# How do I find the integral intsin^-1(x)dx ?

Sep 10, 2014

By integration by parts,
$\int {\sin}^{- 1} x \mathrm{dx} = x {\sin}^{- 1} x + \sqrt{1 - {x}^{2}} + C$

Let us look at some details.
Let $u = {\sin}^{- 1} x$ and $\mathrm{dv} = \mathrm{dx}$.
$R i g h t a r r o w \mathrm{du} = \frac{\mathrm{dx}}{\sqrt{1 - {x}^{2}}}$ and $v = x$

By integration by parts,
$\int {\sin}^{- 1} x \mathrm{dx} = x {\sin}^{- 1} x - \int \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

Let $u = 1 - {x}^{2}$. Rightarrow {du}/{dx}=-2x Rightarrow dx={du}/{-2x}

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

Hence,
$\int {\sin}^{- 1} x \mathrm{dx} = x {\sin}^{- 1} x + \sqrt{1 - {x}^{2}} + C$