Evaluate $\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$ 
Evaluate $$\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$$

My attempt. Taking $\sin x \sin y=t$, $\iint\sin x  \ dx \ dy$.
I am not sure but I think the limits would change to $x=0$ to $\pi/2$ and $t=0$ to $1$.
I do not know how to change the elemental area i.e $dx\, dy$. Can I apply Jacobian where I am only changing one variable (i.e $y$ to $t$)?
Is this the correct way. Is there any easier way to go about it?

Here is an approach using power series expansions.
It is known that for $t\in (-1,1)$
$$\frac{1}{\sqrt{1-t}} =(1-t)^{-1/2}=\sum_{n=0}^\infty 
\binom{-1/2}{n}  (-t)^{n}=
\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n}  t^{n},$$
and for $t\in [-1,1]$,
$$\arcsin(t) =\int_0^t\frac{ds}{\sqrt{1-s^2}}= \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ t^{2n+1}}{2n+1}.$$
Hence, the given integral is equal to
\begin{align*}I&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}\int_0^{\pi/2}\int_0^{\pi/2} \sin x (\sin x \sin y)^{2n+1} \ dx \ dy\\
&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}W_{2n+2} W_{2n+1} \\
&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}\cdot \frac{\pi}{4(n+1)}\\
&=\frac{\pi}{4}\left(2\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}-\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{n+1}\right)\\
&=\frac{\pi}{4}\left(2\arcsin(1)-\int_0^1\frac{dt}{\sqrt{1-t}}\right)\\
&=\frac{\pi(\pi-2)}{4}.
\end{align*}
where $W_r=\int_{0}^{\pi/2}\sin^r(t)dt $ denotes the Wallis integral of order $r$.
