RD Sharma Solutions Class 12 Mathematics Chapter 18 MCQ

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:30 PM IST

The class 12 RD Sharma chapter 18 exercise MCQ Indefinite Integrals course of action is particularly trusted and proposed by students and educators across the entire country The appropriate responses in the RD Sharma class 12th exercise MCQ are handpicked and made via prepared experts, making them exact and sensible enough for students In addition, in RD Sharma Solutions class 12 chapter 18 exercise MCQ, the experts offer response keys and some surprising tips in the book that the students presumably will not find elsewhere RD Sharma class 12 solutions MCQ Chapter 18 has around 43 questions

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

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

Indefinite Integrals Excercise: MCQ

Indefinite Integrals Exercise Multiple Choice Questions Question 1

Answer:
$\frac{1}{4} \tan^{- 1} \frac{x^{2}}{2} + C$
Given:
$\int \frac{x}{4 + x^{4}} d x$
Hint:
Using

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

Explanation:
Let

I = \int \frac{x}{4 + x^{4}} d x

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

[Put x^{2} = t \Rightarrow 2 x d x = dt \Rightarrow x dx = \frac{d t}{2}]

\begin{aligned} = \int \frac{1}{(2)^{2} + t^{2}} \times \frac{d t}{2} \\ = \frac{1}{2} \int \frac{d t}{t^{2} + 2^{2}} \\ = \frac{1}{2} \cdot \frac{1}{2} \tan^{- 1} \frac{t}{2} + C \dots \dots . . {\int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c} \\ = \frac{1}{4} \tan^{- 1} \frac{x^{2}}{2} + C \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 2

Answer:
$\frac{1}{2} \log (\tan (\frac{x}{2} + \frac{π}{12})) + C$
Given:
$\int \frac{1}{\cos x + \sqrt{3} \sin x} d x$
Hint:
You must know about the

\int \cos e c x d x

Explanation:
Let

I = \int \frac{1}{2 (\sin x \cdot \frac{\sqrt{8}}{2} + \cos x \cdot \frac{1}{2})} d x

= \frac{1}{2} \int \frac{d x}{\sin x \cos_{6}^{π} + \cos x \sin \frac{π}{6}}

[∵ \cos \frac{π}{6} = \frac{\sqrt{3}}{2}; \sin \frac{π}{6} = \frac{1}{2}]

= \frac{1}{2} \int \frac{d x}{\sin (x + \frac{π}{6})}

[\sin (A + B) = \sin A \cos B + \cos A \sin B]

= \frac{1}{2} \int cosec (x + \frac{π}{6}) d x

[∵ \sin x = \frac{1}{\cos θ c x}]

= \frac{- 1}{2} \log (cosec (x + \frac{π}{6}) + \cot (x + \frac{π}{6})) + C

[\int cosec (x) d x = - \log (cosec x + \cot x) + c]

Now,

\begin{aligned} \cos e c x + \cot x & = \frac{1}{\sin x} + \frac{\cos x}{\sin x} \\ = \frac{1 + \cos x}{\sin x} \\ = \frac{2 \cos^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos_{2}^{x}} \end{aligned}

[∵ 1 + \cos 2 θ = 2 \cos^{2} θ; \sin θ = 2 \sin \frac{θ}{2} \cos \frac{θ}{2}]

= \cot \frac{x}{2}

∴ From E q \cdot (i)

\begin{aligned} I = \frac{- 1}{2} \log (\cot (\frac{x + \frac{π}{6}}{2})) + C \\ = \frac{1}{2} \log {[\cot (\frac{x}{2} + \frac{π}{12})]}^{- 1} + C \\ = \frac{1}{2} \log (\tan (\frac{x}{2} + \frac{π}{12})) + C \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 3

Answer:
$\frac{1}{2} \log (\sec x^{2} + \tan x^{2}) + C$
Given:
$\int x \sec x^{2} d x$
Hint:
Using

\int \sec θ d θ

Explanation:
Let, $I = \int x \sec x^{2} d x$ $[Put x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2}]$
$\begin{aligned} = \int \sec t \cdot \frac{d t}{2} \\ = \frac{1}{2} \int \sec t d t \\ = \frac{1}{2} \log | \sec t + \tan t | + C \end{aligned}$ ${\int \sec x d x = \log | \sec x + \tan x | + c}$
$= \frac{1}{2} \log | \sec x^{2} + \tan x^{2} | + C$

Indefinite Integrals Exercise Multiple Choice Questions Question 4

Answer:
$A = \frac{2}{3}, B = \frac{5}{3}$
Given:
$\int \frac{1}{5 + 4 \sin x} d x = A \tan^{- 1} (B \tan \frac{x}{2} + \frac{4}{3}) + C$
Hint
Using

\int \frac{1}{1 + t^{2}} d t

Explanation:
Let $I = \int \frac{1}{5 + 4 \sin x} d x$
$\begin{aligned} = \int \frac{1}{5 + 4.2 \sin \frac{x}{2} \cos \frac{x}{2}} d x \\ = \int \frac{1}{5.1 + 8 \sin \frac{x}{2} \cos \frac{x}{2}} d x \end{aligned}$ $[∵ \sin 2 θ = 2 \sin θ \cos θ]$
$= \int \frac{1}{5 (\sin^{2} \frac{x}{2} + \cos^{2} \frac{x^{2}}{2}) + 8 \sin \frac{x}{2} \cos_{2}^{x}} d x$ $[∵ \sin^{2} θ + \cos^{2} θ = 1]$
$= \int \frac{1}{5 \sin^{2} \frac{x}{2} + 5 \cos^{2} \frac{x}{2} + 8 \sin \frac{x}{2} \cos_{2}^{x}} d x$
Divide num and den by

\cos^{2} \frac{x}{2}

\begin{aligned} = \int \frac{\frac{1}{\cos^{2} \frac{x}{2}}}{5 \tan^{2} \frac{x}{2} + 5 + 8 \tan_{2}^{x}} d x \\ = \frac{1}{5} \int \frac{\sec^{2} \frac{x}{2}}{\tan^{2} \frac{x}{2} + \frac{8}{5} \tan_{2}^{x} + 1} d x \end{aligned}

[∵ \sec x = \frac{1}{\cos x}]

= \frac{1}{5} \int \frac{2}{t^{2} + \frac{8}{5} t + 1} d t

[Put \tan \frac{x}{2} = t \Rightarrow \sec^{2} \frac{x}{2} \cdot \frac{1}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t]

\begin{aligned} = \frac{2}{5} \int \frac{d t}{t^{2} + \frac{8 t}{5} + {(\frac{4}{5})}^{2} - {(\frac{4}{5})}^{2} + 1} \\ = \frac{2}{5} \int \frac{d t}{{(t + \frac{4}{5})}^{2} + {(\frac{3}{5})}^{2}} \\ = \frac{2}{5} \times \frac{1}{\frac{3}{5}} \tan^{- 1} (\frac{5 t + 4}{3}) + C \dots \dots \dots \dots \dots {\frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c} \end{aligned}

