Question

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

Answer 1

$I=\int x{\sin }^{-1}xdx$

Let $u={\sin }^{-1}x\Rightarrow du=\frac{dx}{\sqrt{1-x^2}}$

$I=\frac{x^2}{2}{\sin }^{-1}x-\int \frac{x^2}{2\sqrt{1-x^2}}dx$ [using ILATE rule]

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

$I=\frac{x^2}{2}{\sin }^{-1}x-\frac{1}{2}\int \frac{1-(1-x^2)}{\sqrt{1-x^2}}dx$

$I=\frac{x^2}{2}{\sin }^{-1}x-\frac{1}{2}\int \frac{dx}{\sqrt{1-x^2}}+\frac{1}{2}\int \frac{(1-x^2)dx}{\sqrt{1-x^2}}$

$I=\frac{x^2}{2}{\sin }^{-1}x-\frac{1}{2}{\sin }^{-1}x+\frac{1}{2}\int {(1-x^2)}^{1-\frac{1}{2}}dx$

$I=\frac{x^2}{2}{\sin }^{-1}x-\frac{1}{2}{\sin }^{-1}x+\frac{1}{2}\int {(1-x^2)}^{\frac{1}{2}}dx$

$I=\frac{x^2}{2}{\sin }^{-1}x-\frac{1}{2}{\sin }^{-1}x+\frac{1}{2}(\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x)+c$ using $\int \sqrt{1-x^2}dx=\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x$

$I=\frac{x^2}{2}{\sin }^{-1}x-\frac{1}{2}{\sin }^{-1}x+\frac{1}{4}x\sqrt{1-x^2}+\frac{1}{4}{\sin }^{-1}x+c$

$I=\frac{x^2}{2}{\sin }^{-1}x-\frac{2}{4}{\sin }^{-1}x+\frac{1}{4}x\sqrt{1-x^2}+\frac{1}{4}{\sin }^{-1}x+c$

$I=\frac{x^2}{2}{\sin }^{-1}x-\frac{1}{4}{\sin }^{-1}x+\frac{1}{4}x\sqrt{1-x^2}+c$