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

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Answer 1 · 2014-09-10T11:34:56

By integration by parts,
$int sin^{-1}xdx=xsin^{-1}x+sqrt{1-x^2}+C$

Let us look at some details.
Let $u=sin^{-1}x$ and $dv=dx$ .
$Rightarrow du={dx}/sqrt{1-x^2}$ and $v=x$

By integration by parts,
$int sin^{-1}xdx=xsin^{-1}x-intx/sqrt{1-x^2}dx$

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

$intx/sqrt{1-x^2}dx=int x/sqrt{u}{du}/{-2x}=-1/2intu^{-1/2}du$
$=-u^{1/2}+C=-sqrt{1-x^2}+C$

Hence,
$int sin^{-1}xdx=xsin^{-1}x+sqrt{1-x^2}+C$

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