∴ I = \frac{2}{3} \tan^{- 1} (\frac{5}{3} \tan \frac{x}{2} + \frac{4}{3}) + C

[E q \cdot (i)]

Acc to given,

I = A \tan^{- 1} (B \tan \frac{x}{2} + \frac{4}{3}) + C

[E q \cdot (i i)]

From Eq ( i ) and ( ii )

$A \tan^{- 1} (B \tan \frac{x}{2} + \frac{4}{3}) + C = \frac{2}{3} \tan^{- 1} (\frac{5}{3} \tan \frac{x}{2} + \frac{4}{3}) + C$
Comparing both sides, we get
$A = \frac{2}{3}, B = \frac{5}{3}$

Indefinite Integrals Exercise Multiple Choice Questions Question 5

Answer:
$x^{\sin x} + C$
Given:
$\int x^{\sin x} (\frac{\sin x}{x} + \cos x \cdot \log x) d x$
Hint:
You must know about the derivation of

x^{\sin x}

Explanation:
Let $I = \int x^{\sin x} (\frac{\sin x}{x} + \cos x \cdot \log x)$
$= \int t \cdot \frac{d t}{t}$ $[ Put x^{\sin x} = t, Taking log of both sides$
$= \int 1 d t$ $\sin x \cdot \log x = \log t$
$= t + C$ Diff w r t t,

= x^{\sin x} + C

(\log x \cos x + \frac{\sin x}{x}) \frac{d x}{d t} = \frac{1}{t} \Rightarrow (\log x \cos x + \frac{\sin x}{x}) d x = \frac{d t}{t}]

Indefinite Integrals Exercise Multiple Choice Questions Question 6

Answer:
$\tan^{- 1} {(\log_{e} x)}^{2} + C$
Given:
$\int \frac{1}{1 + {(\log_{e} x)}^{2}} d x w.r.t. (\log_{e} x)$
Hint:
Using $\int \frac{1}{1 + t^{2}} d x$
Explanation:
Let $I = \int \frac{1}{1 + {(\log_{e} x)}^{2}} d x (\log x)$
$= \int \frac{1}{1 + {(\log_{e} x)}^{2}} \frac{d x}{x}$ $[\frac{d}{d x} (\log x) = \frac{1}{x} \Rightarrow d (\log x) = \frac{d x}{x}]$
$= \int \frac{1}{1 + t^{2}} \cdot d t$ $[Put \log_{e} x = t \Rightarrow \frac{1}{x} d x = d t]$
$= \tan^{- 1} t + C$ $[∵ \int \frac{1}{1 + x^{2}} d x = \tan^{- 1} x + C]$
$= \tan^{- 1} (\log_{e} x) + C$

Indefinite Integrals Exercise Multiple Choice Questions Question 7

Answer:
$\frac{1}{16}$
Given:
$\int \frac{\cos 8 x + 1}{\tan 2 x - \cot 2 x} d x = a \cos 8 x + C$
Hint
Using

\int \sin x d x

Explanation:
Let $I = \int \frac{\cos 8 x + 1}{\tan 2 x - \cot 2 x} d x$
$Missing or unrecognized delimiter for \left$
$= \int \frac{2 \cos^{2} 4 x \cdot \sin 2 x \cos 2 x}{\sin^{2} 2 x - \cos^{2} 2 x} d x$
$= \int \frac{\cos^{2} 4 x \sin 4 x}{- \cos 4 x} d x$ $[∵ \cos^{2} x - \sin^{2} x = \cos 2 x]$
$= - \int \cos 4 x \sin 4 x d x$
$= - \frac{1}{2} \int \sin 8 x d x$ $[∵ \sin 2 x = 2 \sin x \cos x]$
$= - \frac{1}{2} \int \sin 8 x d x$ $[∵ \sin 2 x = 2 \sin x \cos x]$
$\begin{aligned} = - \frac{1}{2} (- \frac{\cos 8 x}{8}) + C \\ I = \frac{\cos 8 x}{16} + C \end{aligned}$
According to given,

\begin{aligned} I = a \cos 8 x + C \\ ∵ a \cos 8 x + C = \frac{1}{16} \cos 8 x + C \\ ∴ a = \frac{1}{16} \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 8

Answer:
$- \frac{1}{2}$
Given:
$\int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x = a \sin 2 x + C$
Hint:
Using identity

(a^{2} - b^{2}) & \int \cos x d x

Explanation:
Let $I = \int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x$
$= \int \frac{{(\sin^{4} x)}^{2} - {(\cos^{4} x)}^{2}}{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x} d x$ $[∵ \sin^{2} x + \cos^{2} x = 1]$
$\begin{aligned} = \int \frac{(\sin^{4} x - \cos^{4} x) (\sin^{4} x + \cos^{4} x)}{(\sin^{4} x + \cos^{4} x)} d x \\ = \int (\sin^{4} x - \cos^{4} x) d x \\ = - \int (\cos^{4} x - \sin^{4} x) d x \\ = - \int (\cos^{2} x - \sin^{2} x) (\cos^{2} x + \sin^{2} x) d x \end{aligned}$ $[∵ a^{2} - b^{2} = (a + b) (a - b)]$
$= - 1 \int \cos 2 x (1) d x$ $[∵ \sin^{2} x + \cos^{2} x = 1]$
According to given :

I = a \sin 2 x + C

∴ a \sin 2 x + C = - \frac{1}{2} \sin 2 x + C

Comparing both sides, we get

a = - \frac{1}{2}

Indefinite Integrals Exercise Multiple Choice Questions Question 9

Answer:
$- x e^{- x} + C$
Given:
$\int (x - 1) e^{- x} d x$
Hint:
Using integration by parts &

\int e^{- x} d x

Explanation:
Let $I = \int (x - 1) e^{- x} d x$
$\begin{aligned} = \int x e^{- x} d x - \int e^{- x} d x \\ = x \frac{e^{- x}}{- 1} + \int (1) e^{- x} d x - \int e^{- x} d x \end{aligned}$ $[\int u . v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x]$
$= - x e^{- x} + \int e^{- x} d x - \int e^{- x} d x$ $[∵ \int e^{x} d x = e^{x} + C]$
$= - x e^{- x} + C$

Indefinite Integrals Exercise Multiple Choice Questions Question 10

Answer:
$- \frac{1}{\log_{e} 2}$
Given:
$\int \frac{2^{\frac{1}{x}}}{x^{2}} d x = K 2^{\frac{1}{x}} + C$
Hint
Using

