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

1 Answer
Wataru
Sep 10, 2014

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#

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