#### Need solution for RD Sharma Maths Class 12 Chapter 18 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

$\mathrm{I}=\frac{x}{\sqrt{a^{2}-x^{2}}}-\sin ^{-1}\left(\frac{x}{a}\right)+c$