# How do you integrate int arcsinx by integration by parts method?

Sep 26, 2017

$\int {\sin}^{- 1} x \mathrm{dx} = x {\sin}^{- 1} + {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} + C$

#### Explanation:

IBP formula

$I = \int \textcolor{red}{u} v ' \mathrm{dx} = \int \textcolor{red}{u} v - v \textcolor{red}{u '} \mathrm{dx}$

we ahve

$\int {\sin}^{- 1} x \mathrm{dx}$

let$\text{ } \textcolor{red}{u = {\sin}^{- 1} x \implies u ' = \frac{1}{\sqrt{1 - {x}^{2}}}}$

$v ' = 1 \implies v = x$

$I = x \textcolor{red}{{\sin}^{- 1} x} - \int \frac{x}{\textcolor{red}{\sqrt{1 - {x}^{2}}}} \mathrm{dx}$

now
$\int \frac{x}{\textcolor{red}{\sqrt{1 - {x}^{2}}}} \mathrm{dx} = \int x {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \mathrm{dx}$

by inspection we have

$= - {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

this is left as an exercise for the reader to verify

finally we have

$\int {\sin}^{- 1} \mathrm{dx} = x {\sin}^{- 1} + {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} + C$