# How do you integrate int sin^-1x by integration by parts method?

Apr 8, 2018

$x \arcsin x + \sqrt{1 - {x}^{2}} + C$

#### Explanation:

After choosing $u = \arcsin x$ and $\mathrm{dv} = \mathrm{dx}$, $\mathrm{du} = \frac{\mathrm{dx}}{\sqrt{1 - {x}^{2}}}$ and $v = x$

Hence,

$\int \arcsin x \cdot \mathrm{dx} = x \arcsin x - \int x \cdot \frac{\mathrm{dx}}{\sqrt{1 - {x}^{2}}}$

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

=$x \arcsin x + \sqrt{1 - {x}^{2}} + C$