RD Sharma Class 12 Exercise 18.14 Relation Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 18.14 Relation Solutions Maths - Download PDF Free Online

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

RD Sharma class 12th exercise 18.14 is one of the best exercise solutions for CBSE maths textbooks. Students who have used these solutions have greatly benefited from the answers and have even found some common questions in their board exams. Since, students I Class 12 get minimal time for exam preparations, the RD Sharma class 12 chapter 18 exercise 18.14 can be of immense help to them. With the use of RD Sharma solutions, students will be able to practice maths at home and improve their problem-solving skills.

## Indefinite Integrals Excercise:18.14

Indefinite Integrals Excercise 18.14 Question 1

Answer: $\frac{1}{2 a b} \log \left|\frac{a+b x}{a-b x}\right|+c$
Hint: To solve this integral, use the formula of special integral.
Given: $I=\int \frac{1}{a^{2}-b^{2} x^{2}} d x$
Solution: Let
$I=\int \frac{1}{a^{2}-b^{2} x^{2}} d x=\frac{1}{b^{2}} \int \frac{1}{\frac{a^{2}}{b^{2}}-x^{2}} d x$
$=\frac{1}{b^{2}} \int \frac{1}{\left(\frac{a}{b}\right)^{2}-x^{2}} d x$
$=\frac{1}{b^{2}} \cdot \frac{1}{2 \times \frac{a}{b}} \log \left|\frac{\frac{a}{b}+x}{\frac{a}{b}-x}\right|+c \quad \quad \quad \quad \quad\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\right]$
\begin{aligned} &= \frac{1}{2 a b} \log \left|\frac{\frac{a+b x}{b}}{\frac{a-b x}{b}}\right|+c \\\\ &= \frac{1}{2 a b} \log \left|\frac{a+b x}{a-b x}\right|+c \end{aligned}

Indefinite Integrals Excercise 18.14 Question 2

Answer: $\frac{1}{2 a b} \log \left|\frac{a x-b}{a x+b}\right|+c$
Hint: To solve this integral, use special integral formula.
Given: $\int \frac{1}{a^{2} x^{2}-b^{2}} d x$
Solution: Let $I=\int \frac{1}{a^{2} x^{2}-b^{2}} d x \: x=\frac{1}{a^{2}} \int \frac{1}{x^{2}-\frac{b^{2}}{a^{2}}} d x$
$=\frac{1}{a^{2}} \int \frac{1}{x^{2}-\left(\frac{b}{a}\right)^{2}} d x$
$=\frac{1}{a^{2}} \cdot \frac{1}{2 \times \frac{b}{a}} \log \left|\frac{x-\frac{b}{a}}{x+\frac{b}{a}}\right|+c \quad \quad \quad \quad \quad \quad \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right]$
\begin{aligned} &=\frac{1}{2 a b} \log \left|\frac{\frac{a x-b}{a}}{\frac{a x+b}{a}}\right|+c \\ \end{aligned}
$=\frac{1}{2 a b} \log \left|\frac{a x-b}{a x+b}\right|+c$

Indefinite Integrals Excercise 18.14 Question 3

Answer: $\frac{1}{a b} \tan ^{-1}\left|\frac{a x}{b}\right|+c$
Hint: To solve this integral, use special integral formula.
Given:$\int \frac{1}{a^{2} x^{2}+b^{2}} d x$
Solution: Let $I=\int \frac{1}{a^{2} x^{2}+b^{2}} d x=\frac{1}{a^{2}} \int \frac{1}{x^{2}+\frac{b^{2}}{a^{2}}} d x$
$=\frac{1}{a^{2}} \int \frac{1}{x^{2}+\left(\frac{b}{a}\right)^{2}} d x$
$=\frac{1}{a^{2}} \cdot \frac{1}{\frac{b}{a}} \tan ^{-1}\left|\frac{x}{\frac{b}{a}}\right|+c \quad\quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left|\frac{x}{a}\right|+c\right]$
$=\frac{1}{a b} \tan ^{-1}\left|\frac{a x}{b}\right|+c$

