A double integral for $\frac{\pi}{2} \ln 2$. Show that 
  \begin{eqnarray*}
I=\int_0^1 \int_0^1 \frac{dx \,dy}{\sqrt{1-x^2y^2}}  = \frac{\pi}{2} \ln 2.
\end{eqnarray*}


My try ... from this question here we have
\begin{eqnarray*}
\int_0^1 \frac{ \sin^{-1}(x)}{x} \,dx = \frac{\pi}{2} \ln 2 .
\end{eqnarray*}
And from this question here we have
\begin{eqnarray*}
\int_0^1 \ln \left( \frac{1+ax}{1-ax} \right) \frac{dx}{x\sqrt{1-x^2}}=\pi\sin^{-1} a,\qquad |a|\leq 1. 
\end{eqnarray*}
It is easy to show 
\begin{eqnarray*}
\int_0^1  \frac{dy}{1+yz}=\frac{1}{z} \ln(1+z).   
\end{eqnarray*}
So we have (with a little tad of algebra) 
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 &amp;=&amp;  \int_0^1 \frac{ \sin^{-1}(x)}{x}\, dx \\ 
&amp;=&amp; \frac{1}{\pi} \int_0^1  \int_0^1 \ln \left( \frac{1+xt}{1-xt} \right) \frac{dt}{xt\sqrt{1-t^2}} x\, dx \\  
&amp;=&amp; \frac{2}{\pi} \int_0^1  \int_0^1 \int_0^1 \frac{dx \,dy\, dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\  
\end{eqnarray*}
This suggests we should consider the integral (sub $t=\sin(\theta)$) 
\begin{eqnarray*}
\frac{2}{\pi}  \int_0^1 \frac{ dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\  
= \int_0^{\pi/2} \frac{d \theta }{1- x^2  y^2 \sin^2(\theta)}. 
\end{eqnarray*}
Now it is well known (Geometrically expand, integrate term by term &amp; sum the familiar plum) that
\begin{eqnarray*}
\frac{2}{\pi} \int_0^{\pi/2}  \frac{d \theta }{1- \alpha \sin^2(\theta)}=\frac{2}{\pi} \frac{1}{\sqrt{1-\alpha}}    
\end{eqnarray*}
and using this we have
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 =\int_0^1 \int_0^1 \frac{dx\, dy}{\sqrt{1-x^2y^2}}  .
\end{eqnarray*}

The above double integral reminds of
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\int_0^1 \int_0^1 \frac{dx\, dy}{1-x^2y^2}  = \frac{\pi^2}{8}
\end{eqnarray*}
which can be evaluated using the substitution $x= \frac{\sin u}{\cos v}$, $y= \frac{\sin v}{\cos u}$.

My solution above used some pretty heavy machinery to establish the result. So my question is: is there an easier method ?
$$\int_0^1 \int_0^1 \frac{dxdy}{\sqrt{1-x^2y^2}}\overset{xy=t}=\int_0^1 \frac{1}{y}\int_0^y \frac{1}{\sqrt{1-t^2}}dtdy=\int_0^1 \frac{\arcsin y}{y}dy
$$
$$\overset{IBP}=-\int_0^1 \frac{
\ln y}{\sqrt{1-y^2}}dy\overset{y=\sin x}=-\int_0^\frac{\pi}{2}\ln(
\sin x)dx=\frac{\pi}{2}\ln 2$$
See here for the above integral.
Why not go head-on? It seems to work here. That is, write the integral as $$\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{\sqrt{1-x^2y^2}}=\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{y \sqrt{\frac{1}{y^2}-x^2}}.$$ Setting $x=\frac1y\sin\phi$ as usual helps us evaluate the first integral, or you can simply note that it has the form of an $\arcsin.$ If you make this substitution, the first integral becomes $$\int_0^{\arcsin y}\frac1y\mathrm d \phi=\frac{\arcsin y}{y}.$$ Thus, the integral reduces to $$\int_0^1\frac{\arcsin y}{y}\mathrm d y,$$ which may be done by parts. Forgetting limits for now, this begins as $$\arcsin y\log y-\int\frac{\log y}{\sqrt{1-y^2}}\mathrm dy,$$ etc.