We need to evaluate
∫sin−1xdx
Let u=sin−1x,dv=dx
⇒du=1√1−x2dx,v=x
∫sin−1xdx=xsin−1x−∫x√1−x2dx+c...(1)
where c is integration constant.
Consider ∫x√1−x2dx
Let w=1−x2
⇒dw=−2xdx
⇒xdx=−dw2
∫x√1−x2dx=∫−dw2√w
=−12∫dw√w
=−12(2√w)
=−√w
Substitute back w=1−x2 we get
∴∫x√1−x2dx=−√1−x2
Therefore (1) becomes
∫sin−1xdx=xsin−1x+√1−x2+c