Question

$\int \frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$

Answer 1

$$\begin{gathered} \int \frac{x.{\sin }^{-1}x}{\sqrt{1-x^2}}dx \\ \mathrm{Let}{\sin }^{-1}x=\theta \\ x=\sin \theta \\ dx=\cos \theta d\theta \\ \therefore \int \frac{x.{\sin }^{-1}x}{\sqrt{1-x^2}}dx=\int \frac{\left(\sin \mathit{}\theta \right).\theta }{\sqrt{1-{\sin }^2\theta }}.\cos \mathit{}\theta \mathit{}d\theta \\ =\int \frac{\left(\sin \mathit{}\theta \right).\theta }{\cos \theta }.\cos \theta d\theta \\ =\int \underset{\mathrm{I}}{\theta }.\underset{\mathrm{II}}{\sin \theta }d\theta \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\theta \int \sin \theta \mathit{}d\theta \mathit{-}\int \left\{\frac{d}{d\theta }\left(\theta \right)\int \sin \theta \mathit{}d\theta \right\}d\theta \\ =\theta \left(-\cos \theta \right)-\int 1.\left(-\cos \theta \right)d\theta \\ =-\theta \mathit{}\cos \theta +\sin \theta +C \\ =-\theta \sqrt{1-{\sin }^2\theta }+\sin \mathit{}\theta +C \\ =-{\sin }^{-1}x\sqrt{1-x^2}+x+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because {\sin }^{\mathit{-}\mathit{1}}\mathit{}x=\theta \right) \end{gathered}$$