Question

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

Answer 1

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

$u={\sin }^{-1}x\Rightarrow \sqrt{1-x^2}du=dx\to x=v\Rightarrow dv=dx$

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

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

$\therefore -\int \frac{x}{\sqrt{1-x^2}}dx=\sqrt{1-x^2}+C$

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