$\int {\mathrm{xsin}}^{-1}\mathrm{xdx}$ is equal to 

Solution:

Let $\mathrm{I}=\int \mathrm{x} {\sin }^{-1}\mathrm{xdx}$
On taking sin^-1 x as first function and x as second function and integrating by parts, we get
[$\because$ inverse functions comes before,algebraic functions in ILATE]
$\begin{array}{l}\mathrm{I}={\sin }^{-1}\mathrm{x}\int \mathrm{xdx}-\int \left[\frac{\mathrm{d}}{\mathrm{dx}}\left({\sin }^{-1}\mathrm{x}\right)\int \mathrm{xdx}\right]\mathrm{dx}\\ ={\sin }^{-1}\mathrm{x}\cdot \frac{{\mathrm{x}}^2}{2}-\int \left[\frac{1}{\sqrt{1-{\mathrm{x}}^2}}\cdot \frac{{\mathrm{x}}^2}{2}\right]\mathrm{dx}\end{array}$
 $=\frac{{\mathrm{x}}^2}{2}\cdot {\sin }^{-1}\mathrm{x}-\int \left[\frac{1-\left(1-{\mathrm{x}}^2\right)}{\sqrt{1-{\mathrm{x}}^2}}\cdot \frac{1}{2}\right]\mathrm{dx}$
[add and subtract 1 in numerator of second term]
$\begin{array}{l}=\frac{{\mathrm{x}}^2}{2}\cdot {\sin }^{-1}\mathrm{x}-\frac{1}{2}\int \frac{1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}+\frac{1}{2}\int \frac{1-{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\\ =\frac{{\mathrm{x}}^2}{2}\cdot {\sin }^{-1}\mathrm{x}-\frac{1}{2}{\sin }^{-1}\mathrm{x}+\frac{1}{2}\int \sqrt{1-{\mathrm{x}}^2}\mathrm{dx}\end{array}$
$\begin{array}{l}\Rightarrow \mathrm{I}=\frac{{\mathrm{x}}^2}{2}\cdot {\sin }^{-1}\mathrm{x}-\frac{1}{2}{\sin }^{-1}\mathrm{x}+\frac{1}{2}\left[\frac{1}{2}\mathrm{x}\sqrt{1-{\mathrm{x}}^2}+\frac{1}{2}{\sin }^{-1}\mathrm{x}\right]+\mathrm{C}\\ \Rightarrow \mathrm{I}={\sin }^{-1}\mathrm{x}\left[\frac{{\mathrm{x}}^2}{2}-\frac{1}{4}\right]+\frac{1}{4}\mathrm{x}\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}\\ \Rightarrow \mathrm{I}=\frac{{\sin }^{-1}\mathrm{x}}{4}\left(2{\mathrm{x}}^2-1\right)+\frac{1}{4}\mathrm{x}\sqrt{1-{\mathrm{x}}^2}+\mathrm{C}\end{array}$