RD Sharma Class 12 Exercise 18.6 Indefinite Integrals Solutions Maths

Kuldeep MauryaUpdated on 24 Jan 2022, 11:57 AM IST

RD Sharma books are well known for their detailed answers, coverage of syllabus, and exam-oriented answers. A majority of schools and teachers widely use them for setting up exam question papers. Thus, RD Sharma books are the best choice for students when it comes to maths.

RD Sharma solutions deals with the chapter Indefinite Integrals. This chapter contains eight Level 1 questions that are concept-based and easy to understand. It covers concepts like indefinite integration of Sine and Cos functions. The solutions given in Brainly’s material can help you complete this exercise with ease.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.6

Indefinite Integrals Excercise 18.6 Question 1

Answer:
$\frac{x}{2}-\frac{1}{8} \sin (4 x+10)+C$
Hint:
Use,$\sin^{2}x=\left ( \frac{1-\cos 2x}{2} \right )$
Given:
Let $I=\int \sin ^{2}(2 x+5) d x$
Solution:
$I=\int \sin ^{2}(2 x+5) d x$
On Substituting the above formula,
So the equation becomes,
$\Rightarrow \int \frac{1-\cos 2\left ( 2x+5 \right )}{2}dx$
We know,
$\begin{aligned} &\int \cos \mathrm{axdx}=\frac{1}{\mathrm{a}} \sin \mathrm{ax}+\mathrm{C} \\ &\Rightarrow \frac{1}{2} \int \mathrm{dx}-\frac{1}{2} \int \cos (4 \mathrm{x}+10) \mathrm{dx} \end{aligned}$

On integrating we get,

$\Rightarrow \frac{x}{2}-\frac{1}{8} \sin (4 x+10)+C$
So the answer is
$\frac{x}{2}-\frac{1}{8} \sin (4 x+10)+C$

Indefinite Integrals Excercise 18.6 Question 2

Answer:
$-\frac{3}{8} \cos (2 x+1)+\frac{1}{24} \cos (6 x+3)+C$
Hint:
Use: $\sin 3 x=-4 \sin ^{3} x+3 \sin x$
Given:
Let $I=\int \sin ^{3}(2 x+1) d x$
Solution:
$I=\int \sin ^{3}(2 x+1) d x$
The above stated formula can be written as
$\sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}$
The equation becomes,
$\sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}$
On multiplying the above formula to the given question, we get
$\Rightarrow \int \sin ^{3}(2 x+1) d x=\int \frac{3 \sin (2 x+1)-\sin 3(2 x+1)}{4} d x$
We know,
$\int \sin ax=-\frac{1}{a} \cos a x+c$
On substituting the above formula we get ,
$\Rightarrow \frac{3}{4} \int \sin (2 x+1) d x-\frac{1}{4} \int \sin (6 x+3) d x$
On integrating we get ,
$=-\frac{3}{8} \cos (2 x+1)+\frac{1}{24} \cos (6 x+3)+C$
So the answer is,
$=-\frac{3}{8} \cos (2 x+1)+\frac{1}{24} \cos (6 x+3)+C$

Indefinite Integrals Excercise 18.6 Question 3

Answer:
$\frac{24 x+8 \sin 4 x+\sin 8 x}{64}+C$
Hint:
Use $\cos ^{2} x=\left(\frac{1+\cos 2 x}{2}\right)$ and
Consider $\cos ^{4} 2 x d x=\left(\cos ^{2} 2 x\right)^{2}$
Given:
Let $\mathrm{I}=\int \cos ^{4} 2 \mathrm{xdx}$
Solution:
Considering,
$\cos ^{4} 2 x d x=\left(\cos ^{2} 2 x\right)^{2}$
The above equation becomes,
$\begin{aligned} &\Rightarrow\left(\cos ^{2} 2 \mathrm{x}\right)^{2}=\left(\frac{1+\cos 4 \mathrm{x}}{2}\right)^{2} \\ &\Rightarrow\left(\frac{1+\cos 4 \mathrm{x}}{2}\right)^{2}=\left(\frac{1+2 \cos 4 \mathrm{x}+\cos ^{2} 4 \mathrm{x}}{4}\right) \\ &\Rightarrow \cos ^{2} 4 \mathrm{x}=\frac{1+\cos 8 \mathrm{x}}{2} \\ &\Rightarrow\left(\frac{1+2 \cos 4 \mathrm{x}+\cos ^{2} 4 \mathrm{x}}{4}=\frac{1}{4}+\frac{\cos 4 \mathrm{x}}{2}+\frac{1+\cos 8 \mathrm{x}}{8}\right) \end{aligned}$
Then the question becomes,
$\Rightarrow \frac{1}{4} \int \mathrm{dx}+\frac{1}{2} \int \cos 4 \mathrm{xdx}+\frac{1}{8} \int \mathrm{dx}+\frac{1}{8} \int \cos 8 \mathrm{xdx}$
We know,
$\begin{aligned} &\int \operatorname{cosax} d x=\frac{1}{a} \sin a x+c \\ &\Rightarrow \frac{x}{4}+\frac{1}{8} \sin 4 x+\frac{x}{8}+\frac{\sin 8 x}{64}+C \\ &\Rightarrow \frac{24 x+8 \sin 4 x+\sin 8 x}{64} \\ &\frac{24 x+8 \sin 4 x+\sin 8 x}{64}+C \end{aligned}$

Indefinite Integrals Excercise 18.6 Question 4
Answer:

$\frac{x}{2}-\frac{1}{4 b} \sin (2 b x)+C$
Hint:
Use $\sin ^{2} \mathrm{x}=\left(\frac{1-\cos 2 \mathrm{x}}{2}\right)$ and
Given:
Let $\mathrm{I}=\int \sin ^{2} \mathrm{bxdx}$
Solution:
$\mathrm{I}=\int \sin ^{2} \mathrm{bxdx}$
Substituting the above formula, we get
$I=\int \frac{1-\cos(2bx)}{2}dx$
We know,
$\begin{aligned} &\int \cos a x d x=\frac{1}{a} \sin a x+C \\ &\Rightarrow \frac{1}{2} \int d x-\frac{1}{2} \int \cos (2 bx) d x \end{aligned}$

On Integration,

$\Rightarrow \frac{x}{2}-\frac{1}{4 b} \sin (2 b x)+c$
So the answer is
$\frac{x}{2}-\frac{1}{4 b} \sin (2 b x)+C$

Indefinite Integrals Excercise 18.6 Question 5

Answer:
$\frac{1}{2}[\mathrm{x}-\sin \mathrm{x}]+\mathrm{C}$
Hint:
Use $\left[2 \sin ^{2} \frac{x}{2}=1-\cos x\right]$
Given:
Let $\mathrm{I}=\int \sin ^{2} \frac{\mathrm{x}}{2} \mathrm{dx} \\$
Solution:
$\begin{aligned} &\mathrm{I}=\int \sin ^{2} \frac{\mathrm{x}}{2} \mathrm{dx} \\ &=\frac{1}{2} \int 2 \sin ^{2} \frac{\mathrm{x}}{2} \mathrm{dx} \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \int(1-\cos x) d x \\ &=\frac{1}{2}\left[\int d x-\int \cos x d x\right] \\ &=\frac{1}{2}[x-\sin x]+C \end{aligned}$
So the answer is $=\frac{1}{2}[\mathrm{x}-\sin \mathrm{x}]+\mathrm{C}$

Indefinite Integrals Excercise 18.6 Question 6

Answer:
$\frac{1}{2}(\mathrm{x}+\sin \mathrm{x})+\mathrm{c}$
Hint:
$2 \cos ^{2} x / 2=1+\cos x$
Given:
let $\mathrm{I}=\int \cos ^{2} \frac{\mathrm{x}}{2} \mathrm{dx}$
Solution:
$\begin{aligned} &\mathrm{I}=\int \cos ^{2} \frac{\mathrm{x}}{2} \mathrm{dx} \\ &\mathrm{I}=\frac{1}{2} \int 2 \cos ^{2} \frac{\mathrm{x}}{2} \mathrm{dx} \\ &\mathrm{I}=\frac{1}{2} \int(1+\cos \mathrm{x}) \mathrm{dx} \end{aligned}$
$\begin{aligned} &\mathrm{I}=\frac{1}{2} \int \mathrm{d} \mathrm{x}+\frac{1}{2} \int \cos \mathrm{x} \mathrm{d} \mathrm{x} \\ &=\frac{1}{2}(\mathrm{x})+\frac{1}{2} \cdot \sin \mathrm{x}+\mathrm{c} \\ &=\frac{1}{2}(\mathrm{x}+\sin \mathrm{x})+\mathrm{c} \end{aligned}$

So the answer is $=\frac{1}{2}(\mathrm{x}+\sin \mathrm{x})+\mathrm{c}$

Indefinite Integrals Excercise 18.6 Question 7

Answer:
$\frac{\mathrm{x}}{2}+\frac{\sin 2 \mathrm{nx}}{4 \mathrm{n}}+\mathrm{c}$
Hint:
$\cos 2 x=2 \cos ^{2} x-1$
Given:
Let $\mathrm{I}=\int \cos ^{2} \mathrm{n} \mathrm{xdx}$
Solution:
$\begin{aligned} &\mathrm{I}=\int \cos ^{2} \mathrm{n} \mathrm{xdx} \\ &=\frac{1}{2} \int(1+\cos 2 \mathrm{nx}) \mathrm{dx} \\ &=\frac{1}{2}\left[\mathrm{x}+\frac{\sin 2 \mathrm{n} \mathrm{x}}{2 \mathrm{n}}\right]+\mathrm{c} \\ &=\frac{\mathrm{x}}{2}+\frac{\sin 2 \mathrm{nx}}{4 \mathrm{n}}+\mathrm{c} \end{aligned}$
So the answer is $=\frac{\mathrm{x}}{2}+\frac{\sin 2 \mathrm{nx}}{4 \mathrm{n}}+\mathrm{c}$

Indefinite Integrals Excercise 18.6 Question 8

Answer:
$\frac{1}{\sqrt{2}}(x)-\frac{1}{2 \sqrt{2}} \sin 2 x$
Hint:
Use $\left[1-\cos x=2 \sin ^{2} x\right]$
Given:
Let $I=\int \sin x \sqrt{1-\cos 2 x} d x$
Solution:
$I=\int \sin x \sqrt{1-\cos 2 x} d x$
Using the above formula we get,
$\begin{aligned} &\Rightarrow \sqrt{2} \int \sin x(\sin x) d x \\ &\Rightarrow \sqrt{2} \int \sin ^{2} x d x \Rightarrow \frac{+\sqrt{2}}{2} \int 1-\cos 2 x d x \\ &=\frac{1}{\sqrt{2}}(x)-\frac{1}{2 \sqrt{2}} \sin 2 x \end{aligned}$

So the answer is $=\frac{1}{\sqrt{2}}(x)-\frac{1}{2 \sqrt{2}} \sin 2 x+c$

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