RD Sharma Class 12 Exercise 18.13 Indefinite Integrals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 18.13 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:11 PM IST

RD Sharma class 12th exercise 18.13 a students best study guide as an NCERT solution. Practice makes everything perfect and this perfection is required when students will sit for their board exams. Students in class 12 have to finish a huge maths syllabus in a limited time. This might cause them to panic and stress out. However, with the help of RD Sharma class 12 chapter 18 exercise 18.13, they will be able to practice well at home and master the subject quickly. The RD Sharma solutions will soon become their holy grail and help them score high.

## Indefinite Integrals Excercise:18.13

Indefinite Integrals Excercise 18.13 Question 2

Answer: $\frac{1}{8a^{2}}\frac{x^{8}}{\left ( a^{2}-x^{2} \right )^{4}}+c$
Hint: Use substitution method to solve this integral.
Given: $\int \frac{x^{7}}{\left ( a^{2}-x^{2} \right )^{5}} dx$
Solution:
Let $I=\int \frac{x^{7}}{\left ( a^{2}-x^{2} \right )^{5}} dx$
Put $x=a \sin\theta\Rightarrow dx=a\cos\theta d\theta$ …(i) (Differentiating w.r.t to θ)
\begin{aligned} &\text { Then } \mathrm{I}=\int \frac{a^{7} \sin ^{7} \theta}{\left(a^{2}-a^{2} \sin ^{2} \theta\right)^{5}} \cdot a \cos \theta d \theta \\ &=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{\left\{a^{2}\left(1-\sin ^{2} \theta\right)\right\}^{5}} d \theta=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{\left.a^{2 \times 5}\left(\cos ^{2} \theta\right)\right\}^{5}} d \theta \\ &=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{a^{10} \cos ^{10} \theta} d \theta=\frac{1}{a^{2}} \int \frac{\sin ^{7} \theta}{\cos ^{9} \theta} d \theta \end{aligned}
\begin{aligned} &=\frac{1}{a^{2}} \int \frac{\sin ^{7} \theta}{\cos ^{7} \theta} \cdot \frac{1}{\cos ^{2} \theta} d \theta \\ &=\frac{1}{a^{2}} \int \tan ^{7} \theta \cdot \sec ^{2} \theta d \theta \quad \quad\quad\quad\quad\quad\quad\left[\because \frac{\sin \theta}{\cos \theta}=\tan \theta \text { and } \frac{1}{\cos \theta}=\sec \theta\right] \end{aligned}
Again put$\tan\theta=t\Rightarrow \sec2\theta d\theta=dt$ (on differentiating w.r.t to $\theta$)
\begin{aligned} &\text { Then } I=\frac{1}{a^{2}} \int t^{7} d t=\frac{1}{a^{2}} \cdot \frac{t^{7+1}}{7+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &I=\frac{1}{a^{2}} \cdot \frac{t^{8}}{8}+c=\frac{1}{8 a^{2}} \cdot t^{8}+c \\ &I=\frac{1}{8 a^{2}} \cdot t^{8}+c \quad\quad\quad\quad\quad[\because t=\tan \theta] \end{aligned}
From (i)$x=a\sin\theta$
$\Rightarrow \sin \theta=\frac{x}{a}$
and we know that $\sin^{2}\theta+\cos^{2}\theta=1$
\begin{aligned} &\Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta \\ &\Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-\left(\frac{x}{a}\right)^{2}} \\ &\Rightarrow \cos \theta=\sqrt{1-\frac{x^{2}}{a^{2}}}=\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}=\frac{1}{a} \sqrt{a^{2}-x^{2}} \end{aligned}
\begin{aligned} &\therefore \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{x}{a}}{\frac{1}{a} \sqrt{a^{2}-x^{2}}} \\ &\Rightarrow \tan \theta=\frac{x}{a} \cdot \frac{a}{\sqrt{a^{2}-x^{2}}}=\frac{x}{\sqrt{a^{2}-x^{2}}} \end{aligned}

