Prove the following expression by changing order of integration To prove using change of order of integration :
$$\int_0^{\pi/2}\int_0^{\pi/2} \sin(x)\arcsin(\sin(x)\sin(y))\,dx\, dy=\frac{\pi^2}{4}-\frac{\pi}{2}$$
Progress so far:
Let $\sin x \sin y=\sin z$, therefore $\sin x \cos y .dy= \cos z .dz$
When $x=\pi/2$ then $\sin y= \sin z$ $\implies y=z$
and  when $x=0$ then $\sin z=0 \implies z=0$
Therefore I can write the integral as :
$$I=\int_0^{\pi/2}\sin x .dx\int_0^z\sin^{-1}(\sin z).\frac{\cos z . dz}{\sin x . \cos y}$$
$$I=\int_0^{\pi/2}dx\int_0^z\frac{z. \cos z.dz}{\cos y}$$
But how do I proceed from here? Please guide
 A: I honestly do not get your substitutions at all.
Using complete elliptic integrals (according to Mathematica's notation, their arguments stand for the elliptic modulus) and Fubini's Theorem we have
$$\begin{eqnarray*}\iint_{(0,\pi/2)^2}\sin(x)\arcsin(\sin(x)\sin(y))\,dx\,dy &=& \iint_{(0,1)^2}\frac{x\arcsin(xy)}{\sqrt{(1-x^2)(1-y^2)}}\,dx\,dy\\&=&\iiint_{(0,1)^3}\frac{x^2 y}{\sqrt{(1-x^2)(1-y^2)(1-x^2 y^2 z^2)}}\,dx\,dy\,dz\\&=&\iint_{(0,1)^2}\frac{K(y^2 z^2)-E(y^2 z^2)}{z^2 y\sqrt{1-y^2}}\,dy\,dz\\&=&\int_{0}^{1}\frac{E(y^2)-(1-y^2)K(y^2)}{y\sqrt{1-y^2}}\,dy\\&=&\int_{0}^{1}\frac{E(u)-(1-u)K(u)}{2u\sqrt{1-u}}\,du\end{eqnarray*}$$
and we may exploit the Taylor series of $K(u)$ and $E(u)$ to get that the last integral equals
$$ \frac{\pi}{2}+\frac{\pi}{4}\sum_{n\geq 1}\frac{\binom{2n}{n}^2}{16^n}\left[\frac{1}{1-2n}\int_{0}^{1}\frac{u^{n-1}}{\sqrt{1-u}}\,du-\int_{0}^{1}u^{n-1}\sqrt{1-u}\,du\right] $$
which by Euler's Beta function simplifies to:
$$ \frac{\pi}{2}\sum_{n\geq 1}\frac{\binom{2n}{n}}{4^n(2n+1)}=\frac{\pi}{2}\int_{0}^{1}\left[\frac{1}{\sqrt{1-z^2}}-1\right]\,dz=\color{blue}{\frac{\pi}{4}(\pi-2)}.$$
An alternative and pretty efficient way for dealing with integrals involving $E(u)$ or $K(u)$ is to exploit Fourier-Legendre series expansions.