\int a^{x} d x

Explanation:
Let $= \int \frac{2^{\frac{1}{x}}}{x^{2}} d x$ $[Put \frac{1}{x} = t \Rightarrow - \frac{1}{x^{2}} d x = d t \Rightarrow \frac{d x}{x^{2}} = - d t$
$\begin{aligned} = - \int 2^{t} d t \\ = - \frac{2^{t}}{l o g_{e} 2} + 2 \end{aligned}$
$= - \frac{2 \frac{1}{x}}{\log_{e} 2} + 2$ $[∵ \int a^{x} d x = \frac{a^{x}}{\log_{e} a}]$
According to given:

I = K 2^{\frac{1}{x}} + C

∴ K 2^{\frac{1}{x}} + C = - \frac{1}{\log_{e} 2} \cdot 2^{\frac{1}{x}} + C

Comparing both sides,

$K = \frac{- 1}{\log_{e} 2}$

Indefinite Integrals Exercise Multiple Choice Questions Question 11

Answer:
$\frac{1}{2} [x + \log | (\sin x + \cos x) |] + C$
Given:
$\int \frac{1}{1 + \tan x} d x$
Hint:
You must know about the derivation of sin x and cos x and use

\int \frac{d x}{x}

Explanation:
Let

I = \int \frac{1}{1 + \tan x} d x

= \frac{1}{1 + \tan x}

= \frac{1}{1 + \frac{\sin x}{\cos x}}

[∵ \tan x = \frac{\sin x}{\cos x}]

= \frac{\cos x}{\cos x + \sin x}

= \frac{2 \cos x}{2 (\cos x + \sin x)}

= \frac{(\cos x + \sin x) + (\cos x - \sin x)}{2 (\cos x + \sin x)}

[Special step]

\begin{aligned} = \frac{1}{2} [1 + \frac{\cos x - \sin x}{\cos x + \sin x}] \\ ∴ I_{2} = \int \frac{1}{1 + \tan x} d x \\ = \int \frac{1}{2} d x + \frac{1}{2} \int \frac{\cos x - \sin x}{\cos x + \sin x} d x \\ ∴ I_{2} = \frac{1}{2} \int 1 d x + \frac{1}{2} \int \frac{d t}{t} \end{aligned}

[Put \cos x + \sin x = t in I_{2}]

[(- \sin x + \cos x) d x = d t \Rightarrow (\cos x - \sin x) d x = d t]

= \frac{1}{2} x + \frac{1}{2} \log | t | + C

= \frac{1}{2} [x + \log | (\cos x + \sin x) |] + C

Indefinite Integrals Exercise Multiple Choice Questions Question 12

Answer:
None of these
Given:
$\int | x |^{3} d x$
Hint:
Using $\int x^{n} d x$
Explanation:
Let $I = \int | x |^{3} d x$
Here two cases arise

\begin{aligned} I = {- x^{3}, x < 0; x^{3}, x \geq 0 \\ ∴ Case I, when x < 0 \end{aligned}

\begin{aligned} I = \int_{4} - x^{3} d x \\ = \frac{- x^{4}}{4} + C \end{aligned}

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

Case II, when x \geq 0

\begin{aligned} I = \int_{4} x^{3} d x \\ = \frac{x^{4}}{4} + C \end{aligned}

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

Combining these two, we get

I = \pm \frac{x^{4}}{4} + C

Indefinite Integrals Exercise Multiple Choice Questions Question 13

Answer:
$2 \sin \sqrt{x} + C$
Given:

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

Hint
Using

\int \cos x d x

Explanation:
Let $I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$
$= \int \cos t .2 d t$ $[Put \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t]$
$\begin{aligned} = 2 \int \cos t d t \\ = 2 \sin t + C \\ = 2 \sin \sqrt{x} + C \end{aligned}$ $[∵ \int \cos x d x = \sin x + C]$

Indefinite Integrals Exercise Multiple Choice Questions Question 14

Answer:
$- e^{x} \cot x + C$
Given:
$\int e^{x} (1 - \cot x + \cot^{2} x) d x$
Hint:
Using integration by parts
Explanation:
Let

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

\begin{aligned} = \int e^{x} [(1 + \cot^{2} x) - \cot x] d x \\ = \int e^{x} ({cosec}^{2} x - \cot x) d x \end{aligned}

[∵ 1 + \cot^{2} θ = \cos e c^{2} θ]

\begin{aligned} = - \int e^{x} (\cot x - \cos e c^{2} x) d x \\ = & - \int e^{x} \cot x d x + \int e^{x} {cosec}^{2} x d x \\ = & - [\cot x \cdot e^{x} - \int (- \cos e c^{2} x) e^{x} d x] + \int e^{x} \cos e c^{2} x d x \dots . {\int u \cdot v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x} \\ = & - \cot x \cdot e^{x} - \int e^{x} \cos e c^{2} x d x + \int e^{x} \cos e c^{2} x d x \\ = & - \cot x e^{x} + C [Using integration by parts ] \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 15

Answer:
$\frac{\tan^{7} x}{7} + C$
Given:
$\int \frac{\sin^{6} x}{\cos^{8} x} d x$
Hint:
Using

\int x^{n} d x

Explanation:
Let $I = \int \frac{\sin^{6} x}{\cos^{8} x} d x$
$= \int (\frac{\sin^{6} x}{\cos^{6} x} \times \frac{1}{\cos^{2} x}) d x$
$= \int \tan^{6} x \sec^{2} x d x$ $[∵ \tan x = \frac{\sin x}{\cos x}; \sec x = \frac{1}{\cos x}]$
$= \int t^{6} d t$ $Put \tan x = t \Rightarrow \sec^{2} x d x = d t]$
$\begin{aligned} = \frac{t^{7}}{7} + C \\ = \frac{\tan^{7} x}{7} + C \end{aligned}$ $[∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C]$

Indefinite Integrals Exercise Multiple Choice Questions Question 16

Answer:
$\frac{1}{\sqrt{6}} \tan^{- 1} (\frac{1}{\sqrt{6}} \tan \frac{x}{2}) + C$
Given:
$\int (\frac{1}{7 + 5 \cos x}) d x$
Hint:
You must know about the derivation of tan x and using

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

Explanation:
Let

I = \int \frac{1}{7 + 5 \cos x} d x

Put \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}

\begin{aligned} ∴ I & = \int \frac{1}{7 + 5 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x \\ = \int \frac{1 + \tan^{2} \frac{x}{2}}{7 (1 + \tan^{2} \frac{x}{2}) + 5 - 5 \tan^{2} \frac{x}{2}} d x \\ = \int \frac{\sec^{2} \frac{x}{2}}{12 + 2 \tan^{2} \frac{x}{2}} \\ = \frac{1}{2} \int \frac{\sec^{2} \frac{x}{2}}{\tan^{2} \frac{x}{2} + 6} d x \\ = \frac{1}{2} \int \frac{2 d t}{t^{2} + (\sqrt{6})^{2}} \end{aligned}