$\therefore \tan\theta=\frac{x}{\sqrt{a^{2}-x^{2}}}$ …(iii)

Put the value of tanθ from (iii) in (ii) then
\begin{aligned} &I=\frac{1}{8 a^{2}}\left(\frac{x}{\sqrt{a^{2}-x^{2}}}\right)^{8}+c \\ &=\frac{1}{8 a^{2}} \frac{x^{8}}{\left(a^{2}-x^{2}\right)^{\frac{8}{2}}}+c \\ &=\frac{1}{8 a^{2}} \frac{x^{8}}{\left(a^{2}-x^{2}\right)^{4}}+c \end{aligned}

Indefinite Integrals Excercise 18.13 Question 3

Answer: $\frac{x^{2}}{2}+c$
Hint: Use substitution method to solve this integral
Given: $\int \cos \left \{ 2\cot^{-1} \sqrt{\frac{1+x}{1-x}}\right \}dx$
Put$x=\cos2\theta$ …(i)
\begin{aligned} &\text { Then } I=\int \cos \left\{2 \cot ^{-1} \sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}}\right\} d x \\ &=\int \cos \left\{2 \cot ^{-1} \sqrt{\frac{2 \cos ^{2} \theta}{2 \sin ^{2} \theta}}\right\} d x \quad\quad\quad\quad\quad\left\{\begin{array}{l} \because 1+\cos 2 A=2 \cos ^{2} A \\ 1-\cos 2 A=2 \sin ^{2} A \end{array}\right\} \\ &=\int \cos \left\{2 \cot ^{-1} \sqrt{\left(\frac{\cos \theta}{\sin \theta}\right)^{2}}\right\} d x \\ &=\int \cos \left\{2 \cot ^{-1} \sqrt{(\cot \theta)^{2}}\right\} d x \quad\quad\quad\quad\quad\quad\left[\because \frac{\cos A}{\sin A}=\cot A\right] \\ &=\int \cos \left\{2 \cot ^{-1} \cdot \cot \theta\right\} d x \end{aligned}
\begin{aligned} &=\int \cos \{2 \theta\} d x \quad\left[\because \cot \left(\cot ^{-1} x\right)=x\right] \\ &=\int \cos \left\{\cos ^{-1}(x)\right\} d x \\ &=\int x^{1} d x \quad\left[\begin{array}{c} \because x=\cos 2 \theta \\ 2 \theta=\cos ^{-1}(x) \end{array}\right] \\ &=\frac{x^{1+1}}{1+1}+c \quad\left[\because \cos \left(\cos ^{-1} x\right)=x\right] \\ &=\frac{x^{2}}{2}+c \end{aligned}

