Question

Evaluate $\int {\sin }^{-1}xdx$

Answer 1

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

Integrating by parts, we get

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

$dv=dx\Rightarrow v=x$

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

$=x{\sin }^{-1}x+\frac{1}{2}\int \frac{-2xdx}{\sqrt{1-x^2}}$

Let $t=1-x^2\Rightarrow dt=-2xdx$

$=x{\sin }^{-1}x+\frac{1}{2}\int \frac{dt}{\sqrt{t}}$

$=x{\sin }^{-1}x+\frac{1}{2}\int t^{-\frac{1}{2}}dt$

$=x{\sin }^{-1}x+\frac{1}{2}\left[\frac{t^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}\right]+c$

$=x{\sin }^{-1}x+\frac{1}{2}\times \frac{2}{1}\sqrt{t}+c$

$=x{\sin }^{-1}x+\sqrt{1-x^2}+c$ where $t=1-x^2$