RD Sharma Solutions Class 12 Mathematics Chapter 18 VSA

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:44 PM IST

The High-quality of answers and its accuracy in the RD Sharma solutions books has made them popular among educational institutions and their students. However, many students find the class 12 mathematics chapter 18 to be a bit hard. This is due to the lack of practice and clarity in the concepts among the students. Here is where the RD Sharma Class 12th VSA books play a significant role in making the students understand the concepts clearly.

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RD Sharma Class 12 Solutions Chapter 18 VSA Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise: VSA
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter 18 VSA Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise: VSA

Indefinite Integrals Excercise Very Short Answers Question 1

Answer:

\log | x + \log (\sin x) | + c

Hint: You must know about the integration rule of sin, cot and logarithm function
Given:

\int \frac{1 + \cot x}{x + \log \sin x} d x

Solution:
Consider

x + \log (\sin x) = t

and differentiate both sides,
We get

(1 + \cot x) d x = d t

By integrating w.r.t t

\int \frac{d t}{t} = \log t + c

[\int \frac{1}{x} d x = \log x + c]

By substituting the value of t

\int \frac{1 + \cot x}{x + \log \sin x} d x

= \log | x + \log (\sin x) | + c

Indefinite Integrals Excercise Very Short Answers Question 2

Answer:

\frac{x^{8}}{8} + c

Hints: you must know about the integration values of exponential and logarithm function
Given:

\int e^{3 \log x} x^{4} d x

Solution:

\int e^{3 \log x} x^{4} d x

= \int e^{\log x^{3}} x^{4} d x

[∴ a \log x = \log x^{a}]

= \int x^{3} x^{4} d x

[∴ e^{\log m} = m]

= \int x^{7} d x

[\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c]

= \frac{x^{8}}{8} + c

\int e^{3 \log x} x^{4} d x

= \frac{x^{8}}{8} + c

Indefinite Integrals Excercise Very Short Answers Question 3

Answer:

- \frac{1}{3} \cos x^{3} + c

Hint: You must know about the integration value sin function
Given:

\int x^{2} \sin x^{3} d x

Explanation:

\int x^{2} \sin x^{3} d x

Let

x^{3} = t

and differentiate both sides,

3 x^{2} d x = d t

= \frac{1}{3} \int \sin t d t

= - \frac{1}{3} \cos t + c

= - \frac{1}{3} \cos x^{3} + c

[\int \sin x d x = - \cos x + c]

Indefinite Integrals Excercise Very Short Answers Question 4

Answer: $e^{x} (\sin x) + c$
Hint: You must know about the integration rule of sin, cos and exponential function
Given:

\int e^{x} (\sin x + \cos x) d x

Solution:

\int e^{x} (\sin x + \cos x) d x

\int e^{x} \sin x d x + \int e^{x} \cos x d x

= e^{x} (\sin x) - \int \cos x . e^{x} d x + \int e^{x} \cos x d x

= e^{x} (\sin x) + c

Indefinite Integrals Excercise Very Short Answers Question 5

Answer:

e^{x} (\sin x) + c

Hint: You must know about the integration rule of sin, cos and exponential function
Given:

\int e^{x} (\sin x + \cos x) d x

Solution:

\begin{aligned} \int e^{x} (\sin x + \cos x) d x \\ \int e^{x} \sin x d x + \int e^{x} \cos x d x \\ = e^{x} (\sin x) - \int \cos x \cdot e^{x} d x + \int e^{x} \cos x d x \end{aligned}

= e^{x} (\sin x) + c

[\int u v d x = u \int v d x - {\int \frac{d}{d x} u . \int v d x} d x]

Indefinite Integrals Excercise Very Short Answers Question 6

Answer:

\frac{\tan^{7} x}{7} + c

Hint: You must know about the integration rule of tan and sec function
Given:

\int \tan^{6} x \sec^{2} x d x

Solution: Suppose

\tan x = t

differentiate both sides,

[\frac{d}{d x} \tan x = s e c^{2} x]

\begin{aligned} \sec^{2} x d x = d t \\ = \int t^{6} d t \\ = \frac{t^{7}}{7} + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \\ = \frac{\tan^{7} x}{7} + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 7

Answer: $\frac{1}{2} \log | 3 + 2 \sin x | + c$
Hint: You must know about the integration rule of sin and cos function.
Given:

\int \frac{\cos x}{3 + 2 \sin x} d x

Solution:
Let

\sin x = t

differentiate both sides,

[\frac{d}{d x} \sin x = \cos x]

\cos x d x = d t

I = \int \frac{d t}{3 + 2 t}

(\frac{1}{2}) \log (3 + 2 t) + c

[\int \frac{1}{2 x} d x = \frac{1}{2} \log 2 x + c]

= \frac{1}{2} \log | 3 + 2 \sin x | + c

Indefinite Integrals Excercise Very Short Answers Question 8

Answer:

e^{x} \sec x + c

Hint: You must know about the integral values of exponential function and trignometric function.
Given:

\int e^{x} \sec x (1 + \tan x) d x

Solution:

\int e^{x} \sec x (1 + \tan x) d x

= \int e^{x} (\sec x + \sec x \tan x) d x

Let

e^{x} \sec x = t

and differentiate both sides, by using

[\frac{d}{d x} u v = u \frac{d}{d x} v + v \frac{d}{d x} u]

\begin{aligned} (e^{x} \sec x (1 + t an x)) d x = d t \\ I = \int d t \\ = t + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \\ = e^{x} \sec x + c [t = e^{x} \sec x] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 9

Answer:

\frac{n}{2} {(\log x)}^{2} + c

Hint: You must know about the integration rule of logarithm function
Given:

\int \frac{\log x^{n}}{x} d x

Solution:

\int \frac{\log x^{n}}{x} d x

= \int \frac{n \log x}{x} d x

[∵ \log x^{a} = a \log x]

Put

\log x = t

and differentiate both sides,

\frac{d}{d x} \log x = \frac{1}{x}

\begin{aligned} \frac{1}{x} d x = d t \\ I = \int n (t) d t \\ = n (\frac{t^{2}}{2} + c) [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \\ = \frac{n (\log x)^{2}}{2} + n c \end{aligned}

= \frac{n}{2} {(\log x)}^{2} + c

Indefinite Integrals Excercise Very Short Answers Question 10

Answer:

\frac{{(\log x)}^{n + 1}}{n + 1} + c

Hint: You must know about the integration rule of logarithm function
Given:

\int \frac{{(\log x)}^{n}}{x} d x

Solution:

\int \frac{{(\log x)}^{n}}{x} d x

Put

\log x = t

and differentiate both sides,

\frac{d}{d x} \log x = \frac{1}{x}

\frac{1}{x} d x = d t

I = \int t^{n} d t

= \frac{t^{n + 1}}{n + 1} + c

[t = \log x]

= \frac{{(\log x)}^{n + 1}}{n + 1} + c

Indefinite Integrals Excercise Very Short Answers Question 11

Answer:

\frac{\sin^{2} x}{2} + c

Hint: You must know about the integration rule of logarithm, sin and cos function.
Given :

\int e^{\log \sin x} . \cos x d x

Solution :

I = \int e^{\log \sin x} . \cos x d x

I = \int \sin x . \cos x d x

[∵ e \log a = a]

Let

\sin x = t

differentiate both sides,

[\frac{d}{d x} \sin x = \cos x]

c o s x d x = d t

∴ I = \int t d t

= \frac{t^{2}}{2} + c

[\int x^{n} d x = \frac{x^{n + 1}}{n + 1}]

= \frac{\sin^{2} x}{2} + c

Indefinite Integrals Excercise Very Short Answers Question 12

Answer:

\frac{\sin^{4} x}{4} + c

Hint: You must know about the integration rule of trigonometric function.
Given:

\int \sin^{3} x \cos x d x

Solution: put

\sin x = t

differentiate both sides,

[\frac{d}{d x} \sin x = \cos x]

\cos x d x = d t

∴ I = \int t^{3} d t

= \frac{t^{4}}{4} + c

[\int x^{n} d x = \frac{x^{n + 1}}{n + 1}]

= \frac{\sin^{4} x}{4} + c

Indefinite Integrals Excercise Very Short Answers Question 13

Answer:

- \frac{\cos^{5} x}{5} + c

Hint: You must know about the integration rule of trigonometric function.
Given:

\int \cos^{4} x \sin x d x

Solution: put

\cos x = t

differentiate both sides,

[\frac{d}{d x} \cos x = - \sin x]

- \sin x d x = d t

s i n x d x = - d t

∴ I = \int - t^{4} d t

[\int x^{n} d x = \frac{x^{n + 1}}{n + 1}]

= \frac{- t^{5}}{5} + c

= \frac{- \cos^{5} x}{5} + c

Indefinite Integrals Excercise Very Short Answers Question 14

Answer:

\frac{1}{3} \sec^{3} x + c

Hint: You must know about the integration rule of trigonometric function.
Given:

\int \tan x . \sec^{3} x d x

Solution: let

\sec x = u

differentiate both sides,

[\frac{d}{d x} \sec x = \sec x . \tan x]

\sec x . \tan x d x = d u

∴ I = \int \sec^{2} x (\sec x . \tan x) d x

= \int u^{2} d u

[\int x^{n} d x = \frac{^{n + 1}}{n + 1}]

= \frac{1}{3} u^{3} + c

= \frac{1}{3} \sec^{3} x + c

Indefinite Integrals Excercise Very Short Answers Question 15

Answer:

- \log | 1 + e^{- x} | + c

Hint: You must know about the integration rule of exponential function.
Given=

\int \frac{1}{1 + e^{x}} d x

Solution:
$\begin{aligned} \int \frac{1}{1 + e^{x}} d x \\ = \int \frac{e^{- x}}{e^{- x} (1 + e^{x})} d x [m u l t i p l y a n d d i v i d e b y e^{x}] \\ = \int \frac{e^{- x}}{1 + e^{- x}} d x \end{aligned}$
Let

e^{- x} + 1 = t

[\frac{d}{d x} e^{- x} = - e^{- x}]

\begin{aligned} - e^{- x} d x = dt \\ I = \int \frac{- d t}{t} \\ = - \log | t | + c \\ = - \log | e^{- x} + 1 | + c [\int \frac{1}{x} d x = \log x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 16

Answer:

- \log | e^{- x} + 2 | + c

Hint: You must know about the integration rule of exponential function.
Given:

\int \frac{1}{1 + 2 e^{x}} d x

Solution:

\int \frac{1}{1 + 2 e^{x}} d x

= \int \frac{e^{- x}}{e^{- x} (1 + 2 e^{x})} d x

= \int \frac{e^{- x}}{e^{- x} + 2} d x

[ multiply and divide by

e^{x}

]
Let

t = e^{- x} + 2

and differentiate both sides,

[\frac{d}{d x} e^{- x} = - e^{- x}]

d t = - e^{- x} d x

I = \int \frac{- d t}{t}

= - \log | t | + c

= - \log | e^{- x} + 2 | + c

[\int \frac{1}{x} d x = \log x + c]

Indefinite Integrals Excercise Very Short Answers Question 17

Answer:

\frac{{(\tan^{- 1} x)}^{4}}{4} + c

Hint: You must know about the integral rule of tangent function.
Given:

\int \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x

Solution:

I = \int \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x

Let

\tan^{- 1} x = t

and differentiate both sides,

\frac{1}{1 + x^{2}} d x = d t

[\frac{d}{d t} \tan^{- 1} x = \frac{1}{1 + x^{2}}]

\begin{aligned} I = \int (t^{3}) d t \\ = \frac{t^{4}}{4} + c \\ [\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ = \frac{{(\tan^{- 1} x)}^{4}}{4} + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 18

Answer:

\frac{1}{- 3 {(5 + \tan x)}^{3}} + c

Hint: You must know about the integral rule of trigonometric functions.
Given:

\int \frac{\sec^{2} x}{{(5 + \tan x)}^{4}} d x

Solution:

\frac{\sec^{2} x}{{(5 + \tan x)}^{4}}

t = 5 + \tan x

differentiate both sides,

[\frac{d}{d x} \tan x = \sec^{2} x]

d t = \sec^{2} x d x

I = \int \frac{\sec^{2} x}{{(5 + \tan x)}^{4}} d x

= \int \frac{d t}{t^{4}}

= \int t^{- 4} d t

= \frac{t^{- 3}}{- 3} + c

= \frac{- 1}{3 t^{3}} + c

[\int x^{n} d x = \frac{x^{n + 1}}{n + 1}]

∴ I = \frac{1}{- 3 {(5 + \tan x)}^{3}} + c

Indefinite Integrals Excercise Very Short Answers Question 19

Answer:

x + c

Hint: You must know about the integral rule of trigonometric functions.
Given:

\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x

Solution:

\begin{aligned} I = \int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x \\ I = \int \frac{\sin x + \cos x}{\sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x}} d x \\ I = \int \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^{2}}} d x \\ I = \int \frac{\sin x + \cos x}{\sin x + \cos x} d x \end{aligned}

