Evaluate $\int_0^{\frac{π}{2}}\tan (x)\ln (\sin x)\ln (\cos x)dx$ 
Evaluate: 
  $$I=\int_0^{\frac{π}{2}}\tan (x)\ln (\sin x)\ln (\cos x)dx$$

My ideas is to use the Fourier series of log sin and log cos: 
$$\ln (2\sin x)=-\sum_{k=1}^{\infty}\frac{\cos (2kx)}{k}$$
$$\ln (2\cos x)=-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos (2kx)}{k}$$ 
But my problem is that I find difficult integrals like: 
$$\int\tan (x)\cos (2kx)dx$$
My another idea is: 
Use the substation : $y=\tan x$ then $dx=\frac{dy}{1+y^2}$ 
Then where $x=0 \Rightarrow y=0$ and for $x=\frac{π}{2} \Rightarrow y=\infty$ 
So: 
$$I=\frac{1}{2}\int_0^{\infty}\frac{y\ln \left(\frac{y}{\sqrt{1+y^2}}\right)\ln (1+y^2)}{1+y^2}dy$$ 
But now I don't know how to complete.
 A: $I=\int_0^{\frac{π}{2}}\tan (x)\ln (\sin x)\ln (\cos x)dx$
$=-\int_0^{-\infty}t\ln (\sqrt{1-e^{2t}})dt$ 
(substituting $\ln(\cos{x}) = t$, so $\tan{x} dx = -dt$, and $\sin{x} = \sqrt{1-\cos^2{x}} = \sqrt{1-e^{2t}}$)
$=-\int_0^{\infty}u\ln (\sqrt{1-e^{-2u}})du$ (Substituting $t = -u$)
$=-\frac{1}{2} \int_0^{\infty}u\ln (1-e^{-2u})du$
$=-\frac{1}{2}\int_0^{\infty}u\sum_{n=1}^{\infty} (-e^{-2nu}/n)du$
$=\int_0^{\infty}\sum_{n=1}^{\infty} (\frac{(2nu)e^{-2nu}}{8n^3})(2ndu)$
$=\sum_{n=1}^{\infty} \frac{1}{8n^3}\int_0^{\infty} ke^{-k} dk$ (Subtituting $2nu = k$, and $\int_0^{\infty} ke^{-k} dk =1 $)
$=\frac{\zeta{(3)}}{8}$
A: $$\int_0^\frac{\pi}{2} \frac{\ln(\sin x)\ln(\cos x)}{\cos x}\sin xdx\overset{\cos x=t}=\int_0^1 \frac{\ln(\sqrt{1-t^2})\ln t}{t}dt\overset{t^2=x}=\frac18 \int_0^1 \frac{\ln(1- x)\ln x}{x}dx$$
$$=\frac18\int_0^1 \left(-\sum_{n=1}^\infty \frac{x^n}{n}\right)\frac{\ln x}{x}dx=-\frac18 \sum_{n=1}^\infty\frac{1}{n}\int_0^1 x^{n-1}\ln xdx=\frac18\sum_{n=1}^\infty \frac{1}{n^3}=\frac{\zeta(3)}{8}$$
A: We can actually use the Beta Function with this one. Here we will address your integral:
\begin{equation}
I = \int_0^\frac{\pi}{2} \tan(x) \ln(\sin(x))\ln(\cos(x))\:dx
\end{equation}
We note that:
\begin{equation}
\tan(x) = \sin(x)\cos^{-1}(x),\:\:\lim_{a \rightarrow 0^+} \frac{d}{da} \sin^a(x) = \ln(\sin(x)),\:\:\lim_{b \rightarrow 0^+} \frac{d}{db} \cos^b(x) = \ln(\cos(x))
\end{equation}
Thus $I$ becomes:
\begin{equation}
I = \int_0^\frac{\pi}{2} \sin(x)\cos^{-1}(x) \cdot \lim_{a \rightarrow 0^+} \frac{d}{da} \sin^a(x) \cdot \lim_{b \rightarrow 0^+} \frac{d}{db} \cos^b(x)\:dx
\end{equation}
By the Dominated Convergence Theorem and  Leibniz's Integral Rule:
\begin{align}
I &= \lim_{a \rightarrow 0^+} \frac{d}{da}  \lim_{b \rightarrow 0^+} \frac{d}{db}  \int_0^\frac{\pi}{2} \sin(x)\cos^{-1}(x) \sin^a(x) \cos^b(x)\:dx \nonumber \\ 
&= \lim_{(a,b)\rightarrow (0,0)^+} \frac{\partial^2}{\partial a \partial b} \int_0^\frac{\pi}{2} \sin^{a + 1}(x)\cos^{b - 1}(x)\:dx \nonumber \\
&= \lim_{(a,b)\rightarrow (0,0)^+} \frac{\partial^2}{\partial a \partial b} \left[\frac{1}{2}B\left(\frac{a + 2}{2}, \frac{b}{2} \right) \right]
\end{align}
Where $B(\cdot, \cdot)$ is the Beta Function
From here convert to the Gamma Representation, apply the derivatives and the evaluate at $(a,b) = (0,0)$ and you're done!

EDIT - This doesn't work. The Beta Function is undefined for $b = 0$. 
A: $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \tan x \ln (\sin x) \ln (\cos x) d x =&-\frac{1}{2} \int_0^{\frac{\pi}{2}} \ln (\sin x) d\left(\ln ^2(\cos x)\right)\\ \stackrel{IBP}{=}  & \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin x} \ln ^2(\cos x) d x\\ \stackrel{\cos^2x\mapsto x}{=}  & \frac{1}{16} \int_0^1 \frac{\ln ^2 x}{1-x} d x\\=&\frac{\zeta(3)}{8} 
\end{aligned}
$$
for the last integral, please read my post.
