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 ?
 A: $$\int_0^1 \int_0^1 \frac{dxdy}{\sqrt{1-x^2y^2}}\overset{xy=t}=\int_0^1 \frac{1}{y}\int_0^y \frac{1}{\sqrt{1-t^2}}dtdy=\int_0^1 \frac{\arcsin y}{y}dy
$$
$$\overset{IBP}=-\int_0^1 \frac{
\ln y}{\sqrt{1-y^2}}dy\overset{y=\sin x}=-\int_0^\frac{\pi}{2}\ln(
\sin x)dx=\frac{\pi}{2}\ln 2$$
See here for the above integral.
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[15px,#ffd]{I \equiv
\int_{0}^{1}\int_{0}^{1}{\dd x\,\dd y \over \root{1 - x^{2}y^{2}}} =
{\pi \over 2}\ln\pars{2}}:\ {\large ?}}$.
\begin{align}
I & \equiv
\int_{0}^{1}\int_{0}^{1}{\dd x\,\dd y \over \root{1 - x^{2}y^{2}}} =
\sum_{n = 0}^{\infty}{-1/2 \choose n}\int_{0}^{1}\int_{0}^{1}
\pars{-x^{2}y^{2}}^{n}\,\dd x\,\dd y
\\ & =
\sum_{n = 0}^{\infty}{-1/2 \choose n}{\pars{-1}^{n} \over \pars{2n + 1}^{2}} =
\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-1}^{n}\
\overbrace{\bracks{-\int_{0}^{1}\ln\pars{x}x^{2n}\,\dd x}}
^{\ds{1 \over \pars{2n + 1}^{2}}}
\\[5mm] & =
-\int_{0}^{1}\ln\pars{x}
\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-x^{2}}^{n}\,\dd x =
-\int_{0}^{1}{\ln\pars{x} \over \root{1 - x^{2}}}\,\dd x
\\[5mm] & \stackrel{x\ =\ \sin\pars{\theta}}{=}\,\,\,
\underbrace{-\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta}
_{\ds{\ =\ I}}\ =\
-\,{1 \over 2}
\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}\cos\pars{\theta}}\,\dd\theta
\label{1}\tag{1}
\\[5mm] & =
-\,{1 \over 4}\int_{0}^{\pi}\ln\pars{\sin\pars{\theta} \over 2}\,\dd\theta
\\[5mm] & =
{1 \over 4}\,\pi\ln\pars{2} -
{1 \over 4}\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta -
{1 \over 4}\int_{-\pi/2}^{0}\ln\pars{-\sin\pars{\theta}}\,\dd\theta
\\[5mm] & =
{1 \over 4}\,\pi\ln\pars{2}\
\underbrace{- {1 \over 2}\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta}
_{\ds{\ =\ {1 \over 2}\, I}}
\label{2}\tag{2}
\end{align}
See lines \ref{1} and \ref{2}:
$\ds{I = {1 \over 4}\,\pi\ln\pars{2} + {1 \over 2}\,I
\implies I = \bbox[15px,#ffd,border:1px solid navy]{{\pi \over 2}\,\ln\pars{2}}\ \approx 1.0888}$.
A: 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.
