sin x – 4 sin3 x


sin3 x = (3 sin x – sin 3x)/4

We get





So we get

20.

Solution:

It is given that

∫ x cos3 x dx

We know that

cos3 x = (cos 3x + 3 cos x)/4






21.

Solution:

It is given that

∫ x3 cos x2 dx

We can write it as

∫ x x2 cos x2 dx

Take x2 = t

So we get 2x dx = dt

x dx = dt/2





22.

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.

Solution:

It is given that

∫ x sin x cos x dx

We know that

sin 2x = 2 sin x cos x






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

24.

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


∫ 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.

Solution:

It is given that

∫ cosec3 x dx

We can write it as

∫ cosec3 x dx = ∫ cosec x cosec2 x dx

So we get

By integration we get

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


= – cosec x cot x – ∫ cosec x cot2 x dx

Here cot2 x = cosec2 x – 1

We get

= – cosec x cot x – ∫ cosec x (cosec2 x – 1) dx


= – cosec x cot x – ∫ cosec3 x dx + ∫ cosec x dx

We know that ∫ cosec3 x dx = 1

∫cosec3 x dx = – cosec x cot x – ∫ cosec3 x dx + ∫ cosec x dx

On further calculation

2 ∫cosec3 x dx = – cosec x cot x + ∫ cosec x dx


2 ∫cosec3 x dx = – cosec x cot x + log |tan x/2| + c

Here

∫cosec3 x dx = – ½ cosec x cot x + ½ log |tan x/2| + c





27.

Solution:

It is given that

∫ sin x log (cos x) dx

Take cos x = t

So we get

– sin x dx = dt


∫ 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

28.

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.

Solution:

It is given that ∫ log (2 + x2) dx

We can write it as

= ∫1. log (2 + x2) dx

Consider first function as log (2 + x2) and second function as 1





So we get

31.

Solution:

Consider log x = t

Where x = et

So we get

dx = et dt

We can write it as





32.

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.

Solution:

Consider √x = t

We get






dx = 2 √x dt

where dx = 2t dt

We can replace it in the equation

∫e √x dx = ∫et 2t dt

So we get

= 2 ∫t et dt

Consider the first function as t and second function as et

By integration w.r.t. t

= 2 (t et – ∫1. et dt)

We get

= 2 (t et – et) + c

Taking et as common

= 2 et (t – 1) + c

Substituting the value of t we get

= 2 e √x (√x – 1) + c

34.

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


2 ∫ e sin x sin x cos x dx = 2 ∫ e t t dt

Consider first function as t and second function as et





By integrating w.r.t. t

= 2 (t et – ∫1. et dt)

We get

= 2 (t et – et) + c

Here

= 2 et (t – 1) + c

By substituting the value of t

= 2 esin x (sin x – 1) + C

35.

Solution:

Take sin -1 x = t

Here x = sin t

So we get


= t (- cos t) – ∫1. (- cos t) dt

We get

= – t cos t + sin t + c


cos t = √1 – sin2 t

Here we get

= – t (√1 – sin2 t) + sin t + c

By replacing t as sin-1 x

= – sin-1 x (√1 – x2) + x + c



Exercise 13C page: 678

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Transcript of Exercise 13C page: 678