I = \int 1 d x

= x + c

[\int x^{n} d x = \frac{x^{n + 1}}{n + 1}]

Indefinite Integrals Excercise Very Short Answers Question 20

Answer:

x (\log_{e} x - 1) + c

Hint: You must know about the integral rule of logarithm functions.
Given:

\int \log_{e} x d x

Solution:

\int \log_{e} x .1 d x

Using by parts

\begin{aligned} = \log_{e} x \int 1 d x - \int [\frac{d}{d x} \log_{ε} x \int 1 d x] d x \\ = x \log_{e} x - \int \frac{1}{x} x d x \\ = x \log_{ε} x - \int 1 d x [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \\ = x \log_{e} x - x + c \\ = x (\log_{e} x - 1) + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 23

Answer:

\frac{a^{x}}{\log a} + \frac{x^{a + 1}}{a + 1} + c

Hints: You must know about the integral rule of logarithm and exponential functions.
Given:

\int (e^{x \log_{e} a} + e^{a \log_{e} x}) d x

Solution:

\begin{aligned} \int (e^{x \log_{e} a} + e^{a \log_{e} x}) d x \\ = \int e^{x \log_{e} a} d x + e^{a \log_{e} x} d x \\ = \int e^{\log_{e} a x} + e^{\log_{e} x^{a}} d x \\ = \int a^{x} d x + \int x^{a} d x [e^{\log x} = x] \\ = \frac{a^{x}}{\log a} + \frac{x^{a + 1}}{a + 1} + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 24

Answer:

\log | \log (\sin x) | + c

Hints: You must know about the integral rule of logarithm and trigonometric functions
Given:

\int \frac{\cos x}{\sin x \log \sin x} d x

Solution:

\log \sin x = t

and differentiate both sides,

[\frac{d}{d x} \log \sin x = \frac{1}{\sin x} . \frac{d}{d x} \sin x]

\begin{aligned} \frac{1}{\sin x} \cdot \cos x = d t \\ \int \frac{\cos x d x}{\sin x \cdot \log \sin x} = \int \frac{d t}{t} \\ \log | t | + c \\ \log | \log (\sin x) | + c [\int \frac{1}{x} d x = \log x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 25

Answer:

\frac{1}{a^{2} - b^{2}} \log_{e} | a^{2} \sin^{2} x + b^{2} \cos^{2} x | + c

Hints: You must know about the integral rule of trigonometric functions
Given:

\int \frac{\sin 2 x}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x

Solution:

\int \frac{\sin 2 x}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x

Let

a^{2} \sin^{2} x + b^{2} \cos^{2} x = t a^{2} 2 \sin x \cdot \cos x d x + b^{2} 2 \cos x (- \sin x) = d t

[\begin{matrix} \frac{d}{d x} \sin x = \cos x \\ \frac{d}{d x} \cos x = - \sin x \end{matrix}] & [\frac{d}{d x} x^{n} = n x^{n - 1}]

\begin{aligned} \sin 2 x \cdot d x (a^{2} - b^{2}) = d t \\ \sin 2 x d x = \frac{d t}{(a^{2} - b^{2})} \\ [2 \sin x \cos x = \sin 2 x] \\ \int \frac{d t}{(a^{2} - b^{2}) t} \\ \frac{1}{a^{2} - b^{2}} \int \frac{d t}{t} \\ \int \frac{1}{x} d x = \log | x | + c \\ = \frac{1}{a^{2} - b^{2}} \log_{e} | t | + c \\ = \frac{1}{a^{2} - b^{2}} \log_{e} | a^{2} \sin^{2} x + b^{2} \cos^{2} x | + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 26

Answer:

\frac{\log | 3 + a^{x} |}{\ln (a)} + c

Hints: You must know about the integral rules of x
Given:

\int \frac{a^{x}}{3 + a^{x}} d x

Solution:

I = \int \frac{a^{x}}{3 + a^{x}} d x

Let

t = 3 + a^{x} .

and differentiate both sides

d t = 0 + a^{x} \ln (a) d x [\frac{d}{d x} a^{x} = a^{x} \ln | a |]

I = \int \frac{d t}{\ln (a) t} [\int \frac{1}{x} d x = \log x + c]

\begin{aligned} = \frac{1}{\ln a} \int \frac{d t}{t} \\ = \frac{1}{\ln a} \log | t | + c \\ = \frac{\log | 3 + a^{x} |}{\ln (a)} + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 27

Answer:

\log | 3 + x \log x | + c

Hints: You must know about the integral rule of logarithm functions
Given:

\int \frac{1 + \log x}{3 + x \log x} d x

Solution:

\int \frac{1 + \log x}{3 + x \log x} d x

Let

t = 3 + x \log x

and differentiate both sides

d t = 0 + x \frac{d}{d x} (\log x) + \log x \frac{d}{d x} (x)

d t = 0 + x . \frac{1}{x} + (\log x)

Multiplication rule of differentiation

[\frac{d}{d x} u v = u \frac{d (v)}{d x} + v \frac{d}{d x} (u)]

d t = (1 + \log x) d x

I = \int \frac{1 + \log x}{3 + x \log x} d x

I = \int \frac{d t}{t}

= \log | t | + c

∴ I = \log | 3 + x \log x | + c

[\int \frac{1}{x} d x = \log x + c]

Indefinite Integrals Excercise Very Short Answers Question 28

Answer:

\frac{1}{2} \sec^{2} x + c

Hints: You must know about the integral rule of trigonometric functions
Given:

\int \frac{\sin x}{\cos^{3} x} d x

Solution:

I = \int \frac{\sin x}{\cos^{3} x} d x

Let

\cos x = t

differentiate both sides,

[\frac{d}{d x} \cos x - \sin x]

- \sin x d x = d t

I = \int \frac{- d t}{t^{3}} I = \int - t^{- 3} d t I = \frac{- t^{- 2}}{- 3 + 1} + c = \frac{- t^{- 2}}{- 2} + c = \frac{1}{2 t^{2}} + c = \frac{1}{2 \cos^{2} x} + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] = \frac{1}{2} \sec^{2} x + c

Indefinite Integrals Excercise Very Short Answers Question 29

Answer:

- \log | \sin x + \cos x | + c

Hints: You must know about the integral rule of trigonometric functions
Given:

\int \frac{\sin x - \cos x}{\sqrt{1 + \sin 2 x}} d x

Solution:

I = \int \frac{\sin x - \cos x}{\sqrt{1 + \sin 2 x}} d x

\begin{aligned} \int \frac{\sin x - \cos x}{\sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x} \cdot \cos x} d x \\ \int \frac{\sin x - \cos x}{\sqrt{(\sin x + \cos x)^{2}}} d x \\ \int \frac{\sin x - \cos x}{(\sin x + \cos x)} d x \end{aligned}

Let

\sin x + \cos x = t

and differentiate both sides,

[\frac{d}{d x} \sin x d x = \cos x a n d \frac{d}{d x} \cos x = - \sin x]

(\cos x - \sin x) d x = d t

- (\sin x - \cos x) d x = d t

= \int \frac{- d t}{t}

= - \log | t | + c

[\int \frac{1}{x} d x = \log x + c]

= - \log | \sin x + \cos x | + c

Indefinite Integrals Excercise Very Short Answers Question 30

Answer:

\frac{{(\log x)}^{1 - n}}{1 - n} + c

Hints: You must know about the integral rule of trigonometric functions
Given

\int \frac{1}{x {(\log x)}^{n}} d x

Solution: $I = \int \frac{1}{x {(\log x)}^{n}} d x$
Let

\log x = t

differentiate both sides,

[\frac{d}{d x} \log x = \frac{1}{x}]

\frac{1}{x} d x = d t

d x = x . d t

\int \frac{1}{x {(\log x)}^{n}} d x

Put

\log x = t

\begin{aligned} d x = x \cdot d t \\ = \int \frac{1}{x \cdot t^{n}} \cdot x d t \\ = \int \frac{1}{t^{n}} d t \\ = \int t^{- n} d t \\ = \frac{t^{- n + 1}}{- n + 1} + c \\ = \frac{t^{1 - n}}{1 - n} + c \end{aligned}

= \frac{{(\log x)}^{1 - n}}{1 - n} + c

[\int x^{n} d x = \frac{x^{n + 1}}{n + 1}]

Indefinite Integrals Excercise Very Short Answers Question 31

Answer:

\frac{e^{a x}}{a^{2} + b^{2}} (a \sin b x - b \cos b x) + c

Hints: You must know about the integral rule of trigonometric and exponential functions
Given:

\int e^{a x} \sin b x d x

Solution:

I = \int e^{a x} \sin b x d x

and using integration by parts,

\begin{aligned} \sin b x \int e^{a x} d x - \int b \cos b x \frac{e^{a x}}{a} d x \\ = \frac{1}{a} \sin b x e^{a x} - \frac{b}{a} \int \cos b x e^{a x} d x \\ = \frac{1}{a} \sin b x e^{a x} - \frac{b}{a} \cos b x \int [e^{a x} d x + \frac{b}{a} [\int \frac{d}{d x} (\cos b x)] - \int e^{a x} d x] d x \\ I = \frac{1}{a} \sin b x e^{a x} - \frac{b}{a} \cos b x \cdot \frac{e^{a x}}{a} + \frac{b}{a} \int (- \sin b x) \cdot \frac{b e^{a x}}{a} d x \\ I = \frac{1}{a} \sin b x e^{a x} - \frac{b}{a^{2}} \cos b x \cdot e^{a x} - \frac{b^{2}}{a^{2}} I \end{aligned}

\begin{aligned} I (1 + \frac{b^{2}}{a^{2}}) = \frac{1}{a} \sin b x \cdot e^{a x} - \frac{b}{a^{2}} \cos b x \cdot e^{a x} \\ \Rightarrow \int e^{a x} \cdot \sin b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \sin b x - b \cos b x) + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 32

Answer:

\frac{e^{a x}}{a^{2} + b^{2}} [b \cos b x + a \cos b x] + c

Hints: You must know about the integral rule of trigonometric functions
Given

\int e^{a x} \cos b x d x

Solution:

\int e^{a x} \cos b x d x

Integrating by parts

\begin{aligned} I = e^{a x} \cdot \frac{\sin b x}{b} - a \int e^{a x} \frac{\sin b x}{b} d x \\ = \frac{1}{b} e^{a x} \sin b x - \frac{a}{b} \int e^{a x} \sin b x d x \end{aligned}

Again using integration by parts

\begin{aligned} \frac{1}{b} e^{a x} \sin b x - \frac{a}{b} [- e^{a x} \frac{\cos b x}{b} - a \int e^{a x} \frac{\cos b x}{b} d x] \\ \frac{1}{b} e^{a x} \sin b x - \frac{a}{b^{2}} e^{a x} \cos b x - \frac{a^{2}}{b^{2}} \int e^{a x} \cos b x d x \end{aligned}

On computing,

\begin{aligned} I = \frac{e^{a x}}{b^{2}} [b \sin b x + a \cos b x] - \frac{a^{2}}{b^{2}} I + c \\ = \frac{e^{a x}}{a^{2} + b^{2}} [b \sin b x + a \cos b x] + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 33

Answer:

\frac{e^{x}}{x} + c

Hints: You must know about the integral rule of exponential functions
Given:

\int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x

Solution:

\int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x

It is of form

\int e^{x} f (x) + f^{'} (x) d x = e^{x} . f (x) + c

Put

f (x) = \frac{1}{x}

and differentiate both sides,

f^{'} (x) = \frac{- 1}{x^{2}}

Thus,

\int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x

= \frac{e^{x}}{x} + c

Indefinite Integrals Excercise Very Short Answers Question 34

Answer:

e^{a x} f (x) + c

Hints: You must know about the integral rule of exponential functions
Given:

\int e^{a x} [a f (x) + f^{'} (x)] d x

Solution:

a \int e^{a x} f (x) d x + \int e^{a x} f^{'} (x) d x

Now use integration by parts

\begin{aligned} [\int u v d x = u \int v d x - \int [\frac{d}{d x} u \int v d x] d x + c] \\ = a [f (x) \int e^{a x} d x - \int \frac{d}{d x} f (x) \int e^{a x} d x] d x + \int e^{a x} f^{1} (x) d x \\ = a f (x) \frac{e^{a x}}{a} - a \int f^{1} (x) \frac{e^{a x}}{a} d x + a \int e^{a x} f^{1} (x) d x \\ = f (x) \cdot e^{a x} - \int f^{1} (x) e^{a x} d x + \int f^{1} (x) e^{a x} d x \\ = f (x) \cdot e^{a x} + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 35

Answer:

\frac{1}{2} x \sqrt{4 - x^{2}} + 2 \sin^{- 1} (\frac{x}{2}) + c

Hints: You must know about the integral rule of x functions
Given :

\int \sqrt{4 - x^{2}} d x

Solution:

