Question

Solve $\int x{\sin }^{-1}xdx$

Answer 1

$\Rightarrow {\sin }^{-1}\int xdx-\int \left[\frac{d{\sin }^{-1}x}{dx}\int xdx\right]dx\Rightarrow \frac{x^2}{2}{\sin }^{-1}x-\int \frac{1}{\sqrt{1-x^2}}\cdot \frac{x^2}{2}dx\Rightarrow \frac{x^2}{2}{\sin }^{-1}x-\frac{1}{2}\int \frac{x^2-1+1}{\sqrt{1-x^2}}dx\Rightarrow \frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\int \frac{1-x^2}{\sqrt{1-x^2}}dx-\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}dx\Rightarrow \frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\int \sqrt{1-x^2}dx-\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}dx\Rightarrow \frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\left[\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x\right]-\frac{1}{2}{\sin }^{-1}x+C\Rightarrow \frac{x^2}{2}{\sin }^{-1}x+\frac{x}{4}\sqrt{1-x^2}+\frac{1}{4}{\sin }^{-1}x-\frac{1}{2}{\sin }^{-1}x+C\Rightarrow \frac{x^2}{2}{\sin }^{-1}x+\frac{x}{4}\sqrt{1-x^2}-\frac{{\sin }^{-1}x}{4}+C$