Indefinite Integrals Excercise 18.14 Question 4

Answer: $x-\frac{5}{2} \tan ^{-1}\left|\frac{x}{2}\right|+c$
Hint: To solve this integral, use special integral formula.
Given:$\int \frac{x^{2}-1}{x^{2}+4} d x$
Solution:
Let,
\begin{aligned} &I=\int \frac{x^{2}-1}{x^{2}+4} d x=\int \frac{x^{2}-1+4-4}{x^{2}+4} d x\\ \\ &=\int \frac{\left(x^{2}+4\right)-1-4}{x^{2}+4} d x=\int\left(\frac{x^{2}+4}{x^{2}+4}-\frac{5}{x^{2}+4}\right) d x \\\\ &=\int\left(1-\frac{5}{x^{2}+4}\right) d x=\int 1 d x-5 \int \frac{1}{x^{2}+2^{2}} \end{aligned}

$=x-5 \cdot \frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+c \quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left|\frac{x}{a}\right|+c\right]$

$=x-\frac{5}{2} \tan ^{-1}\left|\frac{x}{2}\right|+c$

Indefinite Integrals Excercise 18.14 Question 5

Answer: $\frac{1}{2} \log \left|x+\sqrt{x^{2}+\frac{1}{4}}\right|+c$
Hint: To solve this integral, use special integral formula.
Given:$\int \frac{1}{\sqrt{1+4 x^{2}}} d x$
Solution:
Let
\begin{aligned} &I=\int \frac{1}{\sqrt{1+4 x^{2}}} d x=\int \frac{1}{\sqrt{4\left(\frac{1}{4}+x^{2}\right)}} d x \\ &=\frac{1}{2} \int \frac{1}{\sqrt{x^{2}+\left(\frac{1}{2}\right)^{2}}} d x \end{aligned}
\begin{aligned} &=\frac{1}{2} \log \left|x+\sqrt{x^{2}+\left(\frac{1}{2}\right)^{2}}\right|+c \quad\quad\quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right] \\ \\&=\frac{1}{2} \log \left|x+\sqrt{x^{2}+\frac{1}{4}}\right|+c \end{aligned}

Indefinite Integrals Excercise 18.14 Question 6

Answer: $\frac{1}{b} \log \left|b x+\sqrt{a^{2}+b^{2} x^{2}}\right|+c$
Hint: To solve this integral, use special integral formula.
Given: $\int \frac{1}{\sqrt{a^{2}+b^{2} x^{2}}} d x$
Solution:
Let $I=\int \frac{1}{\sqrt{a^{2}+b^{2} x^{2}}} d x$
$\text { Put } b x=t \Rightarrow b d x=d t \Rightarrow d x=\frac{d t}{b} \text { then }$
\begin{aligned} &I=\int \frac{1}{\sqrt{a^{2}+t^{2}}} \frac{d t}{b}=\frac{1}{b} \int \frac{1}{\sqrt{a^{2}+t^{2}}} d t \\\\ &=\frac{1}{b} \log \left|t+\sqrt{a^{2}+t^{2}}\right|+c \quad\quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right] \end{aligned}
$=\frac{1}{b} \log \left|b x+\sqrt{a^{2}+b^{2} x^{2}}\right|+c \quad\quad\quad\quad\quad\quad\quad[\because t=b x]$

Indefinite Integrals Excercise 18.14 Question 7

Answer:$\frac{1}{b} \sin ^{-1}\left(\frac{b x}{a}\right)+c$
Hint: To solve this integral, use special integral formula.
Given:$\int \frac{1}{\sqrt{a^{2}-b^{2} x^{2}}} d x$
Solution:
Let
$I=\int \frac{1}{\sqrt{a^{2}-b^{2} x^{2}}} d x$
\begin{aligned} &\text { Put } b x=t \Rightarrow b d x=d t \Rightarrow d x=\frac{d t}{b} \text { then } \\ &I=\int \frac{1}{\sqrt{a^{2}-t^{2}}} \frac{d t}{b}=\frac{1}{b} \int \frac{1}{\sqrt{a^{2}-t^{2}}} d t \end{aligned}
\begin{aligned} &=\frac{1}{b} \sin ^{-1}\left(\frac{t}{a}\right)+c \quad\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c\right] \\\\ &=\frac{1}{b} \sin ^{-1}\left(\frac{b x}{a}\right)+c \quad[\because t=b x] \end{aligned}