\int \sqrt{4 - x^{2}} d x

\begin{aligned} x = 2 \sin θ \\ d x = 2 \cos θ d θ \\ \sqrt{4 - x^{2}} = \sqrt{4 - 4 \sin^{2}} θ = 2 \cos θ \\ \int \sqrt{4 - x^{2}} d x = \int 2 \cos θ \cdot 2 \cos θ d θ \\ = 4 \int \cos^{2} θ d θ \end{aligned}

We know that

\frac{\cos 2 θ + 1}{2} = \cos^{2} θ

So,

4 \int \cos^{2} θ d θ = 2 \int (\cos 2 θ + 1) d θ

= 2 \int \cos 2 θ d θ + 2 \int 1 d θ

[\begin{matrix} \int \cos a x d x = \frac{\sin a x}{a} \\ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} \end{matrix}]

= 2 \sin θ \cos θ + 2 θ + c

[\sin 2 x = 2 \sin x \cos x]

Now put the values of

θ, \sin θ, \cos θ

We know

x = 2 \sin θ

θ = \sin^{- 1} x

x = 2 \sin θ

Squaring on both sides

\begin{aligned} x^{2} = 4 \sin^{2} θ \\ x^{2} = 4 (1 - \cos^{2} θ) \\ 4 - x^{2} = 4 \cos^{2} θ \\ \cos θ = \sqrt{\frac{4 - x^{2}}{2}} \\ = 2 \cdot \frac{x}{2} \cdot \frac{\sqrt{4 - x^{2}}}{2} + 2 \sin^{- 1} (\frac{x}{2}) \\ = \frac{x \cdot \sqrt{4 - x^{2}}}{2} + 2 \sin^{- 1} (\frac{x}{2}) + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 36

Answer:

\frac{1}{2} x \sqrt{9 + x^{2}} + \frac{9}{2} \log | x + \sqrt{9 + x^{2}} | + c

Hint: You must know about the integral rule of trigonometric functions
Given:

\int \sqrt{9 + x^{2}} d x

Solution:

\int \sqrt{9 + x^{2}} d x

We know that

\int \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{a^{2} + x^{2}} + \frac{a^{2}}{2} \log | x + \sqrt{x^{2} + a^{2}} | + c

Replace x ->x and a->3

\begin{aligned} \int \sqrt{3^{2} + x^{2}} d x = \frac{x}{2} \sqrt{3^{2} + x^{2}} + \frac{3^{2}}{2} \log | x + \sqrt{x^{2} + 3^{2}} | + c \\ = \frac{1}{2} x \sqrt{9 + x^{2}} + \frac{9}{2} \log | x + \sqrt{x^{2} + 9} | + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 37

Answer:

\frac{x}{2} \sqrt{x^{2} - 9} - \frac{9}{2} \log | x + \sqrt{x^{2} - 9} | + c

Given:

\int \sqrt{x^{2} - 9} d x

Solution:

\begin{aligned} \int \sqrt{x^{2} - 9} d x \\ I = \int \sqrt{x^{2} - a^{2}} d x = \int \sqrt{x^{2} - 3^{2}} d x [\int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \log | x + \sqrt{x^{2} - a^{2}} | + c] \end{aligned}

Given by

∴ I = \frac{x}{2} \sqrt{x^{2} - 9} - \frac{9}{2} \log | x + \sqrt{x^{2} - 9} | + c

Indefinite Integrals Excercise Very Short Answers Question 38

Answer:

\frac{1}{3} \log | 1 + x^{3} | + c

Given:

\int \frac{x^{2}}{1 + x^{3}} d x

Solution:

\int \frac{x^{2}}{1 + x^{3}} d x

Put

1 + x^{3} = t

and differentiate both sides,

[\frac{d}{d x} x^{n} = n x^{n - 1}]

3 x^{2} d x = d t

x^{2} d x = \frac{d t}{3}

\begin{aligned} \int \frac{d t}{3 t} = \frac{1}{3} \int \frac{d t}{t} \\ = \frac{1}{3} \log | t | + c \\ = \frac{1}{3} \log | 1 + x^{3} | + c [\int \frac{1}{x} d x = \log x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 39

Answer:

\frac{1}{3} \log | x^{3} + 6 x^{2} + 5 | + c

Hint: You must know about the integral rule of logarithmic functions
Given:

\int \frac{x^{2} + 4 x}{x^{3} + 6 x^{2} + 5} d x

Solution:

\int \frac{x^{2} + 4 x}{x^{3} + 6 x^{2} + 5} d x

Put

x^{3} + 6 x^{2} + 5 = t

and differentiate both sides,

[\frac{d}{d x} x^{n} = n x^{n - 1}]

\begin{aligned} 3 x^{2} + 12 x d x = d t \\ 3 (x^{2} + 4 x) d x = d t \\ (x^{2} + 4 x) d x = \frac{d t}{3} \\ \int \frac{d t}{3 t} = \frac{1}{3} \int \frac{d t}{t} \\ = \frac{1}{3} \log | t | + c \\ = \frac{1}{3} \log | x^{3} + 6 x^{2} + 5 | + c & [\int \frac{1}{x} d x = \log x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 40

Answer:

2 \tan \sqrt{x} + c

Hint: You must know about the integral rule of trigonometric functions
Given:

\int \frac{\sec^{2} \sqrt{x}}{\sqrt{x}} d x

Solution:

\int \frac{\sec^{2} \sqrt{x}}{\sqrt{x}} d x

Let,

\begin{aligned} t = \tan \sqrt{x} d x \\ d t = \frac{\sec^{2} \sqrt{x}}{2 \sqrt{x}} d x \\ 2 d t = \frac{\sec^{2} \sqrt{x}}{\sqrt{x}} d x \\ = \int 2 d t \\ = 2 t + c \\ = 2 \tan \sqrt{x} + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 41

Answer:

- 2 \cos \sqrt{x} + c

Hint: You must know about the integral rule of trigonometric functions
Given:

\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x

Solution:

\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x

Let

\begin{aligned} t = \cos \sqrt{x} \\ d t = - \frac{\sin \sqrt{x}}{2 \sqrt{x}} d x \\ - 2 d t = \frac{\sin \sqrt{x}}{\sqrt{x}} d x [\frac{d}{d x} \cos x = - \sin x] \end{aligned}

\begin{aligned} \int \frac{\sin \sqrt{x}}{\sqrt{x}} d x = \int - 2 d t \\ = - 2 t + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \\ = - 2 \cos \sqrt{x} + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 42

Answer:

2 \sin \sqrt{x} + c

Hint: You must know about the integral rule of trigonometric functions
Given:

\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x

Solution:

\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x

Let

\begin{aligned} t = \sin \sqrt{x} \\ d t = \frac{\cos \sqrt{x}}{2 \sqrt{x}} d x \\ 2 d t = \frac{\cos \sqrt{x}}{\sqrt{x}} [\frac{d}{d x} \sin x = \cos x] \end{aligned}

\begin{aligned} \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x & = \int 2 d t \\ = 2 t + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \\ = 2 \sin \sqrt{x} + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 43

Answer:

\frac{1}{3} {(1 + \log x)}^{3} + c

Hint: You must know about the integral rule of logarithmic functions.
Given:

\int \frac{{(1 + \log x)}^{2}}{x} d x

Solution:

\int \frac{{(1 + \log x)}^{2}}{x} d x

Let

1 + \log x = t

and differentiate both sides,

\frac{d}{d x} \log x = \frac{1}{x}

\begin{aligned} 0 + \frac{1}{x} d x = d t \\ \frac{1}{x} = \frac{d t}{d x} \\ d x = x d t \end{aligned}

∴ \int \frac{{(1 + \log x)}^{2}}{x} d x

Put

1 + \log x = t

d x = x d t

\begin{aligned} \int \frac{t^{2} x}{x} d t \\ \int t^{2} d t \\ = \frac{t^{2 + 1}}{2 + 1} + c \\ = \frac{t^{3}}{3} + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \\ = \frac{1}{3} (1 + \log x)^{3} + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 44

Answer:

\frac{- 1}{4} \tan (7 - 4 x) + c

Hint: You must know about the integral rule of trigonometric functions.
Given:

\int \sec^{2} (7 - 4 x) d x

Solution:

\int \sec^{2} (7 - 4 x) d x

Put

(7 - 4 x) = t

and differentiate both sides,

[\frac{d}{d x} a x = a]

d x = \frac{- d t}{4}

\begin{aligned} = \int \sec^{2} (t) \frac{d t}{- 4} \\ = \frac{- 1}{4} \int \sec^{2} (t) d t \\ = \frac{- 1}{4} \tan (t) + c \\ = \frac{- 1}{4} \tan (7 - 4 x) + c [∵ \int \sec^{2} x d x = \tan x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 45

Answer:

\frac{1}{2} {(\log x)}^{2} + c

Hint: You must know about the integral rule of logarithmic functions.
Given:

\int \frac{\log x}{x} d x

Solution:

\int \frac{\log x}{x} d x

Put

\log x = t

and differentiate both sides,

\frac{d}{d x} \log x = \frac{1}{x}

\begin{aligned} \int \frac{\log x}{x} d x = \int t d t \\ = \frac{t^{2}}{2} + c \\ \frac{1}{x} d x = d t = \frac{(\log x)^{2}}{2} + c [\int \frac{1}{x} d x = \log x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 47

Answer:

\tan x - \sec x + c

Hint: You must know about the integral rule of trigonometric functions
Given:

\int \frac{1 - \sin x}{\cos^{2} x} d x

Solution:

\int \frac{1 - \sin x}{\cos^{2} x} d x

\begin{aligned} \int \frac{1}{\cos^{2} x} - \frac{\sin x}{\cos^{2} x} d x \\ = \int \sec^{2} x d x - \int \sec x \cdot \tan x \\ = \tan x - \sec x + c [∵ \int \sec^{2} x d x = \tan x and \int \sec x \cdot \tan x d x = \sec x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 48

Answer:

\frac{x^{2}}{2} + \frac{1}{x} + c

Hints: You must know about the integral rule of functions
Given:

\int \frac{x^{3} - 1}{x^{2}} d x

Solutions:

\int \frac{x^{3} - 1}{x^{2}} d x

\int \frac{x^{3}}{x^{2}} + \frac{1}{x^{2}} d x

\begin{aligned} \int (x - \frac{1}{x^{2}}) d x \\ \int x d x - \int \frac{1}{x^{2}} d x \\ \int x d x - \int x^{- 2} d x \\ [\frac{x^{1 + 1}}{1 + 1}] - [\frac{x^{- 2 + 1}}{- 2 + 1}] + c \\ \frac{x^{2}}{2} - \frac{x^{- 1}}{- 1} + c \\ \frac{x^{2}}{2} + \frac{1}{x} + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 49

Answer:

\frac{x^{3}}{3} + x + c

Hints: You must know about the integral rule of functions.
Given:

\int \frac{x^{3} - x^{2} + x - 1}{x - 1} d x

Solution:

\begin{aligned} \int \frac{x^{3} - x^{2} + x - 1}{x - 1} d x \\ \int \frac{x^{2} (x - 1) + 1 (x - 1)}{x - 1} d x \\ \int \frac{(x^{2} + 1) (x - 1)}{x - 1} d x \\ \int (x^{2} + 1) d x \end{aligned}

\begin{aligned} \int x^{2} d x + \int 1 d x \\ = \frac{x^{3}}{3} + x + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 50

Answer:

e^{\tan^{- 1} x} + c

Hints: You must know about the integral rule of exponential and trigonometric functions.
Given:

\int \frac{e^{\tan^{- 1} x}}{1 + x^{2}} d x

Let

\tan^{- 1} x = t

and differentitate on both sides,

\begin{aligned} \frac{1}{1 + x^{2}} = \frac{d t}{d x} \\ d x = (1 + x^{2}) d t \\ \int \frac{e^{t}}{1 + x^{2}} \cdot (1 + x^{2}) d t \\ \int e^{t} d t = e^{t} + c \\ = e^{\tan^{- 1} x} + c [\int e^{x} d x = e^{x}] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 51

Answer:

\sin^{- 1} x + c

Hints: You must know about the integral rule of trigonometric functions.
Given:

\int \frac{1}{\sqrt{1 - x^{2}}} d x

Solution:

\int \frac{1}{\sqrt{1 - x^{2}}} d x

Let

x = \sin (u)

\begin{aligned} d x = \cos u d u \\ \int \frac{1}{\sqrt{1 - x^{2}}} d x = \int \frac{1}{1 - \sin^{2} u} \cos u d u \\ \sin^{2} u + \cos^{2} u = 1 \\ \cos^{2} u = 1 - \sin^{2} u \end{aligned}

\begin{aligned} = \int \frac{\cos (u)}{\sqrt{\cos^{2} (u)}} d u \\ = \int \frac{\cos u}{\cos u} d u \\ = \int 1 d u \\ = u + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \end{aligned}

Where

x = \sin u

and

u = \sin^{- 1} x + c

= \sin^{- 1} x + c

Indefinite Integrals Excercise Very Short Answers Question 52

Answer:

\tan x + \sec x + c

Hints: You must know about the integral rule of trigonometric functions.
Given:

\int \sec x (\sec x + \tan x) d x

\int \sec x (\sec x + \tan x) d x

Solution:

\begin{aligned} \int \sec^{2} x + \sec x \tan x d x \\ \int \sec^{2} x d x + \int \sec x \cdot \tan x d x \\ \tan x + \sec x + c [∵ \int \sec^{2} x d x = \tan x and \int \sec x \cdot \tan x d x = \sec x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 53

Answer:

\frac{1}{4} \tan^{- 1} (\frac{x}{4}) + c

Hints: You must know about rules of integration
Given:

\int \frac{1}{x^{2} + 16} d x

Solution:

\begin{aligned} \int \frac{1}{x^{2} + 16} d x \\ = \int \frac{1}{x^{2} + (4)^{2}} d x \\ \int \frac{1}{x^{2} + 4^{2}} d x = \frac{1}{4} \tan^{- 1} (\frac{x}{4}) + c \int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 54

Answer:

\frac{2 x^{\frac{3}{2}}}{3} - \frac{2 x^{\frac{5}{2}}}{5} + c

Hints: You must know about rules of integration
Given:

\int (1 - x) \sqrt{x} d x

Solution:

\begin{aligned} \int (1 - x) \sqrt{x} d x \\ \int \sqrt{x} - x \sqrt{x} d x \\ \int x^{\frac{1}{2}} - x^{\frac{3}{2}} d x \\ \int x^{\frac{1}{2}} d x - \int x^{\frac{3}{2}} d x \end{aligned}

\begin{aligned} \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} + c \\ \frac{2 x^{\frac{3}{2}}}{3} - \frac{2 x^{\frac{5}{2}}}{5} + c [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 55

Answer:

\frac{1}{6} \log | 3 x^{2} + \sin 6 x | + c

Hints: You must know about the integral rule of trigonometric functions.
Given:

\int \frac{x + \cos 6 x}{3 x^{2} + \sin 6 x} d x

Solution:

\int \frac{x + \cos 6 x}{3 x^{2} + \sin 6 x} d x

Let

3 x^{2} + \sin 6 x = t

and differentiate both sides,

[\frac{d}{d x} \sin x = \cos x]

(6 x + 6 \cos 6 x) d x = d t

(x + \cos 6 x) d x = \frac{d t}{6}

\begin{aligned} I = \int \frac{d t}{6 t} => \frac{1}{6} \log (t) + c \\ = \frac{1}{6} \log | 3 x^{2} + \sin 6 x | + c [\int \frac{1}{x} d x = \log x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 56

Answer:

\frac{e^{x}}{x} + c

Hints: You must know about the rules of exponential function of integration.
Given:

\int \frac{(x - 1)}{x^{2}} e^{x} d x = f (x) e^{x} + c

Solution:

\begin{aligned} \int \frac{(x - 1)}{x^{2}} e^{x} d x \\ \int e^{x} f (x) + f^{1} (x) d x \\ e^{x} f (x) + c {\int e^{x} (f (x) + f^{1} (x)) d x = e^{x} f (x) + c} \end{aligned}

\begin{aligned} \frac{x - 1}{x^{2}} = \frac{x}{x^{2}} - \frac{1}{x^{2}} \\ = \frac{1}{x} - \frac{1}{x^{2}} \\ f (x) = \frac{1}{x}, f^{1} (x) = \frac{- 1}{x^{2}} \end{aligned}

\begin{aligned} f (x) + f^{1} (x) = \frac{1}{x} - \frac{1}{x^{2}} \\ ∴ I = \int e^{x} (f (x) + f^{1} (x)) d x if f (x) = \frac{1}{x} \\ I = e^{x} f (x) + c \\ ∴ f (x) = \frac{1}{x} \\ I = e^{x} / x + c \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 57

Answer:

f (x) = \sec x

Hints: You must know about the integral rule of trigonometric functions.
Given:

\int e^{x} (\tan x + 1) \sec x d x = e^{x} f (x) + c

, find f(x)
Solution:

I = \int e^{x} (\tan x + 1) \sec x d x

Consider,

\begin{aligned} I = \int e^{x} (\tan x + 1) \sec x d x \\ = \int e^{x} (\sec x \tan x + \sec x) d x \\ = \int e^{x} \sec x d x + \int e^{x} \sec x \tan x d x \end{aligned}

In second integral, apply integration by parts

\begin{aligned} = \int e^{x} \sec x d x + e^{x} \int \sec x \tan x d x - \int \frac{d}{d x} e^{x} [\int (\sec x \tan x) d x] d x \\ = e^{x} \sec x - \int e^{x} \sec x d x + \int e^{x} \sec x d x \\ \begin{aligned} = e^{x} \sec x + c \\ e^{x} \sec x + c = e^{x} f (x) + c \end{aligned} \\ ∴ f (x) = secx \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 58

Answer:

- \cot x + c

Hints: You must know about the integral rule of trigonometric functions.
Given:

\int \frac{2}{1 - \cos 2 x} d x

Solution:

\begin{aligned} \int \frac{2}{1 - \cos 2 x} d x \\ = \int \frac{2}{2 \sin^{2} x} d x \\ = \int \frac{1}{\sin^{2} x} d x \\ = \int {cosec}^{2} x d x \\ = - \cot x + c [\int {cosec}^{2} x d x = - \cot x + c] \end{aligned}

Indefinite Integrals Excercise Very Short Answers Question 59

Answer:

2 x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + c

Hints: You must know about the integration rules
Given: Write Anti derivative of

\begin{aligned} (3 \sqrt{x} + \frac{1}{\sqrt{x}}) \\ \int (3 \sqrt{x} + \frac{1}{\sqrt{x}}) d x \\ \int 3 (x)^{\frac{1}{2}} d x + \int (x)^{\frac{- 1}{2}} d x \end{aligned}

\begin{aligned} \frac{3 x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + \frac{x^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} + c \\ 3 . \frac{2}{3} x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + c \end{aligned}

2 x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + c

Indefinite Integrals Excercise Very Short Answers Question 60

Answer:

\frac{π x}{2} - \frac{x^{2}}{2} + c

Hints: You must know about the rules of integration of trigonometric and inverse trigonometric functions.
Given:

\int \cos^{- 1} (\sin x) d x

Solution:

\int \cos^{- 1} (\sin x) d x

Let

\cos^{- 1} (\sin x) = θ

\begin{aligned} \sin x = \cos θ \\ \sin x = \sin (\frac{π}{2} - θ) \\ x = (\frac{π}{2} - θ) \\ θ = \frac{π}{2} - x \\ ∴ \int \cos^{- 1} (\sin x) d x = \int (\frac{π}{2} - x) d x \\ = \int \frac{π}{2} d x - \int x d x \end{aligned}

= \frac{π x}{2} - \frac{x^{2}}{2} + c

Indefinite Integrals Excercise Very Short Answers Question 62

Answer:

\log | (1 + \log x) | + c

Hints: You must know about the rules of integration of trigonometric and inverse trigonometric functions.
Given:

\int \frac{1}{x (1 + \log x)} d x

Solution:

\begin{aligned} \int \frac{1}{x (1 + \log x)} d x \\ Put (1 + \log x) = t \\ \frac{1}{x} d x = d t \\ \int \frac{1}{x (1 + \log x)} d x = \int \frac{d t}{t} \\ = \log | t | + c \\ ∴= \log | (1 + \log x) | + c \end{aligned}

Indefinite Integrals Excercise Very Short answers Question 61

Answer:

\tan x - \cot x + c

Hints: You must know about the rules of integration of trigonometric and inverse trigonometric functions.
Given:

\int \frac{1}{\sin^{2} x \cdot \cos^{2} x} d x

Solution:

\begin{aligned} \int \frac{1}{\sin^{2} x \cdot \cos^{2} x} d x \\ \int \frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x \cdot \cos^{2} x} d x \end{aligned}

\begin{aligned} \int \frac{\sin^{2} x}{\sin^{2} x \cdot \cos^{2} x} d x + \int \frac{\cos^{2} x}{\sin^{2} x \cdot \cos^{2} x} d x \\ = \int \frac{1}{\cos^{2} x} d x + \int \frac{1}{\sin^{2} x} d x \\ = \int \sec^{2} x d x + \int {cosec}^{2} x d x \\ = \tan x - \cot x + c \end{aligned}

Indefinite Integrals Excercise Very Short answers Question 22

Answer:

\frac{1}{4} e^{2 x^{2}} + c

Hint: You must know about the integral rule of logarithm and exponential functions.
Given:

\int e^{2 x^{2} + \ln x} d x

Solution:

I = \int e^{2 x^{2} + \ln x} d x

= \int x . e^{2 x^{2}} d x

[e^{\log x} = x]

Let

x^{2} = t

and differentiate both sides,

\frac{d}{d x} x^{n} = n x^{n - 1}

2 x d x = d t

I = \frac{1}{2} \int e^{2 t} d t

= \frac{e^{2 t}}{4} + c

[\int e^{a x} d x = \frac{e^{a x}}{a}]

= \frac{1}{4} e^{2 x^{2}} + c

Indefinite Integrals Excercise Very Short answers Question 46

Answer:

\frac{2^{x}}{\ln 2} + c

Hint: You must know about the integral rule of logarithmic functions.
Given:

\int 2^{x} d x

Solution:

\begin{aligned} \int 2^{x} d x \\ e^{\ln 2} = 2 \\ \int 2^{x} d x = \int {(e^{\ln 2})}^{x} \\ = \int e^{\ln 2} d x \\ Let u = x \ln 2, \\ \frac{d u}{d x} = \ln 2, \\ d x = \frac{d u}{\ln 2} \\ \int e^{x \ln 2} d x = \int e^{u} \cdot \frac{d u}{\ln 2} \end{aligned}

\begin{aligned} = \frac{1}{\ln 2} \int e^{u} d u \\ = \frac{1}{\ln 2} e^{u} + c \\ = \frac{1}{\ln 2} e^{x \ln 2} + c \\ = \frac{1}{\ln 2} {(e^{\ln 2})}^{x} + c ∴ e^{\ln 2} = 2 \\ = \frac{1}{\ln 2} 2^{x} + c \end{aligned}

= \frac{2^{x}}{\ln 2} + c

Students who study Class 12 chapter 18, Indefinite Integrals, tend to lose marks in the Very Short Answers (VSA) part. This is mostly due to the absence of confidence that they might not arrive at the right solution even if tried. And also due to insufficient time to work out and check the sums. There are 62 VSA questions in this chapter. It takes loads of time and effort to be put in by the students. The main concepts in this portion are evaluating the integrals, writing derivatives, and finding the f(x) value at the given condition.

The essential part of solving the VSAs is knowing the tricks and shortcuts to arrive at the solution. When the students have no idea of applying shortcut methods, they can refer to the RD Sharma Class 12 Chapter 18 VSA book. RD Sharma solutions This book follows the NCERT pattern; hence it becomes easier for the CBSE board school students to adapt. The RD Sharma Class 12th VSA book consists of various practice questions that the students can work out before their exams.

If you are weak in understanding the concepts in chapter 18, jump to the Class 12 RD Sharma Chapter 18 VSA Solution material to know how very short answers must be solved. Integrations would be no more difficult for you if the preparation and practice take place in the right way. You can refer to this book while doing your homework, making assignments, and preparing for tests and exams.

Another special advantage of using the RD Sharma Class 12 Solutions Indefinite Integrals VSA is that you need not spend money to use it. These helpful solution books are available on the Career360 website costless. No kind of deposit or monetary charge is required to own the RD Sharma Class 12th VSA book.

Teachers use these authorized books to prepare questions for the tests and exams too. Hence, the students who prepare for their exams using the RD Sharma Class 12 Solutions Chapter 18 VSA tend to score higher than the other students.

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. Which is the prescribed book for the class 12 students to refer to the VSA solutions of chapter 18, mathematics?

The RD Sharma Class 12th VSA solutions book is the most recommended reference material for the CBSE students to clarify their doubts on Indefinite Integrals VSA.

2. How can I utilize the RD Sharma solution book for class 12, chapter 18 mathematics for free?

It is a stroke of luck for the students to access the RD Sharma Class 12th VSA solution book for free of cost. This can be found on the Career 360 website.

3. Are the solved sums present in the RD Sharma books verified?

A group of accomplished faculty has provided the solutions in the RD Sharma books. These solutions are then verified to check the accuracy of the answers. Hence, the students need not hang back to use them.

4. Is there any particular method for solving the VSAs?

Students need not solve the Very Short Answers (VSA) in an elaborated manner. Instead, shortcuts and tricks have to be used to save time and answer accordingly.

5. Can I download the RD Sharma solution books?

Anyone who can access the RD Sharma books at the Career 360 website will download them to their device without paying any monetary charge.

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RD Sharma Solutions Class 12 Mathematics Chapter 18 VSA

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