Indefinite Integrals Excercise 18.13 Question 4

Answer: $\frac{-1}{3} \frac{\left(1+x^{2}\right)^{\frac{3}{2}}}{x^{3}}+c \$
Hint: Use substitution method to solve this integral
Given:$\int \frac{\sqrt{1+x^{2}}}{x^{4}}dx$
Solution:
Let $I=\int \frac{\sqrt{1+x^{2}}}{x^{4}}dx$
Putting $x=\tan\theta$ … (i)
$dx=\sec 2\theta d\theta$
Then
\begin{aligned} &I=\int \frac{\sqrt{1+\tan ^{2} \theta}}{\tan ^{4} \theta} \sec ^{2} \theta d \theta \\ &=\int \frac{\sqrt{\sec ^{2} \theta}}{\tan ^{4} \theta} \cdot \sec ^{2} \theta d \theta \quad\quad\quad\quad\quad\quad\quad\left[\begin{array}{l} \because \sec ^{2} \theta-\tan ^{2} \theta=1 \\ 1+\tan ^{2} \theta=\sec ^{2} \theta \end{array}\right] \\ &=\int \frac{\sec \theta \cdot \sec ^{2} \theta}{\tan ^{4} \theta} d \theta=\int \frac{\sec ^{3} \theta}{\tan ^{4} \theta} d \theta \\ &=\int \frac{1}{\cos ^{3} \theta \cdot \frac{\sin ^{4} \theta}{\cos ^{4} \theta}} d \theta=\int \frac{\cos ^{4} \theta}{\sin ^{4} \theta \cdot \cos ^{3} \theta} d \theta \\ &=\int \frac{\cos \theta}{\sin ^{4} \theta} d \theta \end{aligned}
Again putting $\sin\theta=t$
\begin{aligned} &\Rightarrow \cos \theta d \theta=d t \text { then } \\ &I=\int \frac{1}{t^{4}} \cdot d t \\ &=\int t^{-4} d t=\frac{t^{-4+1}}{-4+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{t^{-3}}{-3}+c \\ &=\frac{-1}{3} t^{-3}+c \\ &=\frac{-1}{3 t^{3}}+c \quad \text { u. (ii) } \end{aligned}
Also from (i), $x=\tan\theta$
\begin{aligned} &x^{2}=\tan ^{2} \theta\quad\quad\quad\quad \quad \text { [Squaring both sides] }\\ &\Rightarrow x^{2}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \quad\quad\quad\quad\quad\quad\quad\quad\left[\tan \theta=\frac{\sin \theta}{\cos \theta}\right]\\ &\Rightarrow x^{2} \cos ^{2} \theta=\sin ^{2} \theta\\ &\Rightarrow x^{2}\left(1-\sin ^{2} \theta\right)=\sin ^{2} \theta \quad\quad\quad\quad\quad\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \Rightarrow \sin ^{2} \theta=1-\cos ^{2} \theta \end{array}\right] \end{aligned}
\begin{aligned} &\Rightarrow x^{2}-x^{2} \sin ^{2} \theta=\sin ^{2} \theta\\ &\Rightarrow x^{2}=\sin ^{2} \theta+x^{2} \sin ^{2} \theta=\sin ^{2} \theta\left(1+x^{2}\right)\\ &\Rightarrow \sin ^{2} \theta=\frac{x^{2}}{1+x^{2}}\\ &\Rightarrow \sin \theta=\sqrt{\frac{x^{2}}{1+x^{2}}}\\ &\Rightarrow \sin \theta=\frac{x}{\sqrt{1+x^{2}}} \end{aligned}
Putting the value of sinθ from (iii) in (ii), we get
$I=\frac{-1}{3}\frac{1}{\left ( \sin\theta \right )^{3}}+c$
\begin{aligned} &=\frac{-1}{3} \frac{1}{\left(\frac{x}{\sqrt{1+x^{2}}}\right)^{3}}+c \\ &=\frac{-1}{3} \frac{1}{\frac{x^{3}}{\left(\sqrt{1+x^{2}}\right)^{3}}}+c \\ &=-\frac{1}{3} \frac{\left(1+x^{2}\right)^{\frac{3}{2}}}{x^{3}}+c \end{aligned}

