Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^2.{\sin }^{-1}xdx}{{\left(1-x^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ \mathrm{Putting}x=\sin \theta \\ \Rightarrow dx=\cos \theta d\theta \\ \&\theta ={\sin }^{-1}x \\ \therefore I=\int \frac{{\sin }^2\theta .\theta .\cos \theta d\theta }{{\left(1-{\sin }^2\theta \right)}^{{\displaystyle \frac{3}{2}}}} \\ =\int \frac{{\sin }^2\theta .\theta .\cos \theta d\theta }{{\left({\cos }^2\theta \right)}^{{\displaystyle \frac{3}{2}}}} \\ =\int \frac{{\sin }^2\theta .\theta .\cos \theta d\theta }{{\cos }^3\theta } \\ =\int {\tan }^2\theta .\theta .d\theta \\ =\int \left({\sec }^2\theta -1\right)\theta .d\theta \\ =\int \underset{\mathrm{I}}{\theta }.\underset{\mathrm{II}}{{\sec }^2\theta }d\theta -\int \theta .d\theta \\ =\theta \int {\sec }^2\theta d\theta -\int \left\{\frac{d}{d\theta }\left(\theta \right)\int {\sec }^2\theta d\theta \right\}d\theta -\int \theta .d\theta \\ =\theta \tan \theta -\int 1.\tan \theta d\theta -\frac{{\theta }^2}{2} \\ =\theta .\tan \theta -\ln \left|\sec \theta \right|-\frac{{\theta }^2}{2}+C \\ =\theta .\frac{\sin \theta }{\cos \theta }+\ln \left|\cos \theta \right|-\frac{{\theta }^2}{2}+C \\ =\theta .\frac{\sin \theta }{\cos \theta }+\ln \left|\sqrt{1-{\sin }^2\theta }\right|-\frac{{\theta }^2}{2}+C \\ =\frac{\theta .\sin \theta }{\sqrt{1-{\sin }^2\theta }}+\frac{1}{2}\ln \left|1-{\sin }^2\theta \right|-\frac{{\theta }^2}{2}+C \\ =\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}+\frac{1}{2}\ln \left(1-x^2\right)-\frac{1}{2}{\left({\sin }^{-1}x\right)}^2+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \mathit{}\mathit{}\theta ={\sin }^{-1}\mathit{}x\right] \end{gathered}$$