Question

Integrate the function $x{\sin }^{-1}x$

Answer 1

Let $I=\int x{\sin }^{-1}xdx$
Taking ${\sin }^{-1}x$ as first function and $x$ as second function and integrating by parts, we obtain
$I={\sin }^{-1}x\int xdx-\int \{\left(\frac{d}{dx}si{n}^{-1}x\right)\int xdx\}dx$
$={\sin }^{-1}x\left(\frac{x^2}{2}\right)-\int \frac{1}{\sqrt{1-x^2}}\cdot \frac{x^2}{2}dx$
$=\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\int \frac{-x^2}{\sqrt{1-x^2}}dx$
$=\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\int \{\frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\}dx$
$=\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\int \{\sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}}\}dx$
$=\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\int \{\int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx\}$
$=\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\{\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x-{\sin }^{-1}x\}+C$
$=\frac{x^2{\sin }^{-1}x}{2}+\frac{x}{4}\sqrt{1-x^2}+\frac{1}{4}{\sin }^{-1}x-\frac{1}{2}{\sin }^{-1}x+C$
$=\frac{1}{4}(2x^2-1){\sin }^{-1}x+\frac{x}{4}\sqrt{1-x^2}+C$