How to calculate $I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$? How do I integrate this guy? I've been stuck on this for hours..
$$I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$$

Let $x = \sin^2y$.
(It seems we've started like @Sasha here and like @sos440.)
Then 
$$\begin{eqnarray*}
I &=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x(1-x)} \\
    &=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x}
    + \underbrace{\frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{1-x}}_{x\to 1-x}
    \qquad (\textrm{partial fractions}) \\
    &=& \frac{1}{8} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x}
    \qquad (\textrm{integral linked above}) \\
    &=& \frac{1}{8} \int_0^1 dx\, \frac{\log(x)}{x} \left(-\sum_{k=1}^\infty \frac{x^k}{k}\right)
    \qquad (\textrm{Taylor expansion for }\log(1-x) ) \\
    &=& -\frac{1}{8} \sum_{k=0}^\infty \frac{1}{k+1} \underbrace{\int_0^1 dx\, x^k \log x}_{-1/(k+1)^2} 
    \qquad (\textrm{standard integral involving log}) \\
    &=& \frac{1}{8} \sum_{k=0}^\infty \frac{1}{(k+1)^3} \\
    &=& \frac{\zeta(3)}{8}. 
\end{eqnarray*}$$
Addendum: 
Note that $\csc(y)\sec(y) = \cot y + \tan y$. 
Then 
$$\begin{eqnarray*}
I &=& \frac{1}{2} \int_{0}^{\pi/2} dy\,
    \frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y} \\
    &=& \frac{1}{2} \int_{0}^{\pi/2} dy\,
    \frac{\ln(\sin y)\ln(\cos y)}{\tan y}
    + \underbrace{\frac{1}{2} \int_{0}^{\pi/2} dy\,
    \frac{\ln(\sin y)\ln(\cos y)}{\cot y}}_{y\to \pi/2-y} \\
    &=& \int_{0}^{\pi/2} dy\,
    \frac{\ln(\sin y)\ln(\cos y)}{\tan y}.
\end{eqnarray*}$$
This is exactly the integral linked above. 