[ Put \tan \frac{x}{2} = t \Rightarrow \sec^{2} \frac{x}{2} \cdot \frac{1}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t]

\begin{aligned} = \int \frac{d t}{t^{2} + (\sqrt{6})^{2}} \\ = \frac{1}{\sqrt{6}} \tan^{- 1} (\frac{t}{\sqrt{6}}) + C \dots \dots \dots \frac{1}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + c \\ = \frac{1}{\sqrt{6}} \tan^{- 1} (\frac{\tan \frac{x}{2}}{\sqrt{6}}) + C \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 17

Answer:
$\log | 1 - \cot \frac{x}{2} | + C$
Given:
$\int \frac{1}{1 - \cos x - \sin x} d x$
Hint:
Using partial fraction and

\int \frac{1}{x} d x

Explanation:
Let

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

$\begin{aligned} Put \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan \frac{x^{x}}{2}}; \sin x = \frac{2 \tan_{2}^{x}}{1 + \tan^{\frac{x}{2}}} \\ ∴ I = \int \frac{1}{1 - (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) - (\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x \end{aligned}$
$\begin{aligned} = \int \frac{1 + \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} - 1 + \tan^{2} - 2 \tan \frac{x}{2}} d x \\ = & \int \frac{\sec^{2} \frac{x}{2}}{2 \tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2}} d x \\ = & \int \frac{\sec^{2} \frac{x}{2} \cdot \frac{1}{2}}{\tan \frac{x}{2} (\tan_{2}^{x} - 1)} d x [∵ 1 + \tan^{2} θ = \sec^{2} θ] \end{aligned}$
$= \int \frac{d t}{t (t - 1)}$ $[ Put \tan \frac{x}{2} = t \Rightarrow \sec^{2} \frac{x}{2} \cdot \frac{1}{2} d x = d t]$
Now,

\frac{1}{t (t - 1)} = \frac{A}{t} + \frac{B}{t - 1}

Multiplying by t (t - 1)

\Rightarrow 1 = A (t - 1) + B (t)

Putting t = 1

$\begin{aligned} 1 = A (0 - 1) + B (0) \Rightarrow A = - 1 \\ ∴ \frac{1}{t (t - 1)} = \frac{- 1}{t} + \frac{1}{t - 1} \\ ∴ \int \frac{1}{t (t - 1)} d t = - \int \frac{1}{t} d t + \int \frac{1}{t - 1} d t \end{aligned}$
$\begin{aligned} = - \log | t | + \log | t - 1 | + C \\ = - \log | \tan \frac{x}{2} | + \log | \tan \frac{x}{2} - 1 | + C \\ = \log | \frac{\tan_{\frac{x}{2}} - 1}{\tan_{2}^{\frac{x}{x}}} | + C \\ = \log | 1 - \cot \frac{x}{2} | + C \end{aligned}$ $[∵ \cot x = \frac{1}{\tan x}]$

Indefinite Integrals Exercise Multiple Choice Questions Question 18

Answer:
$\frac{e^{x}}{x + 4} + C$
Given:
$\int \frac{x + 3}{(x + 4)^{2}} e^{x} d x$
Hint:
Using

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

Explanation:
Let $I = \int \frac{x + 3}{(x + 4)^{2}} e^{x} d x$
$= \int \frac{(x + 3 + 1 - 1)}{(x + 4)^{2}} e^{x} d x$
$= \int e^{x} [\frac{x + 4}{(x + 4)^{2}} - \frac{1}{(x + 4)^{2}}] d x$
$\begin{aligned} = \int e^{x} [\frac{1}{x + 4} - \frac{1}{(x + 4)^{2}}] d x \\ = \int e^{x} \frac{1}{x + 4} d x - \int e^{x} \frac{1}{(x + 4)^{2}} d x \\ = \frac{1}{x + 4} \cdot e^{x} - \int \frac{- 1}{(x + 4)^{2}} \cdot e^{x} d x - \int \frac{1}{(x + 4)^{2}} \cdot e^{x} d x \dots \dots \dots . . {\int u . v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x} \\ = \frac{e^{x}}{x + 4} + \int \frac{1}{(x + 4)^{2}} e^{x} d x - \int \frac{1}{(x + 4)^{2}} e^{x} d x \\ = \frac{e^{x}}{x + 4} + C \end{aligned}$

Indefinite Integrals Exercise Multiple Choice Questions Question 19

Answer:
$\frac{- 1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C$
Given:
$\int \frac{\sin x}{3 + 4 \cos^{2} x} d x$
Hint:
You must know about the derivative of cos x and

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

Explanation:
Let

I = \int \frac{\sin x}{3 + 4 \cos^{2} x} d x

= \int \frac{- d t}{3 + 4 t^{2}}

[Put \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t]

\begin{aligned} = \frac{- 1}{4} \int \frac{d t}{t^{2} + \frac{3}{4}} \\ = & \frac{- 1}{4} \int \frac{d t}{(t)^{2} + {(\frac{\sqrt{3}}{2})}^{2}} \\ = & \frac{- 1}{4} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} (\frac{t}{\frac{\sqrt{3}}{2}}) + c \dots \dots \int \frac{1}{x^{2} + a^{2}} d x = \tan^{- 1} \frac{x}{a} + c \\ = & \frac{- 1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 t}{\sqrt{3}}) + C \\ = & \frac{- 1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 20

Answer:
$- e^{x} \cot \frac{x}{2} + C$
Given:
$\int e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x$
Hint:
You must know about the derivation of cot x and

\int e^{x} d x

Explanation:
Let

I = \int e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x

= \int e^{x} (\frac{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} - 2 \sin \frac{x}{2} \cos_{2}^{x}}{2 \sin^{2} \frac{x}{2}}) d x

[\begin{array}{l} ∵ \sin^{2} θ + \cos^{2} θ = 1; \sin 2 θ = 2 \sin θ \cos θ; \\ 1 - \cos θ = 2 \sin^{2} \frac{θ}{2} \end{array}]

\begin{aligned} = \frac{1}{2} \int e^{x} {(\frac{\sin \frac{x}{2} - \cos \frac{x}{2}}{\sin \frac{x}{2}})}^{2} d x \\ = \frac{1}{2} \int e^{x} {(1 - \cot \frac{x}{2})}^{2} d x \\ = \frac{1}{2} \int e^{x} ((1 + \cot^{2} \frac{x}{2}) - 2 \cot \frac{x}{2}) d x \end{aligned}

