Show that $\int_{0}^{\pi/4}\int_{0}^{\pi/4}\left(\sec(x+y)+\sec(x-y)\right)\mathrm{d}x\mathrm{d}y=2G$ Show that
$$
\int_{0}^{\pi/4}\int_{0}^{\pi/4}
\left[\vphantom{\large A}\sec\left(x + y\right) +
\sec\left(x - y\right)\right]\,\mathrm{d}x\mathrm{d}y =
2G
$$
where $G$ is Catalan's constant.
I am not sure where to start or how to begin.
 A: Essentially we only need to find the rather easy inner integral (namely the one depending on $x$). Starting to do so we get
$$\small\begin{align*}
\int\sec(x+y)+\sec(x-y)\mathrm dx=&-\log\left(\cos\left(\frac{x-y}2\right)-\sin\left(\frac{x-y}2\right)\right)+\log\left(\cos\left(\frac{x-y}2\right)+\sin\left(\frac{x-y}2\right)\right)\\&-\log\left(\cos\left(\frac{x+y}2\right)-\sin\left(\frac{x+y}2\right)\right)+\log\left(\cos\left(\frac{x+y}2\right)+\sin\left(\frac{x+y}2\right)\right)
\end{align*}$$
Plugging in the values for $x$, $0$ and $\frac\pi4$, we obtain after some messy algebra (including the substitutions $\frac y2-\frac\pi8\mapsto y$ and $\frac y2+\frac\pi8\mapsto y$) that the integrals equals
$$\int_0^\frac\pi4\int_0^\frac\pi4\sec(x+y)+\sec(x-y)\mathrm dx\mathrm dy=2\int_0^\frac\pi4\log\left(\frac{1+\tan y}{1-\tan y}\right)\mathrm dy$$
Now enforcing $\tan y\mapsto y$ followed up by $\frac{1-y}{1+y}\mapsto y$ we obtain
$$2\int_0^\frac\pi4\log\left(\frac{1+\tan y}{1-\tan y}\right)\mathrm dy=-2\int_0^1\frac{\log\left(\frac{1-y}{1+y}\right)}{1+y^2}\mathrm dy=-2\int_0^1\frac{\log(y)}{1+y^2}\mathrm dx$$
The latter one is a standard integral well-known to equal the negative of Catalan's Constant $-\mathrm G$. 

$$\therefore~\int_0^\frac\pi4\int_0^\frac\pi4\sec(x+y)+\sec(x-y)\mathrm dx\mathrm dy~=~2\mathrm G$$

