<s>[INST] 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 &=&  \int_0^1 \frac{ \sin^{-1}(x)}{x}\, dx \\ 
&=& \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 \\  
&=& \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 & 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 ?
 [/INST] Through series expansions:
$$ \iint_{(0,1)^2}\frac{dx\,dy}{\sqrt{1-x^2 y^2}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\iint_{(0,1)^2}x^{2n}y^{2n}\,dx\,dy=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)^2}=\int_{0}^{1}\frac{\arcsin(x)}{x}\,dx $$
and this is
$$ \int_{0}^{\pi/2}x\cot(x)\,dx\stackrel{\text{IBP}}{=}\int_{0}^{\pi/2}\log(\sin x)\,dx $$
which is well known to be $\frac{\pi}{2}\log(2)$. It is possible to exploit symmetry, derivatives of the Beta function, Fourier (or Fourier-Legendre) series and probably much more. For instance, the identity
$$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n} $$
and just Riemann sums.
</s>