[∵ \cot x = \frac{\cos x}{\sin x}]

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

[∵ 1 + \cot^{2} θ = \cos e^{2} θ]

\begin{aligned} = \frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x - \int \cot \frac{x}{2} e^{x} d x \\ = & - \int \cot \frac{x}{2} \cdot e^{x} d x + \int \cos e c^{2} \frac{x}{2} e^{x} \cdot \frac{1}{2} d x \\ = & - [\cot \frac{x}{2} \cdot e^{x} - \int - {cosec}^{2} \frac{x}{2} \cdot \frac{1}{2} \cdot e^{x} d x] + \frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x \dots . . [\int u . v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x] \\ = & - \cot \frac{x}{2} \cdot e^{x} - \frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x + \frac{1}{2} \int e^{x} \cos e c^{2} \frac{x}{2} d x \\ = & - \cot \frac{x}{2} \cdot e^{x} + C \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 21

Answer:
$\frac{- e^{- x}}{e^{x} + e^{- x}} + C$
Given:
$\int \frac{2}{{(e^{x} + e^{- x})}^{2}} d x$
Hint:
You must know about the

\int x^{n} d x

Explanation:
$\begin{aligned} I = \int \frac{2 d x}{{(e^{x} + e^{- x})}^{2}} \\ I = \int \frac{2 e^{x}}{e^{x} {(e^{x} + e^{- x})}^{2}} d x \\ Put e^{x} = t \Rightarrow e^{x} d x = d t \\ I = \int \frac{2 d t}{t {(t + \frac{1}{t})}^{2}} \\ I = \int \frac{2 d t}{\frac{t {(t^{2} + 1)}^{2}}{t^{2}}} \\ = \int \frac{2 t d t}{{(t^{2} + 1)}^{2}} \\ Put z = t^{2} + 1 \Rightarrow d z = 2 t d t \end{aligned}$
$\begin{aligned} I = \int \frac{d z}{z^{2}} \\ = \int z^{- 2} d z \\ = \frac{z^{- 2 + 1}}{- 2 + 1} + C [∵ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C] \\ = - \frac{1}{z} + C \\ = \frac{- 1}{(t^{2} + 1)} + C [∵ z = t^{2} + 1] \\ = \frac{- 1}{e^{2 x} + 1} + C [∵ t = e^{x}] \\ = \frac{- 1}{e^{x} (e^{x} + \frac{1}{e^{x}})} + C \\ = \frac{- e^{- x}}{(e^{x} + e^{- x})} + C \end{aligned}$
Hence

I = \frac{- e^{- x}}{e^{x} + e^{- x}} + C

Option (a) is correct

Indefinite Integrals Exercise Multiple Choice Questions Question 22

Answer:
$\tan (x e^{x}) + C$
Given:
$\int \frac{e^{x} (1 + x)}{\cos^{2} (x e^{x})} d x$
Hint:
You must know about the

\int \sec^{2} x d x

Explanation:
Let

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

= \int \frac{d t}{\cos^{2} t}

[ Put x e^{x} = t \Rightarrow (x e^{x} + e^{x} (1)) d x = d t \Rightarrow e^{x} (x + 1) d x = d t]

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

[∵ \sec θ = \frac{1}{\cos θ}]

\begin{aligned} = \tan t + C \\ = \tan (x e^{x}) + C \end{aligned}

Hence,

\int \frac{e^{x} (1 + x) d x}{\cos^{2} (x e^{x})} = \tan (x e^{x}) + C

Indefinite Integrals Exercise Multiple Choice Questions Question 23

Answer:
$\frac{1}{3} \tan^{3} x + C$
Given:
$\int \frac{\sin^{2} x}{\cos^{4} x} d x$
Hint:
You must know about the derivation of tan x and

\int x^{n} d x

Explanation:
Let

I = \int \frac{\sin^{2} x}{\cos^{4} x} d x

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

[∵ \frac{\sin x}{\cos x} = \tan x; \frac{1}{\cos x} = \sec x]

Put

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

\begin{aligned} = \int t^{2} d t \\ = \frac{(t)^{2 + 1}}{2 + 1} + C \\ = \frac{1}{3} \tan^{3} x + C \end{aligned}

Hence,

\int \frac{\sin^{2} x}{\cos^{4} x} = \frac{1}{3} \tan^{3} x + C .

Indefinite Integrals Exercise Multiple Choice Questions Question 24

Answer:
$\frac{a^{x + \frac{1}{x}}}{\log_{e} a}$
Given:
$f (x) = (1 - \frac{1}{x^{2}}) a^{x + \frac{1}{x}}, a > 0$
Hint:
You must know about the

\int a^{x} d x

Explanation:
Let

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

= \int a^{t} d t

[Put x + \frac{1}{x} = t \Rightarrow (1 - \frac{1}{x^{2}}) d x = d t]

\begin{aligned} = \frac{a^{t}}{\log_{e} a} \\ = \frac{a^{x + \frac{1}{x}}}{\log_{e} a} \end{aligned}

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

Hence primitive of

f (x) is \frac{a^{x + \frac{1}{x}}}{\log_{e} a}

Indefinite Integrals Exercise Multiple Choice Questions Question 25

Answer:
$\log | 1 + \log x | + C$
Given:
$\int \frac{1}{x + x \log x} d x$
Hint:
Using

\int \frac{1}{x} d x

Explanation:
Let

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

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

[ Put 1 + \log x = t \Rightarrow \frac{1}{x} d x = d t]

\begin{aligned} = \log | t | + C \\ = \log | 1 + \log x | + C \end{aligned}

Hence,

\int \frac{1}{x + x \log x} d x = \log | 1 + \log x | + C

Indefinite Integrals Exercise Multiple Choice Questions Question 26

Answer:
$\sin^{- 1} \sqrt{x} - \sqrt{x (1 - x)} + C$
Given:
$\int \sqrt{\frac{x}{1 - x}} d x$ is equal to:
Hint:
You must know about the

\int \sin^{2} θ d θ .

