Question

$\int Si{n}^{-1}xdx$

Answer 1

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

$\int f{g}^{\prime }=fg-\int f^{\prime }g$

$f={\sin }^{-1}x$ $g^{\prime }=1$

$f^{\prime }=\frac{1}{\sqrt{1-x^2}}$ $g=x$

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

Substitute $u=1-x^2$

$\Rightarrow dx=\frac{-1}{2x}du$

$=x{\sin }^{-1}x-\left[\frac{1}{2}\int \frac{1}{\sqrt{u}}du\right]$

$=x{\sin }^{-1}x-[-\sqrt{u}]$

$=x{\sin }^{-1}x-[-\sqrt{1-x^2}]$

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