<s>[INST] Decoupling a double integral I came across this question while making some calculations.

QUESTION. Can you find some transformation to "decouple" the double integral as follows?
$$\int_0^{\frac{\pi}2}\int_0^{\frac{\pi}2}\frac{d\alpha\,d\beta}{\sqrt{1-\sin^2\alpha\sin^2\beta}}
=\frac14\int_0^{\frac{\pi}2}\frac{d\theta}{\sqrt{\cos\theta\,\sin\theta}}\int_0^{\frac{\pi}2}\frac{d\omega}{\sqrt{\cos\omega\,\sin\omega}}.$$

 [/INST] (Thanks go to Etanche and Jandri)
\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2(\theta)\sin^2 \varphi}}d\varphi d\theta\\
 &\overset{z\left(\varphi\right)=\arcsin\left(\sin(\theta)\sin \varphi\right)}=\int_0^{\frac{\pi}{2}} \left(\int_0^ \theta\frac{1}{\sqrt{\sin(\theta-z)\sin(\theta+ z)}}dz\right)d\theta\tag1\\
 &=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du \tag2\\
 &=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du+\underbrace{\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{\frac{\pi}{2}}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du}_{w=\pi-v}\\
 &=\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\
&\overset{u\longleftrightarrow v}=\int_0^{\frac{\pi}{2}} \left(\int_{0}^{u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\
 &=\boxed{\frac{1}{2}\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv}
 \end{align}
and to obtain the form in the OP, finally substitute $u=2\theta $, $v=2\omega $.
$(1)$: $\displaystyle dz=\dfrac{\sqrt{\sin^2\theta-\sin^2 z}}{\sqrt{1-\sin^2 z}}d\varphi$, $\sin^2 a-\sin^2 b=\sin(a-b)\sin(a+b)$
$(2)$: Change of variable $u=\theta-z,v=\theta+z$
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