Explanation:
Let I

= \int \sqrt{\frac{x}{1 - x}}

= \int \sqrt{\frac{\sin^{2} θ}{1 - \sin^{2} θ} \times} 2 \sin θ \cos θ d θ

[ Put x = \sin^{2} θ \Rightarrow d x = 2 \sin θ \cos θ d θ]

= \int \sqrt{\frac{\sin^{2} θ}{\cos^{2} θ}} \cdot 2 \sin θ \cos θ d θ

[∵ 1 - \sin^{2} θ = \cos^{2} θ]

\begin{aligned} = \int \sqrt{\tan^{2} θ} \cdot 2 \sin θ \cos θ d θ \\ = \int \frac{\sin θ}{\cos θ} \times 2 \sin θ \cos θ d θ \\ = \int 2 \sin^{2} θ d θ \\ = \int (1 - \cos 2 θ) d θ \end{aligned}

[∵ 2 \sin^{2} θ = 1 - \cos 2 θ]

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

Indefinite Integrals Exercise Multiple Choice Questions Question 27

Answer:
$e^{x} f (x) + C$
Given:
$\int e^{x} [f (x) + f^{'} (x)] d x$
Hint:
You must know about the rule to integrate by parts
Explanation:
Let

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

\begin{aligned} = \int e^{x} f (x) d x + \int e^{x} f^{'} (x) d x \\ = f (x) e^{x} - \int e^{x} f^{'} (x) d x + \int e^{x} f^{'} (x) d x \dots {\int u . v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x} \\ = f (x) e^{x} + C \end{aligned}

Hence,

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

Indefinite Integrals Exercise Multiple Choice Questions Question 28

Answer:
$\pm \log (\sin x - \cos x) + C$
Given:
$\int \frac{\sin x + \cos x}{\sqrt{1 - \sin 2 x}}$
Hint:
You must know about the derivation of sin and cos function and

\int \frac{1}{x} d x

Explanation:
Let

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

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

[∵ \sin^{2} x + \cos^{2} x = 1]

= \int \frac{\sin x + \cos x}{\sqrt{(\cos x - \sin x)^{2}}} d x

[∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b]

\begin{aligned} = \int \frac{\sin x + \cos x}{(\cos x - \sin x)} d x \\ = \int \frac{d t}{t} \end{aligned}

[Put \cos x - \sin x = t \Rightarrow (\cos x + \sin x) d x = d t]

\begin{aligned} = \log | t | + C \\ = \log (\cos x - \sin x) + C \end{aligned}

Hence,

\int \frac{\sin x + \cos x}{\sqrt{1 - \sin 2 x}} = \log (\cos x - \sin x) + C .

Indefinite Integrals Exercise Multiple Choice Questions Question 29

Answer:
$\sin x + C$
Given:
$\int x \sin x d x = - x \cos x + α$
Hint:
You must know the derivation of sin x and rule to integrate by parts
Explanation:
Let

I = \int x \sin x d x

= x (- \cos x) + \int (1) \cos x d x

[ : :Integration by parts \int u . v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x]

= - x \cos x + \sin x + C

Acc to given

\begin{aligned} I = - x \cos x + α \\ \Rightarrow - x \cos x + \sin x + C = - x \cos x + α \end{aligned}

Comparing both sides
We get,

α = \sin x + C

Indefinite Integrals Exercise Multiple Choice Questions Question 30

Answer:
$x - \tan x + C$
Given:
$\int \frac{\cos 2 x - 1}{\cos 2 x + 1} d x$
Hint:
Using

\int \tan^{2} x d x

Explanation:
Let

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

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

[∵ \cos 2 x = \cos^{2} x - \sin^{2} x]

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

[\begin{array}{l} ∵ \sin^{2} x + \cos^{2} x = 1 \\ \Rightarrow 1 - \sin^{2} x = \cos^{2} x \end{array}]

\begin{aligned} = \int \frac{- 2 \sin^{2} x}{2 \cos^{2} x} d x \\ = - \int \tan^{2} x d x \end{aligned}

[∵ \tan x = \frac{\sin x}{\cos x}]

\begin{aligned} = - \int (\sec^{2} x - 1) d x \\ = \int (1 - \sec^{2} x) d x \\ = \int 1 d x - \int \sec^{2} x d x \\ = x - \tan x + C \end{aligned}

Hence,

\int \frac{\cos 2 x - 1}{\cos 2 x + 1} d x = x - \tan x + C

Indefinite Integrals Exercise Multiple Choice Questions Question 31

Answer:
$2 (\sin x + x \cos θ) + C$
Given:
$\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$
Hint:
You must know about the integration of cos x
Explanation:
Let

I = \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x

= \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} θ - 1)}{\cos x - \cos θ} d x

[∵ \cos 2 θ = 2 \cos^{2} θ - 1]

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

[∵ a^{2} - b^{2} = (a + b) (a - b)]

\begin{aligned} = 2 \int \cos x d x + 2 \int \cos θ d x \\ = 2 \int \cos x d x + 2 \cos θ \int 1 d x \\ = 2 \sin x + 2 \cos θ \cdot x + C \\ = 2 (\sin x + x \cos θ) + C \end{aligned}

Hence,

\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x = 2 (\sin x + x \cos θ) + C .

Indefinite Integrals Exercise Multiple Choice Questions Question 32

Answer:
$\frac{1}{10} {[4 + \frac{1}{x^{2}}]}^{- 5} + C$
Given:
$\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$
Hint:
Using

\int x^{n} d x

Explanation:
Let

I = \int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x

= \int \frac{x^{9}}{x^{12} {(4 + \frac{1}{x^{2}})}^{6}} d x \dots \dots \dots taking x^{2} common from denominator

\begin{aligned} = \int \frac{1}{x^{3} {(4 + \frac{1}{x^{2}})}^{6}} d x \\ = \int \frac{\frac{1}{x^{3}}}{{(4 + \frac{1}{x^{2}})}^{6}} d x \end{aligned}

[ Put 4 + \frac{1}{x^{2}} = t \Rightarrow \frac{- 2}{x^{3}} d x = d t]

\begin{aligned} = - \frac{1}{2} \int \frac{d t}{t^{6}} \\ = - \frac{1}{2} [\frac{t^{- 6 + 1}}{- 6 + 1}] + C \\ = - \frac{1}{2} [\frac{(t)^{- 5}}{- 5}] + C \\ = \frac{1}{10} \times \frac{1}{t^{5}} + C \\ = \frac{t^{- 5}}{10} + C \\ = \frac{1}{10} {[4 + \frac{1}{x^{2}}]}^{- 5} + C \end{aligned}

Hence,

\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x = \frac{1}{10} {[4 + \frac{1}{x^{2}}]}^{- 5} + C

Indefinite Integrals Exercise Multiple Choice Questions Question 33

Answer:
$a = \frac{1}{3}; b = - 1$
Given:
$\int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x = a {(1 + x^{2})}^{\frac{3}{2}} + b \sqrt{1 + x^{2}} + C$
Hint:
Using