Indefinite Integrals Excercise 18.13 Question 1

Answer:$\frac{x}{\sqrt{a^{2}-x^{2}}}-\sin ^{-1}\left(\frac{x}{a}\right)+c$
Hint: Use substitution method to solve this integral.
Given:$\int \frac{x^{2}}{\left(a^{2}-x^{2}\right)^{\frac{3}{2}}} d x$
Solution:
Let $I=\int \frac{x^{2}}{\left(a^{2}-x^{2}\right)^{\frac{3}{2}}} d x$
Put$x = a \sin u$ …(i)
$dx = a \cos u du$ (on differentiating w.r.t to u)
Then
\begin{aligned} &I=\int \frac{a^{2} \sin ^{2} u}{\left(a^{2}-a^{2} \sin ^{2} u\right)^{\frac{3}{2}}} \cdot a \cos u d u=\int \frac{a^{3} \sin ^{2} u \cdot \cos u}{\left(a^{2}\left(1-\sin ^{2} u\right)\right\}^{\frac{3}{2}}} d u \\ &=\int \frac{a^{3} \sin ^{2} u \cdot \cos u}{\left(a^{2}\right)^{\frac{3}{2}}\left(1-\sin ^{2} u\right)^{\frac{3}{2}}} d u=\int \frac{a^{3} \sin ^{2} u \cdot \cos u}{\left(a^{3}\right)\left(\cos ^{2} u\right)^{\frac{3}{2}}} d u \\ &=\int \frac{a^{3} \sin ^{2} u \cdot \cos u}{\left(a^{3}\right)\left(\cos ^{2} u\right)^{\frac{3}{2}}} d u \\ &=\int \frac{\sin ^{2} u \cdot \cos u}{(\cos u)^{2 \times \frac{3}{2}}} d u \end{aligned}
\begin{aligned} &=\int \frac{\sin ^{2} u \cdot \cos u}{(\cos u)^{3}} d u \\ &=\int \frac{\sin ^{2} u}{(\cos u)^{2}} d u=\int\left(\frac{\sin u}{\cos u}\right)^{2} d u \\ &=\int(\tan u)^{2} d u=\int \tan ^{2} u d u \quad\left[\because \frac{\sin \theta}{\cos \theta}=\tan \theta\right] \end{aligned}
\begin{aligned} &=\int\left(\sec ^{2} u-1\right) d u=\int \sec ^{2} u d u-\int 1 d u \\ &{\left[\because 1+\tan ^{2} \theta=\sec ^{2} \theta \Rightarrow \tan ^{2} \theta=\sec ^{2} \theta-1\right]} \\ &I=\tan u-u+c \quad \text { ... (ii) } \quad\quad\quad\quad\left[\because \int 1 d x=x+c\right] \end{aligned}
From (i) $x = a \sin u$
$\Rightarrow \operatorname{\sin} \mathrm{u}=\frac{x}{a} \Rightarrow u=\sin ^{-1}\left(\frac{x}{a}\right)$ …(iii)
And we know that $1=\sin ^{2} u+\cos ^{2} u$ then
\begin{aligned} &\Rightarrow \cos ^{2} u=1-\sin ^{2} u \\ &\Rightarrow \cos u=\sqrt{1-\sin ^{2} u}=\sqrt{1-\left(\frac{x}{a}\right)^{2}} \end{aligned}

Now

\begin{aligned} &\operatorname{tan} u=\frac{\sin u}{\cos u}=\frac{\frac{x}{a}}{\sqrt{1-\frac{x^{2}}{a^{2}}}}=\frac{\frac{x}{a}}{\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}}=\frac{\frac{x}{a}}{\frac{1}{a} \sqrt{a^{2}-x^{2}}} \\ &\tan u=\frac{x}{a} \cdot \frac{a}{\sqrt{a^{2}-x^{2}}}=\frac{x}{\sqrt{a^{2}-x^{2}}} \\ &\operatorname{tan} u=\frac{x}{\sqrt{a^{2}-x^{2}}} \end{aligned} …(iv)
Now put the value of tan u and u from equation (iv) and (iii) in equation (ii), we get
$\mathrm{I}=\frac{x}{\sqrt{a^{2}-x^{2}}}-\sin ^{-1}\left(\frac{x}{a}\right)+c$

