Show that $\int_{[0,1]\times [0,1]}\frac{dx \, dy}{1-xy} = \frac{\pi^2}{6}$ $$
\int_{[0,1] \times [0,1]} \frac{dx \, dy}{1-xy}=\frac{\pi^2}{6},
$$
there is a hint for substitution given but I can't seem to get anywhere with it.
$$
\begin{cases}
x = \frac{u-v}{\sqrt{2}} \\
y= \frac{u+v}{\sqrt{2}}
\end{cases}
$$
and evaluate by letting $u = \sqrt{2}\sin t$. There should also encounter the term $(1-\sin t)/\cos t$ (Which I couldn't obtain)
I have managed to compute the jacobian of transformation, which is exactly 1, and did the substitution for u and v, obtaining:
$$
\int_{[0,\sqrt{2}]\times [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]} \frac{2}{2-u^2 +v^2} \,du\, dv
$$
doing the subtitution for $u$, we get:
$$
\int \frac{2\sqrt{2}\cos t}{2\cos^2 t+v^2}\,dt\, dv
$$
It is then not obvious how to integrate with respect to $t$ now, so change order of integration and integrate with respect to v.
which will be equal to $\int 4\arctan \frac{1}{2\cos t} \, dt$
Now this is a rather complicated integral, how to proceed from here? or is there an alternative method which I overlooked in the previous steps?
Could I have some insight on this problem please??
 A: As far as I understand you question, you would like to solve the integral via substitution. (otherwise did's comment to your question might provide a simpler solution)
You are right that the Jacobian is 1 however you did some mistakes with the integration bounds after the substitution. The correct solution is
$$\int_{[0,1] \times [0,1]} \frac{dx \, dy}{1-xy} = \underbrace{\int_0^{\sqrt{2}/2}\!du
\int_{-u}^u\!dv \, \frac{2}{2-u^2+v^2}}_{I_1}+ \int_{\sqrt{2}/2}^\sqrt{2}\!du
\int_{-\sqrt{2}+u}^{\sqrt{2}-u}\!dv \, \frac{2}{2-u^2+v^2}
$$
which you obtain by rewriting the conditions $0\leq x,y \leq 1$ in terms of $u$ and $v$.
Now, I show you how to integrate the first of the two integrals, $I_1$ (the other is very similar). Calculating the integral over $u$, we obtain
$$ I_1= \int_{0}^{\sqrt{2}/2}\!du\, \frac{4\arctan(u/\sqrt{2-u^2})}{\sqrt{2-u^2}}.$$
Substituting $x=\arctan(u/\sqrt{2-u^2})$ yields
$$ I_1 = 4 \int_0^{\pi/6} dx\,x= \frac{\pi^2}{18}.  $$
The last step is equivalent to the maybe more intuitive substitution $u=\sqrt{2} \sin t$ as indicated in your post.
A: $$\int_0^1 \int_0^1 \frac{1}{1-xy} \, dy \, dx = \int_0^1 \frac{\ln(1-x)}{-x} \, dx$$
since $\ln(1-x)=-\sum_{n=1}^\infty \frac{x^n}{n}$, then $$\int_0^1 \frac{\ln(1-x)}{-x} \, dy \, dx=\int_0^1 \sum_{n=1}^\infty \frac{x^{n-1}}{n} \, dx = \sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}}{n} \, dx = \sum_{n=1}^\infty\frac{1}{n^2}=\cdots.$$
A: Instead of doing the $u$ substitution before evaluating the integral in $v$, it's easier to do it after computing that integral. Here is a detailed computation.
(I've  assumed that you want to evaluate this integral and not simply to show that it is equal to $\zeta(2)$. This exercise appears in several references:  Tom Apostol's A Proof that Euler Missed: Evaluating $\zeta(2)$ the Easy Way, Martin Aigner and Günter Ziegler's Proofs from The BOOK and as an exercise in a number theory text by LeVeque .)
By the substitution $x=\frac{\sqrt{2}}{2}\left( u-v\right) ,y=\frac{\sqrt{2}}{2}\left( u+v\right) $, whose Jacobian $J=\frac{\partial (x,y)}{\partial
(u,v)}=1$, the region of integration becomes the blue square in the $u,v$-plane with vertices
$$(0,0),\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right),(\sqrt{2},0),\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right), $$ as shown in the following figure.

Observing that
$$
\frac{1}{1-xy}=\frac{2}{2-u^2+v^2}
$$
is symmetric in $v$, we get
$$\begin{eqnarray*}
I &=&\int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy & x=\frac{\sqrt{2}}{2} (u-v) ,\quad y=\frac{\sqrt{2}}{2} (u+v)  \\
&=&2\int_{u=0}^{\sqrt{2}/2}\int_{v=0}^{u} \frac{2}{2-u^{2}+v^{2}}\times
1\,du\,dv \\
&&{}+2\int_{u=\sqrt{2}/2}^{\sqrt{2}}\int_{v=0}^{\sqrt{2}-u}\frac{2}{
2-u^2+v^2}\times 1\,du\,dv \\
&=&4\int_0^{\sqrt{2}/2}\left( \int_{0}^{u}\frac{dv}{2-u^{2}+v^{2}}\right)
\,du\, \\
&&{}+4\int_{\sqrt{2}/2}^{\sqrt{2}}\left( \int_{0}^{\sqrt{2}-u}\frac{dv}{
2-u^{2}+v^{2}}\right) \,du \\
&=&4\int_{0}^{\sqrt{2}/2}\frac{1}{\sqrt{2-u^{2}}}\arctan \frac{u}{\sqrt{
2-u^{2}}}\,du, & u=\sqrt{2}\sin t \\
&&{}+4\int_{\sqrt{2}/2}^{\sqrt{2}}\frac{1}{\sqrt{2-u^{2}}}\arctan \frac{\sqrt{2
}-u}{\sqrt{2-u^{2}}}\,du\, & u=\sqrt{2}\cos \theta \\
&=&4\int_{0}^{\pi /6}\arctan(\tan t)\, dt\\&&{}+4\int_0^{\pi
/3}\arctan \left( \frac{1-\cos \theta}{\sin \theta}\right) d\theta, \\
&=&4\int_{0}^{\pi /6}t\,dt+4\int_0^{\pi /3}\arctan \left( \tan \frac{\theta}{2} \right) \, d\theta \\
&=&\frac{\pi^2}{18}+\frac{\pi^2}{9}=\frac{\pi^2}{6}.
\end{eqnarray*}
$$
One of the substitutions is a new one.  After the first pair of substitutions $x,y$, we have done two additional ones in the resulting integrals, as indicated above: in the 1st, $u=\sqrt{2}\sin t,du=\sqrt{2}\cos t\,dt$, and in the 2nd, $u=\sqrt{2}\cos \theta,du=-\sqrt{2}\sin \theta\,d\theta$.
