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?

Perform integration by parts,
$\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arcsin(\sin x \sin y) \ dx \ dy\\
&=\int_0^{\frac{\pi}{2}}\left(\Big[-\cos x\arcsin(\sin x \sin y)\Big]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\cos^2x\sin y}{\sqrt{1-\sin^2 x\sin^2 y }}\ dx\right)\ dy\\
&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\frac{\cos^2x\sin y}{\sqrt{1-\sin^2 x\sin^2 y }}\ dx \ dy\\
&=\int_0^{\frac{\pi}{2}}\left(\Big[-\frac{\cos x\text { arcsinh} \left(\tan x\cos y\right)}{\tan x}\Big]_{y=0}^{y=\frac{\pi}{2}}\right)\ dx\\
&=\int_0^{\frac{\pi}{2}} \frac{\cos x\text{ arcsinh}(\tan x)}{\tan x}\ dx
\end{align}$
Perform the change of variable $u=\tan x$,
$\begin{align}J&=\int_0^{\infty}\frac{\text{ arcsinh } x}{x\left(1+x^2\right)^{\frac{3}{2}}}\ dx
\end{align}$
Perform integration by parts,
$\begin{align}J&=\left[\left(\frac{1}{\sqrt{1+x^2}}-\text{arcsinh}\left(\frac{1}{x}\right)\right)\text{arcsinh }x\right]_0^{\infty}-\int_0^{\infty}\left(\frac{1}{\sqrt{1+x^2}}-\text{arcsinh}\left(\frac{1}{x}\right)\right)\frac{1}{\sqrt{1+x^2}}\ dx\\
&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx-\int_0^{\infty}\frac{1}{1+x^2}\ dx\\
&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx-\frac{\pi}{2}
\end{align}$
In the following integral perform the change of variable $u=\frac{1}{x}$,
$\begin{align}
K&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx\\
&=\int_0^{\infty}\frac{\text{arcsinh }x}{x\sqrt{1+x^2}}\ dx\\
\end{align}$
Perform the change of variable $u=\text{arcsinh }x$,
$\begin{align}
K&=\int_0^{\infty}\frac{x}{\sinh x}\ dx
\end{align}$
and according to Contour integral of $\int_{0}^{\infty}\frac{x}{\sinh x}\operatorname{dx}$ ,
$\displaystyle K=\frac{\pi^2}{4}$
Therefore,
$\boxed{\displaystyle J=\frac{\pi^2}{4}-\frac{\pi}{2}}$
Addenda:
1)Another way to compute $K$.
Performing integration by parts,
$\begin{align}K&=\left[x\ln\left(\tanh\left(\frac{x}{2}\right)\right)\right]_0^{\infty}-\int_0^{\infty}\ln\left(\tanh\left(\frac{x}{2}\right)\right)\ dx\\
&=-\int_0^{\infty}\ln\left(\tanh\left(\frac{x}{2}\right)\right)\ dx\\
\end{align}$
Perform the change of variable $y=\tanh\left(\frac{x}{2}\right)$,
$\begin{align}K&=2\int_0^1 \frac{\ln x}{x^2-1}\ dx\\
&=\int_0^1 \frac{\ln x}{x-1}\ dx-\int_0^1 \frac{\ln x}{x+1}\ dx
\end{align}$
$\begin{align}L&=\int_0^1 \frac{\ln x}{x-1}\ dx+\int_0^1 \frac{\ln x}{x+1}\ dx\\
&=\frac{1}{2}\int_0^1 \frac{2x\ln\left(x^2\right)}{x^2-1}\ dx
\end{align}$
Perform the change of variable $u=x^2$,
$\begin{align}L&=\frac{1}{2}\int_0^1 \frac{\ln x}{x-1}\ dx
\end{align}$
Therefore,
$\begin{align}\int_0^1 \frac{\ln x}{x+1}\ dx=-\frac{1}{2}\int_0^1 \frac{\ln x}{x-1}\ dx
\end{align}$
Therefore,
$\begin{align}K&=\frac{3}{2}\int_0^1 \frac{\ln x}{x-1}\ dx
\end{align}$
Knowing that,
$\begin{align}\int_0^1 \frac{\ln x}{x-1}\ dx=\dfrac{\pi^2}{6}\end{align}$
Therefore,
$\boxed{\displaystyle K=\dfrac{\pi^2}{4}}$
2)Playing around with lindep, intnum, functions of GP-PARI, i suspect that the following formulae are true:
$\begin{align}
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arcsin(\cos x \cos y) \ dx \ dy&=\frac{1}{8}\pi^2-\ln 2\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arccos(\sin x \sin y) \ dx \ dy&=\frac{1}{2}\pi\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arccos(\cos x \cos y) \ dx \ dy&=\frac{1}{8}\pi^2+\ln 2\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arctan(\sin x \sin y) \ dx \ dy&=\frac{1}{8}\pi^2-\frac{1}{4}\pi\ln 2\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arctan^2(\sin x \sin y) \ dx \ dy&=\frac{1}{32}\pi^3+\frac{1}{8}\pi^2\ln 2-\frac{1}{2}\text{G}\pi\\
\end{align}$
$\text{G}$ is the Catalan constant.
