Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{x{\sin }^{-1}x}{{\left(1-x^2\right)}^{{\displaystyle \frac{3}{2}}}}dx \\ \mathrm{Putting}{\sin }^{-1}x=\theta \\ \Rightarrow x=\sin \theta \\ \Rightarrow dx=\cos \theta d\theta \\ \therefore I=\int \frac{\sin \theta \theta \cos \theta d\theta }{{\left(1-{\sin }^2\theta \right)}^{{\displaystyle \frac{3}{2}}}} \\ =\int \frac{\theta \sin \theta \cos \theta d\theta }{{\left({\cos }^2\theta \right)}^{{\displaystyle \frac{3}{2}}}} \\ =\int \theta \frac{\sin \theta }{{\cos }^2\theta }d\theta \\ =\int \underset{\mathrm{I}}{\theta }\underset{\mathrm{II}}{\sec \theta \tan \theta }d\theta \\ =\theta \times \sec \theta -\int 1\times \sec \theta d\theta \\ =\theta \times \sec \theta -\int \sec \theta d\theta \\ =\theta \times \sec \theta -\log \left|\sec \theta +\tan \theta \right|+C \\ =\frac{\theta }{\cos \theta }-\log \left|\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }\right|+C \\ =\frac{\theta }{\sqrt{1-{\sin }^2\theta }}-\log \left|\frac{1+\sin \theta }{\cos \theta }\right|+C \\ =\frac{\theta }{\sqrt{1-{\sin }^2\theta }}-\log \left|\frac{1+\sin \theta }{\sqrt{1-{\sin }^2\theta }}\right|+C \\ =\frac{\theta }{\sqrt{1-{\sin }^2\theta }}-\log \left|\frac{\sqrt{1+\sin \theta }}{\sqrt{1-\sin \theta }}\right|+C \\ =\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}-\log \left|\frac{\sqrt{1+x}}{\sqrt{1-x}}\right|+C \\ =\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}-\frac{1}{2}\log \left|\frac{1+x}{1-x}\right|+C \end{gathered}$$