\int x^{n} d x

Explanation:
Let

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

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

[ Put 1 + x^{2} = t \Rightarrow x^{2} = t - 1 \Rightarrow 2 x d x = d t]

\begin{aligned} = \frac{1}{2} \int \frac{t - 1}{\sqrt{t}} d t \\ = \frac{1}{2} \int \sqrt{t} d t - \frac{1}{2} \int \frac{1}{\sqrt{t}} d t \\ = \frac{1}{2} \int (t)^{\frac{1}{2}} d t - \frac{1}{2} \int (t)^{\frac{- 1}{2}} d t \\ = \frac{1}{2} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] - \frac{1}{2} [\frac{(t)^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1}] + C \\ = \frac{1}{2} \times \frac{t^{\frac{3}{2}}}{\frac{3}{2}} - \frac{1}{2} \times \frac{(t)^{\frac{1}{2}}}{\frac{1}{2}} + C \\ = \frac{1}{3} t^{\frac{3}{2}} - t^{\frac{1}{2}} + C \end{aligned}

= \frac{1}{3} {(1 + x^{2})}^{\frac{3}{2}} - \sqrt{1 + x^{2}} + C

Acc to given

I = a {(1 + x^{2})}^{\frac{8}{2}} + b \sqrt{1 + x^{2}} + C

\Rightarrow \frac{1}{3} {(1 + x^{2})}^{\frac{3}{2}} - \sqrt{1 + x^{2}} + C = a {(1 + x^{2})}^{\frac{3}{2}} + b \sqrt{1 + x^{2}} + C

On comparing, we get

a = \frac{1}{3}, b = - 1

Indefinite Integrals Exercise Multiple Choice Questions Question 34

Answer:
$x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log | 1 + x | + C$
Given:
$\int \frac{x^{3}}{x + 1} d x$
Hint:
You must know how to integrate

\int x^{n} d x

and

\int \frac{1}{x} d x

Explanation:
Let

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

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

[∵ x^{3} + 1 = (x + 1) (x^{2} + 1 - x)]

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

Indefinite Integrals Exercise Multiple Choice Questions Question 35

Answer:
$a = \frac{- 1}{10}, b = \frac{2}{5}$
Given:
$\int \frac{1}{(x + 2) (x^{2} + 1)} d x = a \log | 1 + x^{2} | + b \tan^{- 1} x + \frac{1}{5} \log | x + 2 | + C$
Hint:
Using partial fraction and

\int \frac{1}{x} d x

and

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

Explanation:
Let

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

Let

\frac{1}{(x + 2) (x^{2} + 1)} = \frac{A}{x + 2} + \frac{B x + C}{x^{2} + 1}

[E q . (i))]

Multiplying by

(x + 2) (x^{2} + 1)

, we get

I = A (x^{2} + 1) + (B x + C) (x + 2)

Putting x = -2

\begin{aligned} 1 = A (5) + (- 2 B + C) (- 2 + 2) \\ \Rightarrow 1 = 5 A + 0 \\ \Rightarrow A = \frac{1}{5} \end{aligned}

Putting x = 0

\begin{aligned} 1 = A (1) + (B (0) + C) (0 + 2) \\ \Rightarrow 1 = \frac{1}{5} + 2 C \\ \Rightarrow C = \frac{2}{5} \end{aligned}

Putting x = +1
$\begin{aligned} 1 = A (1 + 1) + (B + C) (1 + 2) \\ \Rightarrow 1 = 2 (\frac{1}{5}) + (B + \frac{2}{5}) (3) \\ \Rightarrow 1 = \frac{2}{5} + 3 B + \frac{6}{5} \\ \Rightarrow 3 B = 1 - \frac{8}{5} \\ \Rightarrow 3 B = \frac{- 3}{5} \\ \Rightarrow B = \frac{- 1}{5} \end{aligned}$
$∴ From [E q \cdot (i)]$
$\begin{aligned} \frac{1}{(x + 2) (x^{2} + 1)} = \frac{\frac{1}{5}}{x + 2} + \frac{(\frac{- 1}{5}) x + (\frac{2}{5})}{x^{2} + 1} \\ ∴ \int \frac{1}{(x + 2) (x^{2} + 1)} d x = \frac{1}{5} \int \frac{1}{x + 2} d x + (\frac{- 1}{5}) \cdot \frac{1}{2} \int \frac{2 x}{x^{2} + 1} d x + \frac{2}{5} \int \frac{1}{x^{2} + 1} d x \\ ∴ I = \frac{1}{5} \log | x + 2 | - \frac{1}{10} \log | 1 + x^{2} | + \frac{2}{5} \tan^{- 1} x + C \end{aligned}$ $[E q . (i i))]$
Acc to given
$I = a \log | 1 + x^{2} | + b \tan^{- 1} x + \frac{1}{5} \log | x + 2 |$ $[E q . (i i i))]$
$\begin{aligned} ∴ From E q \cdot (i i) and E q \cdot (iii) we get \\ a = \frac{- 1}{10}, b = \frac{2}{5} \end{aligned}$

Indefinite Integrals Exercise Multiple Choice Questions Question 36

Answer:
$e^{x} \cos x + C$
Given:
$\int e^{x} (\cos x - \sin x) d x$
Hint:
You must know the rule to integrate by parts
Explanation:
Let

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

\begin{aligned} = \int e^{x} \cos x d x - \int e^{x} \sin x d x \\ = \cos x e^{x} + \int e^{x} \sin x d x - \int e^{x} \sin x d x \dots . . {\int u \cdot v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x} \\ = \cos x e^{x} + C \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 37

Answer:
$a = \frac{- 1}{8}, b = \frac{+ 7}{8}$
Given:
$\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x = a x + b \log_{e} | 4 e^{x} + 5 e^{- x} | + C$
Hint:
Using

\int e^{x} d x

Explanation:
We are given that

\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x = a x + b \log_{e} | 4 e^{x} + 5 e^{- x} | + C

Differentiate both sides

\frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} = a + b \cdot \frac{1}{(4 e^{x} + 5 e^{- x})} (4 e^{x} - 5 e^{- x})

[∵ \int e^{x} d x = e^{x} + C]

\begin{aligned} \Rightarrow \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} = \frac{a (4 e^{x} + 5 e^{- x}) + b (4 e^{x} - 5 e^{- x})}{4 e^{x} + 5 e^{- x}} \\ \Rightarrow 3 e^{x} - 5 e^{- x} = 4 a e^{x} + 5 a e^{- x} + 4 b e^{x} - 5 b e^{- x} \\ \Rightarrow 3 e^{x} - 5 e^{- x} = (4 a + 4 b) e^{x} + (5 a - 5 b) e^{- x} \end{aligned}

Comparing co eff of like terms, we get

\begin{aligned} 4 a + 4 b = 3; 5 (a - b) = - 5 \\ \Rightarrow a - b = - 1 \Rightarrow a = b - 1 \\ \Rightarrow 4 (b - 1) + 4 b = 3 \\ \Rightarrow 8 b - 4 = 3 \\ \Rightarrow 8 b = 7 \Rightarrow b = \frac{7}{8} \\ ∴ a = \frac{7}{8} - 1 \Rightarrow a = \frac{- 1}{8} \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 38