Indefinite Integrals Excercise 18.13 Question 5

Answer: $\frac{1}{54}\left[\tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{3(x+1)}{x^{2}+2+10}\right]+c$
Hint: Use substitution method to solve this integral
Given:$\int \frac{1}{\left(x^{2}+2 x+10\right)^{2}} d x \\$
Solution:
\begin{aligned} &\text { Let } I=\int \frac{1}{\left(x^{2}+2 x+10\right)^{2}} d x \\ &=\int \frac{1}{\left(x^{2}+2 x+1+9\right)^{2}} d x \\ &=\int \frac{1}{\left\{\left(x^{2}+2 x \cdot 1+1\right)+9\right\}^{2}} d x \\ &=\int \frac{1}{\left\{(x+1)^{2}+9\right\}^{2}} d x \end{aligned}

Putting $x+1=3\tan\theta$ …(i)

$d x=3 \sec ^{2} \theta d \theta$
Then
\begin{aligned} &I=\int \frac{1}{\left\{(3 \tan \theta)^{2}+9\right\}^{2}} 3 \sec ^{2} \theta d \theta \\ &=\int \frac{3 \sec ^{2} \theta}{\left\{9\left(1+\tan ^{2} \theta\right)\right\}^{2}} d \theta=\int \frac{3 \sec ^{2} \theta}{9^{2}\left(\sec ^{2} \theta\right)^{2}} d \theta \end{aligned}

\begin{aligned} &=\int \frac{3 \sec ^{2} \theta}{81 \sec ^{4} \theta} d \theta \quad\left[\begin{array}{l} \because \sec ^{2} \theta-\tan ^{2} \theta=1 \\ \Rightarrow 1+\tan ^{2} \theta=\sec ^{2} \theta \end{array}\right] \\ &=\int \frac{1}{27} \frac{1}{\sec ^{2} \theta} d \theta=\int \frac{1}{27} \cos ^{2} \theta d \theta \quad\left[\because \cos \theta=\frac{1}{\sec \theta}\right] \\ &=\frac{1}{27} \int\left\{\frac{1+\cos 2 \theta}{2}\right\} d \theta \quad\left[\begin{array}{l} \because 2 \cos ^{2} A=1+\cos 2 A \\ \cos ^{2} A=\frac{1+\cos 2 A}{2} \end{array}\right] \end{aligned}

\begin{aligned} &=\frac{1}{27}\left\{\int \frac{1}{2} d \theta+\int \frac{1}{2} \cos 2 \theta d \theta\right\} \\ &=\frac{1}{27} \times \frac{1}{2} \int \theta^{0} d \theta+\frac{1}{27} \times \frac{1}{2} \int \cos 2 \theta d \theta \\ &=\frac{1}{54} \cdot \frac{\theta^{0+1}}{0+1}+\frac{1}{54} \frac{\sin 2 \theta}{2}+c\quad\quad\quad\quad[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \text { and } \int \cos a x d x=\frac{\sin a x}{a}+c ]\\ \end{aligned}
$=\frac{1}{54} \theta+\frac{1}{108} \sin 2 \theta+c$ …(ii)
Also from (i)
\begin{aligned} &x+1=3 \tan \theta \\ &\tan \theta=\frac{x+1}{3} \Rightarrow \theta=\tan ^{-1}\left(\frac{x+1}{3}\right) \end{aligned} … (iii)
And$\tan\theta=\frac{x+1}{3}$
$\begin{array}{ll} \Rightarrow \tan ^{2} \theta=\left(\frac{x+1}{3}\right)^{2} & \text { [Squaring both sides] } \\\\ \Rightarrow \frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{(x+1)^{2}}{9} \quad & {\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]} \\ \\\Rightarrow \sin ^{2} \theta=\frac{(x+1)^{2}}{9} \cdot \cos ^{2} \theta & {\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]} \end{array}$
\begin{aligned} &\Rightarrow \sin ^{2} \theta=\frac{\left(x^{2}+2 x+1\right) \cos ^{2} \theta}{9} \\ &\Rightarrow \sin ^{2} \theta=\frac{\left(x^{2}+2 x+1\right)}{9}\left(1-\sin ^{2} \theta\right) \quad\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta \end{array}\right] \\ &\Rightarrow \sin ^{2} \theta=\frac{\left(x^{2}+2 x+1\right)}{9}-\left(\frac{\left(x^{2}+2 x+1\right)}{9}\right) \sin ^{2} \theta \end{aligned}
\begin{aligned} &\Rightarrow \frac{x^{2}+2 x+1}{9}=\sin ^{2} \theta+\left(\frac{x^{2}+2 x+1}{9}\right) \sin ^{2} \theta \\ &\Rightarrow \frac{x^{2}+2 x+1}{9}=\sin ^{2} \theta\left\{1+\left(\frac{x^{2}+2 x+1}{9}\right)\right\} \\ &\Rightarrow \frac{x^{2}+2 x+1}{9}=\left(\frac{9+x^{2}+2 x+1}{9}\right) \sin ^{2} \theta=\left(\frac{x^{2}+2 x+10}{9}\right) \sin ^{2} \theta \end{aligned}
\begin{aligned} &\Rightarrow \sin ^{2} \theta=\frac{x^{2}+2 x+1}{9} \times \frac{9}{x^{2}+2 x+10}=\frac{(x+1)^{2}}{\left(x^{2}+2 x+10\right)} \quad\quad\quad\quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\ &\Rightarrow \sin \theta=\sqrt{\frac{(x+1)^{2}}{x^{2}+2 x+10}}=\frac{(x+1)}{\sqrt{x^{2}+2 x+10}} \quad \ldots \text { (iv) } \\ &\quad \text { Since } \sin ^{2} \theta+\cos ^{2} \theta=1 \end{aligned}

