How do I find the integral ∫sin−1(x)dx ?

1 Answer
Wataru
Sep 10, 2014

By integration by parts,
∫sin−1xdx=xsin−1x+√1−x2+C

Let us look at some details.
Let u=sin−1x and dv=dx.
⇒du=dx√1−x2 and v=x

By integration by parts,
∫sin−1xdx=xsin−1x−∫x√1−x2dx

Let u=1−x2. ⇒dudx=−2x⇒dx=du−2x

∫x√1−x2dx=∫x√udu−2x=−12∫u−12du
=−u12+C=−√1−x2+C

Hence,
∫sin−1xdx=xsin−1x+√1−x2+C

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