Question

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

Answer 1

We need to evaluate $\int si{n}^{-1}xdx$

Let $u=si{n}^{-1}x,dv=dx$

$\Rightarrow du=\frac{1}{\sqrt{1-x^2}}dx,v=x$

$\int si{n}^{-1}xdx=xsi{n}^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx+c...\left(1\right)$

where $c$ is integration constant.

Consider $\int \frac{x}{\sqrt{1-x^2}}dx$

Let $w=1-x^2$

$\Rightarrow dw=-2xdx$

$\Rightarrow xdx=-\frac{dw}{2}$

$\int \frac{x}{\sqrt{1-x^2}}dx=\int -\frac{dw}{2\sqrt{w}}$

$=-\frac{1}{2}\int \frac{dw}{\sqrt{w}}$

$=-\frac{1}{2}\left(2\sqrt{w}\right)$

$=-\sqrt{w}$

Substitute back $w=1-x^2$ we get

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

Therefore $\left(1\right)$ becomes

$\int si{n}^{-1}xdx=xsi{n}^{-1}x+\sqrt{1-x^2}+c$