RD Sharma Class 12 Exercise 18.6 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.6 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 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.

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

RD Sharma Class 12th Exercise 18.6 is designed by a group of subject experts with years of experience. Moreover, this material complies with the CBSE syllabus, which means that students can refer to it directly to check their class progress.

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Frequently Asked Questions (FAQs)

1. Does this material cover all concepts?

Yes, RD Sharma Class 12 Chapter 18 Exercise 18.6 material is designed by experts to cover the entire syllabus. Therefore, students can refer to it and get a good hold of all the concepts from the chapter.

2. Can I refer to this material to score good marks?

Brinley has provided RD Sharma Class 12th Exercise 18.6 material to help students better understand the concepts. It is the best means for preparation as it has solved questions that will help students cover the syllabus efficiently.

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RD Sharma Class 12 Solutions Indefinite Integrals Ex 18.6 material is available on Brainly’s website for absolutely free. It is made to serve as a medium for exam preparation for CBSE students

4. What do you mean by indefinite integration?

Indefinite integration is the process of finding an integral of a value where the upper and lower limits are not given.

5. What are the limits in integration?

Limits in integration are used to calculate the integral of a value under fixed parameters