Answer:
$\frac{e^{x}}{(1 + x^{2})} + C$
Given:
$\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x$
Hint:
Using integration by parts
Explanation:
Let

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

[∵ (a - b)^{2} = a^{2} - 2 a b + b^{2}]

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

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

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

= \frac{1}{1 + x^{2}} \cdot e^{x} - \int e^{x} (\frac{(1 + x^{2}) (0) - (1) (2 x)}{{(1 + x^{2})}^{2}}) d x - \int e^{x} \frac{2 x}{{(1 + x^{2})}^{2}} d x {\int u . v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x}

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

Indefinite Integrals Exercise Multiple Choice Questions Question 39

Answer:
$x \tan \frac{x}{2} + C$
Given:
$\int \frac{x + \sin x}{1 + \cos x} d x$
Hint:
Using integration by parts and

\int \sec^{2} θ d θ

Explanation:
Let

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

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

[∵ 1 + \cos 2 θ = 2 \cos^{2} θ; \sin 2 θ = 2 \sin θ \cos θ]

= \frac{1}{2} \int x \sec^{2} \frac{x}{2} d x + \int \tan \frac{x}{2} d x

[Put \frac{x}{2} = t \Rightarrow x = 2 t \Rightarrow d x = 2 d t]

\begin{aligned} = \frac{1}{2} \int 2 t \sec^{2} t .2 d t + \int \tan t \cdot 2 d t \\ = 2 \int t \sec^{2} t d t + 2 \int \tan t d t \\ = 2 [t . t a n t - \int \tan t d t] + 2 \int \tan t d t \dots {\int u . v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x} \\ = 2 t . \tan t - 2 \int \tan t d t + 2 \int \tan t d t \\ = 2 t \tan t + C \\ = 2 (\frac{x}{2}) \tan \frac{x}{2} + C \\ = x \tan \frac{x}{2} + C \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 40

Answer:
$(x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C$
Given:
$\int \tan^{- 1} \sqrt{x} d x$
Hint:
Using integration by parts and

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

Explanation:
Let

I = \int \tan^{- 1} \sqrt{x} d x

\Rightarrow I = \int \tan^{- 1} t .2 t d t

[Put \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow d x = 2 \sqrt{x} d t \Rightarrow d x = 2 t d t]

= 2 \int \tan^{- 1} t \cdot t d t

= 2 [\tan^{- 1} t \cdot \frac{t^{2}}{2} - \int \frac{1}{1 + t^{2}} \cdot \frac{t^{2}}{2} d t] \dots {\int u . v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x}

\begin{aligned} = 2 \tan^{- 1} t \cdot \frac{t^{2}}{2} - \int (\frac{t^{2} + 1 - 1}{t^{2} + 1}) d t \\ = \tan^{- 1} t \cdot t^{2} - \int 1 d t + \int \frac{1}{1 + t^{2}} d t \\ = t^{2} \tan^{- 1} t - t + \tan^{- 1} t + C \\ = x \tan^{- 1} \sqrt{x} - \sqrt{x} + \tan^{- 1} \sqrt{x} + C \\ = (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C \end{aligned}

Indefinite Integrals Exercise Multiple Choice Questions Question 41

Answer:
$cosec (b - a) \log | \frac{\sin (x - b)}{\sin (x - a)} |$
Given:
$\int \frac{1}{\sin (x - a) \sin (x - b)} d x$
Hint:
Using

\sin (x - y) = \sin x \cos y - \cos x \sin y and \int \frac{1}{x} d x

Explanation:
Let

I = \int \frac{1}{\sin (x - a) \sin (x - b)} d x

= \frac{1}{\sin (b - a)} \int \frac{\sin (b - a)}{\sin (x - a) \sin (x - b)} d x \dots \dots multiplying by \sin (b - a) in num.\& den.

= \frac{1}{\sin (b - a)} \int \frac{\sin ((x - a) - (x - b))}{\sin (x - a) \sin (x - b)} d x

[\sin (x - y) = \sin x \cos y - \cos x \sin y]

\begin{aligned} = \frac{1}{\sin (b - a)} \int \frac{\sin (x - a) \cos (x - b) - \cos (x - a) \sin (x - b)}{\sin (x - a) \sin (x - b)} d x \\ = \frac{1}{\sin (b - a)} [\int \frac{\cos (x - b)}{\sin (x - b)} d x - \int \frac{\cos (x - a)}{\sin (x - a)}] d x \\ = \frac{1}{\sin (b - a)} [\log | \sin (x - b) | - \log | \sin (x - a) |] + C \end{aligned}

= cosec (b - a) \log | \frac{\sin (x - b)}{\sin (x - a)} |

[∵ \log a - \log b = \log \frac{a}{b}]

Indefinite Integrals Exercise Multiple Choice Questions Question 42

Answer:
$\frac{1}{3} e^{x^{3}} + C$
Given:
$\int x^{2} e^{x^{3}} d x$
Hint:
Using

\int e^{x} d x

Explanation:
Let

I = \int x^{2} e^{x^{3}} d x

= \int e^{t} \cdot \frac{d t}{3}

[Put x^{3} = t \Rightarrow 3 x^{2} d x = d t \Rightarrow x^{2} d x = \frac{d t}{3}]

\begin{aligned} = \frac{1}{3} e^{t} + C \\ = \frac{1}{3} e^{x^{3}} + C \end{aligned}

[∵ \int e^{x} d x = e^{x} + C]

Indefinite Integrals Exercise Multiple Choice Questions Question 43

Answer:
$e^{x} \log (x + 1) + C$
Given:
$\int \frac{e^{x}}{x + 1} [1 + (x + 1) \log (x + 1)] d x$
Hint:
Using integration by parts and

\int e^{x} d x

Explanation:
Let

I = \int \frac{e^{x}}{x + 1} [1 + (x + 1) \log (x + 1)] d x

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

= \log (x + 1) e^{x} - \int \frac{1}{x + 1} e^{x} d x + \int \frac{1}{x + 1} e^{x} d x \dots {\int u \cdot v d x = u \int v d x - \int \frac{d u}{d x} \int v d x d x}

= \log (x + 1) e^{x} + C

Hence,

\int \frac{e^{x}}{x + 1} [1 + (x + 1) \log (x + 1)] d x = \log (x + 1) e^{x} + C

In this class 12 RD Sharma chapter 18 exercise MCQ. At Career360 Class 12 Maths, RD Sharma Class 12 Solutions Indefinite Integrals Ex MCQ helps students get a fair academic score in the exam.

This chapter of RD Sharma class 12th exercise MCQ Indefinite Integrals primary based on the possibility of soundness. Students can download the RD Sharma class 12 solutions MCQ Chapter 18 Indefinite Integrals to look into this theme.

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