So we can write

\begin{aligned} &\Rightarrow \cos ^{2} \theta=1-\sin ^{2} \theta \\ &\Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta} \\ &\begin{aligned} \Rightarrow \cos \theta &=\sqrt{1-\left[\frac{(x+1)}{\sqrt{x^{2}+2 x+10}}\right]^{2}} \\ &=\sqrt{\frac{x^{2}+2 x+10-\left(x^{2}+2 x+1\right)}{x^{2}+2 x+10}} \end{aligned} \quad\quad\quad\quad\quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \end{aligned}
\begin{aligned} &\Rightarrow \cos \theta=\sqrt{\frac{x^{2}+2 x+10-x^{2}-2 x-1}{x^{2}+2 x+10}}\\ &\Rightarrow \cos \theta=\sqrt{\frac{9}{x^{2}+2 x+10}}=\frac{3}{\sqrt{x^{2}+2 x+10}}\quad\quad\quad\quad....(v) \end{aligned}

Also, we know $\sin 2 \theta=2 \sin \theta \cos \theta$

\begin{aligned} &\Rightarrow \sin 2 \theta=2 \frac{(x+1)}{\sqrt{x^{2}+2 x+10}} \cdot \frac{3}{\sqrt{x^{2}+2 x+10}} \quad\quad \quad \quad [\text { from (iv) and (v)] }\\ &\Rightarrow \sin 2 \theta=\frac{2.3(x+1)}{x^{2}+2 x+10}=\frac{6(x+1)}{x^{2}+2 x+10}\quad \quad \quad \quad \cdot \cdot \cdot \cdot (vi) \end{aligned}
Now put the value of θ and sin2θ from equation (iii) and (vi) in (ii) we get
\begin{aligned} &I=\frac{1}{54} \theta+\frac{1}{108} \sin 2 \theta+c \\ &=\frac{1}{54} \tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{1}{108} \cdot \frac{6(x+1)}{\left(x^{2}+2 x+10\right)}+c \\ &=\frac{1}{54} \tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{1}{54} \frac{3(x+1)}{\left(x^{2}+2 x+10\right)}+c \\ &=\frac{1}{54}\left[\tan ^{-1}\left(\frac{x+1}{3}\right)+\frac{3(x+1)}{\left(x^{2}+2 x+10\right)}\right]+c \end{aligned}

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