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

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Answer 1 · 2018-04-08T23:30:56

$xarcsinx+sqrt(1-x^2)+C$

After choosing $u=arcsinx$ and $dv=dx$ , $du=(dx)/sqrt(1-x^2)$ and $v=x$

Hence,

$int arcsinx*dx=xarcsinx-int x*(dx)/sqrt(1-x^2)$

= $xarcsinx-int (xdx)/sqrt(1-x^2)$

= $xarcsinx+sqrt(1-x^2)+C$

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