TITLE: A Sine and Inverse Sine integral
QUESTION [20 upvotes]: A demonstration of methods
While reviewing an old text book an integral containing sines and sine inverse was encountered, namely,
$$\int_{0}^{\pi/2} \int_{0}^{\pi/2} \sin(x) \, \sin^{-1}(\sin(x) \, \sin(y)) \, dx \, dy = \frac{\pi^{2}}{4} - \frac{\pi}{2}.$$
One can express $\sin^{-1}(z)$ as a series and integrate, but are there other methods nearly as efficient?

REPLY [5 votes]: To complement the existing answers, here is a geometrical way.
Consider the following integral on the surface of a unit sphere (positive octant):
$$\mathcal{I}=\int\limits_{x^2+y^2+z^2=1,~x,y,z\ge0}d\Omega~\sin^{-1}(y).$$
where $\Omega$ is the solid angle (or area).  By using the spherical coordinate system: $x=\sin\theta\cos\phi,~y=\sin\theta\sin\phi,~z=\cos\theta,$ where $0\le\theta,\phi\le\pi/2,$ we see this is the original integral that we need to evaluate.
Note $\mathcal{I}$ is invariant under rotation, so we also have
$$\mathcal{I}=\int\limits_{x^2+y^2+z^2=1,~x,y,z\ge0}d\Omega~\sin^{-1}(z).$$
Again using the spherical coordinates, we get
$$\mathcal{I}=\int_0^{\pi/2}d\phi\int_0^{\pi/2}d\theta~\sin\theta~\sin^{-1}(\cos\theta)=\frac{\pi}{2}\int_0^{\pi/2}d\theta~\sin\theta\left(\frac{\pi}{2}-\theta\right)=\frac{\pi}{2}\left(\frac{\pi}{2}-1\right)$$
as desired, where we have used integration by parts for the last step.

For the general case that David H. considered, we have when $r\le1$:
$$\mathcal{I}(r)=\int\limits_{x^2+y^2+z^2=r^2,~x,y,z\ge0}d\Omega~\sin^{-1}(y).$$
Similarly we get
$$\mathcal{I}(r)=\int_0^{\pi/2}d\phi\int_0^{\pi/2}d\theta~\sin\theta~\sin^{-1}(r\cos\theta)=\frac{\pi}{2r}\int_0^rdt\sin^{-1}(t)~=\frac{\pi}{2r}\left(r\sin^{-1}(r)+\sqrt{1-r^2}-1\right)=\frac{\pi}{2}\left(\sin^{-1}(r)+\frac{\sqrt{1-r^2}-1}{r}\right).$$
Again we have used integration by parts for the last integral.