Indefinite Integrals Excercise 18.14 Question 8

Answer: $-\log \left|(2-x)+\sqrt{(2-x)^{2}+1}\right|+c$
Hint: To solve this integral, use special integral formula.
Given:$\int \frac{1}{\sqrt{(2-x)^{2}+1}} d x$
Solution:
Let
$I=\int \frac{1}{\sqrt{(2-x)^{2}+1}} d x$
$\text { Put } 2-x=t \Rightarrow-d x=d t \Rightarrow d x=-d t \text { then }$
$I=\int \frac{1}{\sqrt{t^{2}+1}}(-d t)=-\int \frac{1}{\sqrt{t^{2}+1^{2}}} d t$
$=-\log \left|t+\sqrt{t^{2}+1}\right|+c \quad\quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{a^{2}+x^{2}}} d x=\log \left|x+\sqrt{a^{2}+x^{2}}\right|+c\right]$
$=-\log \left|(2-x)+\sqrt{(2-x)^{2}+1}\right|+c$

Indefinite Integrals Excercise 18.14 Question 9

Answer: $-\log \left|(2-x)+\sqrt{(2-x)^{2}-1}\right|+c$
Hint: To solve this integral, use special integral formula.
Given: $\int \frac{1}{\sqrt{(2-x)^{2}-1}} d x \\$
Solution:
Let
\begin{aligned} &I=\int \frac{1}{\sqrt{(2-x)^{2}-1}} d x \\ \end{aligned}
$\text { Put } 2-x=t \Rightarrow-d x=d t \Rightarrow d x=-d t \text { then }$
\begin{aligned} &I=\int \frac{1}{\sqrt{t^{2}-1}}(-d t)=-\int \frac{1}{\sqrt{t^{2}-1^{2}}} d t \\ & \end{aligned}
$=-\log \left|t+\sqrt{t^{2}-1}\right|+c \quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{x^{2}-a^{2}}} d x=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right]$
$=-\log \left|(2-x)+\sqrt{(2-x)^{2}-1}\right|+c \quad \quad \quad \quad \quad[\because t=2-x]$

Indefinite Integrals Excercise 18.14 Question 10

Answer:$\frac{x^{3}}{3}-x+2 \tan ^{-1}(x)+c$
Hint: To solve this integral, use special integral formula.
Given: $\int \frac{x^{4}+1}{x^{2}+1} d x$
Solution:
Let
\begin{aligned} &I=\int \frac{x^{4}+1}{x^{2}+1} d x=\int \frac{x^{4}+1+2 x^{2}-2 x^{2}}{x^{2}+1} d x \\\\ &=\int \frac{\left\{\left(x^{2}\right)^{2}+1+2 x^{2}\right\}-2 x^{2}}{x^{2}+1} d x \quad\left[\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right] \end{aligned}

$=\int \frac{\left(x^{2}+1\right)^{2}-2 x^{2}}{x^{2}+1} d x=\int \frac{\left(x^{2}+1\right)^{2}}{x^{2}+1}-\frac{2 x^{2}}{x^{2}+1} d x$

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

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

$=\frac{x^{2+1}}{2+1}+x-2 x+2 \cdot \frac{1}{1} \tan ^{-1}\left(\frac{x}{1}\right)+c \quad \quad \quad \quad \quad \quad\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c, \int 1 d x=x+c \\\\ \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \end{array}\right]$
$=\frac{x^{3}}{3}-x+2 \tan ^{-1}(x)+c$

The RD Sharma class 12 solutions Indefinite Integrals 18.14 covers an important part of the maths syllabus that has 32 Exercises in total. Chapter 18 of the NCERT maths book will teach them the basic concepts of indefinite integrals. Some areas covered in this chapter are Graphs of indefinite integrals and Indefinite integrals of common functions. Exercise 18.14 has 10 questions with subparts. These questions are based on the basic concepts of Indefinite Integrals. The RD Sharma class 12th exercise 18.14 will help students solve these exercise questions and build their skills.

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