$\int_0^\frac{\pi}{2}\frac{\ln(\sin(x))\ln(\cos(x))}{\tan(x)}dx$ I have the problem below:
$$\int_0^\frac{\pi}{2}\frac{\ln(\sin(x))\ln(\cos(x))}{\tan(x)}dx$$
I have tried $u=\ln(\sin(x))$ so $dx=\tan(x)du$
so the integral becomes:
$$\int_{-\infty}^0u\ln(\cos(x))du$$
but I cannot find a simple way of getting rid of this $\ln(\cos(x))$
I also tried using the substitution $v=x-\frac{\pi}{2}$
so the integral becomes:
$$\int_{-\frac{\pi}{2}}^0\frac{\ln(\cos(v))\ln(\cos(v+\frac{\pi}{2}))}{\tan(v+\frac{\pi}{2})}dv$$
but this does not seems to lead anywhere useful
EDIT to follow up
$$B(\alpha,\beta)=\int_0^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)dx$$
$$\frac{\partial^2}{\partial_\alpha\partial_\beta}B(\alpha,\beta)=\int_0^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)\ln(\sin(x))\ln(\cos(x))dx$$
so I see that when $\alpha\to1$ and $\beta\to-1$ that this is the form that we want, so do we now have to take partial derivates of the non-integral form of the beta function then take the double integral for our chosen values?
 A: You may consider that
$$ f(\alpha,\beta) = \int_{0}^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)\,dx = \frac{\Gamma\left(\frac{a}{2}\right)\Gamma\left(1+\frac{b}{2}\right)}{2\,\Gamma\left(1+\frac{a+b}{2}\right)}$$
by Euler's Beta function. By applying $\frac{\partial^2}{\partial\alpha\,\partial\beta}$ to both sides, then considering the limit as $\beta\to 0$ and $\alpha\to 0^+$, we have
$$ \int_{0}^{\pi/2}\frac{\log\sin(x)\log\cos(x)}{\tan(x)}\,dx = -\frac{1}{12}\psi''(1) = \color{red}{\frac{\zeta(3)}{8}}.$$
By enforcing the substitutions $x\mapsto\arctan(x)$ or $x\mapsto 2\arctan(x)$ in the original integral we get interesting identities for nasty (poly)logarithmic integrals.
A: \begin{equation}
 \int_0^\frac{\pi}{2}\frac{\ln(\sin(x))\ln(\cos(x))}{\tan(x)}dx
 =
 \int_0^\frac{\pi}{2}
 \cos x
 \frac{\ln(\sin(x))\ln(\sqrt{1 - \sin^2 x})}{\sin(x)}dx
\end{equation}
Take
\begin{equation}
 u = \sin x
\end{equation}
so
\begin{equation}
 du = \cos x dx
\end{equation}
You get
\begin{equation}
 \int_0^1 \ln u \ln(\sqrt{1- u^2}) \frac{1}{u} du
\end{equation}
Take $u = e^v$, you get
\begin{equation}
 \frac{1}{2}
 \int_{-\infty}^0
 v \ln(1 - e^{2v}) dv
\end{equation}
Integration by parts will give you 
\begin{equation}
 \frac{1}{2}
 [\frac{1}{2}v^2 \ln(1 - e^{2v})]_{-\infty}^0
 -
 \frac{1}{2}
 \int_{-\infty}^0
 \frac{v^2}{1- e^{2v}}
 (-2e^{2v}) dv
\end{equation}
The term $ [\frac{1}{2}v^2 \ln(1 - e^{2v})]_{-\infty}^0$ is zero so
\begin{equation}
 -
 \frac{1}{2}
 \int_{-\infty}^0
 \frac{v^2}{1- e^{2v}}
 (-2e^{2v}) dv
\end{equation}
Take the following change of variable
\begin{equation}
 k = -v
\end{equation}
So we get
\begin{equation}
 \frac{1}{2}
 \int_0^{\infty}
 \frac{k^2}{e^{2k} - 1} dK
\end{equation}
Take now 
\begin{equation}
 n = 2k 
\end{equation}
as change of variable, you get
\begin{equation}
 \frac{1}{2}
 \int_0^{\infty}
 \frac{k^2}{e^{2k} - 1} dk
 =
 \frac{1}{16}
 \int_0^{\infty}
 \frac{n^2}{e^n - 1}
\end{equation}
Using the Riemann Zeta function
\begin{equation}
 \zeta(x)
 =
 \frac{1}{\Gamma(x)}
 \int_0^{\infty}
 \frac{u^{x-1}}{e^u - 1}
 du
\end{equation}
or
\begin{equation}
 \zeta(x)\Gamma(x)
 =
 \int_0^{\infty}
 \frac{u^{x-1}}{e^u - 1}
 du
\end{equation}
Using the above integral for $u = n$ and $x = 3$ you will get that your integral evaluates to 
\begin{equation}
 \frac{1}{2}
 \int_0^{\infty}
 \frac{k^2}{e^{2k} - 1} dk
 =
 \frac{1}{16}
 \int_0^{\infty}
 \frac{n^2}{e^n - 1}
 =
 \frac{\overbrace{\Gamma(3)}^{2!}\zeta(3)}{16}
 =
 \frac{\zeta(3)}{8}
\end{equation}
