A Sine and Inverse Sine integral 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?  
 A: 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.
A: As mickep has observed 
 \begin{equation*}
 I_{1} = \int_0^{\pi/2}\int_0^{\pi/2}\sin(x)\arcsin(\sin(x)\sin(y))\,dxdy = \dfrac{1}{2} \int_{0}^{\pi/2}\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\dfrac{\cos^{2}x}{\sin x}\, dx.
\end{equation*}
We intend to finish the solution without using series expansion. Instead we will use complex integration and Cauchy integral theorem.
After integration by parts we have
\begin{gather*}
I_{1} = \dfrac{1}{2}\int_{0}^{\pi/2}\left(\dfrac{1}{\sin x}-\sin x\right)\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\, dx = \dfrac{1}{2}\left[\left(\ln\left(\tan\dfrac{x}{2}\right)+\cos x\right)\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\right]_{0}^{\pi/2}\\
- \int_{0}^{\pi/2}\left(\ln\left(\tan\dfrac{x}{2}\right)+\cos x\right)\dfrac{1}{\cos x}\, dx = - \underbrace{\int_{0}^{\pi/2}\dfrac{\ln\left(\tan\dfrac{x}{2}\right)}{\cos x}\, dx}_{I_{2}} - \dfrac{\pi}{2}.
 \end{gather*}
 Finally we will prove that $I_2 = -\dfrac{\pi^{2}}{4}.$ Since $\displaystyle \tan^{2}\dfrac{x}{2} = \dfrac{1-\cos x}{1+\cos x}$ we get
 \begin{gather*}
I_{2}= \dfrac{1}{2}\int_{0}^{\pi/2}\dfrac{1}{\cos x}\ln\left(\dfrac{1-\cos x}{1+\cos x}\right)\, dx = \dfrac{1}{2}\int_{0}^{\pi/2}\dfrac{1}{\sin x}\ln\left(\dfrac{1-\sin x}{1+\sin x}\right)\, dx = \left[t = \tan\dfrac{x}{2}\right]\\
= \int_{0}^{1}\dfrac{1}{t}\ln\left(\dfrac{1-t}{1+t}\right)\, dt = \left[\ln(t)\ln\left(\dfrac{1-t}{1+t}\right)\right]_{0}^{1} - 2\int_{0}^{1}\dfrac{\ln(t)}{t^{2}-1}\, dt.
\end{gather*}
But
\begin{equation*}
\int_{0}^{1}\dfrac{\ln(t)}{t^{2}-1}\, dt = \int_{1}^{\infty}\dfrac{\ln(t)}{t^{2}-1}\, dt .
\end{equation*}
Thus $
\displaystyle I_{2} = -\int_{0}^{\infty}\dfrac{\ln(t)}{t^{2}-1}\, dt.$ If $\log $ is the principal branch of the logarithm then the function $ f(z) = \dfrac{\log(z)}{z^{2}-1}$ has a removable singularity at $z=1$. If
we integrate $f(z)$ ''around quadrant 1'' and use Cauchy integral theorem we get
\begin{equation*}
\int_{0}^{\infty}\dfrac{\ln(x)}{x^{2}-1}\, dx - \int_{0}^{\infty}\dfrac{\ln(|iy|)+i\pi/2}{(iy)^{2}-1}i\, dy = 0.
\end{equation*}
 Consequently
 \begin{equation*}
-I_{2} + i\int_{0}^{\infty}\dfrac{\ln(y)}{y^{2}+1}\, dy -\dfrac{\pi}{2}\underbrace{\int_{0}^{\infty}\dfrac{1}{y^{2}+1}\, dy}_{ = \pi/2}  = 0.\tag{1}
\end{equation*}
From the real part in (1) we finally get $I_2 = -\dfrac{\pi^2}{4}$.
