Integrating arcsin x
\[\int^1_0 sin^{-1} xdx\]
consider the following graph.
\[A+B= \frac{\pi}{2}\]
, but \[A=\int^1_0 sin^{-1} y dy, \; B=\int^{\pi /2}_0 sinxdx\]
so \[\int^1_0 sin^{-1} y dy= \frac{\pi}{2} -\int^{\pi /2}_0 sinxdx=\frac{\pi}{2}- \frac{\pi}{2} - [-cosx]^{\pi /2}_0 =\frac{\pi}{2}-1\]
