#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 61 Maths Textbook Solution.

Answer: $\frac{x\sin ^{-1}x}{\sqrt{1-x^{2}}}+log\left ( \sqrt{1-x^{2}} \right )-\frac{\left ( \sin ^{-1}x \right )^{2}}{2}+c$

Hint:Put $\sin ^{-1}x=u$

Given:$\int \frac{x^{2}\sin ^{-1}x}{\left ( 1-x^{2} \right )^{\frac{3}{2}}}dx$

Solution: Put

$\sin ^{-1}x=u$$\Rightarrow \frac{1}{\sqrt{1-x^{2}}}dx=du$

and

$x=\sin u$

\begin{aligned} &\therefore \int \frac{x^{2} \sin ^{-1} x}{\left(1-x^{2}\right)^{\frac{3}{2}}} d x=\int \frac{x^{2} \sin ^{-1} x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}} d x \\ &\Rightarrow \int\left(\frac{u \sin ^{2} u}{1-\sin ^{2} u}\right) d u \\ &=\int \frac{u \sin ^{2} u}{\cos ^{2} u} d u=\int u \tan ^{2} u d u=\int u\left(\sec ^{2} u-1\right) d u \end{aligned}

\begin{aligned} &=\int u \sec ^{2} u d u-\int u d u \\ &=u \times \int \sec ^{2} u d u-\int \frac{d}{d u} u \int \sec ^{2} u d u d u-\frac{u^{2}}{2}+c \\ &=u \tan u-\int 1 \times \tan u d u-\frac{u^{2}}{2}+c \\ &=u \tan u-\int \frac{\sin u}{\cos u} d u-\frac{u^{2}}{2}+c \end{aligned}

Put       $\cos u=v\Rightarrow -\sin udu=dv\Rightarrow \sin udu=-dv$

\begin{aligned} &=u \tan u+\int \frac{d v}{v}-\frac{u^{2}}{2}+c \\ &=u \tan u+\ln v-\frac{u^{2}}{2}+c \\ &=u \frac{\sin u}{\cos u}+\ln \left(\sqrt{1-\sin ^{2} u}\right)-\frac{u^{2}}{2}+c \\ &=u \frac{\sin u}{\sqrt{1-\sin ^{2} u}}+\ln \left(\sqrt{1-\sin ^{2} u}\right)-\frac{u^{2}}{2}+c \\ &=\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}+\log \left(\sqrt{1-x^{2}}\right)-\frac{\left(\sin ^{-1} x\right)^{2}}{2}+c \end{aligned}