How do you integrate $\int arcsin x$ $int arcsinx$ by integration by parts method?

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Answer 1 · 2017-09-26T08:07:34

$\int {sin}^{- 1} x d x = x {sin}^{- 1} + {(1 - x^{2})}^{\frac{1}{2}} + C$ $intsin^(-1)xdx=xsin^(-1)+(1-x^2)^(1/2)+C$

IBP formula

$I=intcolor(red)(u)v'dx=intcolor(red)(u)v-vcolor(red)(u')dx$

we ahve

$intsin^(-1)xdx$

let $" "color(red)( u=sin^(-1)x=>u'=1/sqrt(1-x^2))$

$v'=1=>v=x$

$I=xcolor(red)(sin^(-1)x)-intx/color(red)(sqrt(1-x^2))dx$

now
$intx/color(red)(sqrt(1-x^2))dx=intx(1-x^2)^(-1/2)dx$

by inspection we have

$=-(1-x^2)^(1/2)$

this is left as an exercise for the reader to verify

finally we have

$intsin^(-1)dx=xsin^(-1)+(1-x^2)^(1/2)+C$

How do you integrate int arcsinx∫arcsinxint arcsinx by integration by parts method?