Catalan's constant and $\int_{0}^{2 \pi} \int_{0}^{2 \pi} \ln(\cos^{2} \theta + \cos^{2} \phi) ~d \theta~ d \phi$ According to my book (The Nature of computation, page 691):
$$\int_{0}^{2 \pi} \int_{0}^{2 \pi} \ln(\cos^{2} \theta + \cos^{2} \phi) ~d \theta ~d \phi= 16 \pi^2  \left(\frac{C}{\pi}- \frac{\ln2}{2}\right),$$
where $C$ is Catalan's constant. I have tried to derive this expression by looking at integral representations of $C$, but I have not been able to perform the integral. Any help? 
Thank you.
 A: $I=\int_{0}^{2 \pi} \int_{0}^{2 \pi} ln(\frac{cos2\theta+1}{2} + \frac{cos2\phi+1}{2}) d \theta d \phi $,
At first $0<\theta+x<2\pi$
$(\int_{0}^{\frac \pi4}\frac{cos}2  - \int_{\frac \pi4}^{\frac \pi2}\frac{cos}2-\int_{\frac \pi2}^{\frac {3\pi}4}\frac{sin}2+\int_{\frac {3\pi}4}^{ {\pi}}\frac{sin}2)*2$
$J=\int_{0}^{2 \pi}ln(\frac{cos2\theta+1+x}2)$
$=\int_{0}^{2 \pi}ln(\frac{sin2\theta+1+x}2)$
$=\int_{0}^{2 \pi}ln2+\int_{0}^{2 \pi}ln(sin{\theta+1+x})+\int_{0}^{2 \pi}lncos{\theta+1+x}$
$=4{\pi}ln2+2J$
$J=-4{\pi}ln2$
as same
$I=\int_{0}^{2 \pi}ln(\frac{cos2\theta+1+\phi}2)  d \phi $
$=-8{\pi}ln2$
A: Let $\cos^2\phi=\frac{(1-r)^2}{4r}$, or
${\sqrt r}= \sqrt{1+\cos^2\phi}-\cos \phi$. Then, substitute $x=2\theta$
\begin{align}
I=& 
\int_{0}^{2 \pi} \int_{0}^{2 \pi} \ln(\cos^{2} \theta + \cos^{2} \phi) ~d \theta~ d \phi\\
 =& \ 8\int_{0}^{\pi/2} \int_{0}^{\pi}  \ln\frac{1+2r\cos x+ r^2}{4r}~dx \ d\phi
\end{align}
Note that
$\ln r= -2\sinh^{-1}\cos \phi$ and
\begin{align} 
&\int_0^\pi \ln( 1 + 2r\cos x + r^2) dx
 =-2\sum_{k\ge1}\frac{(-r)^k}k  \int_0^\pi \cos kx \ dx =0
\end{align}
Then, with $\int_{0}^{\pi/2}
\sinh^{-1}\cos \phi \ d\phi=G$
\begin{align}
I=8\pi\int_{0}^{\pi/2} \ln \frac1{4r}d\phi=16\pi \int_{0}^{\pi/2}
(\sinh^{-1}\cos \phi - \ln2)d\phi=16\pi G-8\pi^2\ln2
\end{align}
