Evaluating $\int_0^{\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx$ I need to solve
$$
\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx
$$
I tried to use symmetric properties of the trigonometric functions as is commonly used to compute
$$
\int_0^{\Large\frac\pi2}\ln\sin x\ dx = -\frac{\pi}{2}\ln2
$$
but never succeeded. (see this for example)
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$\ds{\int_{0}^{\pi/2}{\ln\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}}
     \over \tan\pars{x}}\,\dd x:\ {\large ?}}$

\begin{align}
&\int_{0}^{\pi/2}{\ln\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}}\over \tan\pars{x}}\,\dd x
\,\,\,\stackrel{x\ =\ \arcsin\pars{t}}{=}\,\,\,
\int_{0}^{1}{\ln\pars{t}\ln\pars{\root{1 - t^{2}}} \over t\,/\root{1 - t^{2}}}\,
{\dd t \over \root{1 - t^{2}}}
\\[5mm]= &\
\half\int_{0}^{1}{\ln\pars{t}\ln\pars{1 - t^{2}} \over t}\,\dd t =
\half\int_{0}^{1}{\ln\pars{t^{1/2}}\ln\pars{1 - t}\over
t^{1/2}}\,\half\,t^{-1/2}\,\dd t
\\[5mm]
= &\
{1 \over 8}\int_{0}^{1}{\ln\pars{t}\ln\pars{1 - t} \over t}\,\dd t
=-\,{1 \over 8}\int_{0}^{1}{{\rm Li}_{1}\pars{t} \over t}\,\ln\pars{t}\,\dd t
=-\,{1 \over 8}\int_{0}^{1}{\rm Li}_{2}'\pars{t}\ln\pars{t}\,\dd t
\\[5mm]&={1 \over 8}\int_{0}^{1}{{\rm Li}_{2}\pars{t} \over t}\,\dd t
={1 \over 8}\int_{0}^{1}{\rm Li}_{3}'\pars{t}\,\dd t
={1 \over 8}\,{\rm Li}_{3}\pars{1} =
\bbox[10px,border:1px dotted navy]{\ds{\zeta\pars{3}}}
\approx 0.1503
\end{align}

$\ds{{\rm Li}_{\rm s}\pars{z}}$ is a
  PolyLogarithm Function: I  used some properties of them as reported in the cited link.
  $\ds{\zeta\pars{s}}$ is the Riemann Zeta Function.

A: Let's start out with the substitution $ \displaystyle \ln(\sin x) = u $ and get:
$$\ln(\cos x)=\frac{\ln(1-e^{2u})}{2}$$
$$\displaystyle\frac{1}{\tan x} \ dx =du$$
that further yields
$$\int_0^{\pi/2}\frac{(\ln{\sin x})(\ln{\cos x})}{\tan x}dx= \frac{1}{2} \int_{-\infty}^{0}  \ln(1-e^{2u}) u \ du$$
According to Taylor expansion we have 
$$\ln(1-e^{2u})= \sum_{k=1}^{\infty} (-1)^{2k+1} \frac{e^{2 k u}}{k}$$
then
$$\frac{1}{2} \int_{-\infty}^{0}  \ln(1-e^{2u}) u \ du=$$
$$\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^{2k+1}}{k} \int_{-\infty}^{0} u e^{2ku} \ du =$$
$$\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^{2k+1}}{k} \frac{-1}{4k^2} = \frac{1}{8} \sum_{k=1}^{\infty} \frac{1}{k^3}=\frac{1}{8} \zeta(3).$$
Remark: the value of $\zeta(3)\approx1.2020569$ is called Apéry's Constant - see here.
Q.E.D. (Chris)
A: Let $u = \sin^2(x)$. Then $\frac{\mathrm{d}x}{\tan(x)} = \frac{\cos(x)}{\sin(x)} \mathrm{d}x = \frac{d\sin(x)}{\sin(x)} = \frac{1}{2}\frac{\mathrm{d}u}{u}$, $\ln(\sin(x)) = \frac{1}{2}\ln(u)$ and $\ln(\cos(x)) = \frac{1}{2} \ln(1-u)$:
$$
  \int_0^{\pi/2} \frac{\ln(\sin(x)) \ln(\cos(x))}{\tan(x)} \mathrm{d} x = \frac{1}{8} \int_0^{1} \frac{\ln u}{u} \cdot \ln(1-u) \mathrm{d} u = \left.\frac{1}{8} \frac{\mathrm{d}^2}{\mathrm{d} s \mathrm{d} t} \int_0^1 u^{s-1} (1-u)^{t-1} \mathrm{d}u \right|_{s\to 0^+,t=1} = \left.\frac{1}{8} \frac{\mathrm{d}^2}{\mathrm{d} s \mathrm{d} t} \frac{\Gamma(s) \Gamma(t)}{\Gamma(s+t)} \right|_{s\to 0^+,t=1} 
$$
First differentiate with respect to $t$ and substitute $t=1$:
$$
 \left.\frac{1}{8} \frac{\mathrm{d}}{\mathrm{d} s} \frac{\Gamma(s)}{\Gamma(s+1)}\left( \psi(1) - \psi(s+1)\right)  \right|_{s\to 0^+} = \left.\frac{1}{8} \frac{\mathrm{d}}{\mathrm{d} s} \frac{\left( \psi(1) - \psi(s+1)\right) }{s} \right|_{s\to 0^+}
$$
Using Taylor series expansion for the digamma function $\psi(s)$ we have:
$$
   \frac{\left( \psi(1) - \psi(s+1)\right) }{s} = -\zeta(2) +  \zeta(3) s + \mathcal{o}(s)
$$
Hence the value of the integral is:
$$
  \int_0^{\pi/2} \frac{\ln(\sin(x)) \ln(\cos(x))}{\tan(x)} \mathrm{d} x = \frac{1}{8} \zeta(3)
$$

Alternatively you could use $$\frac{\ln(1-u)}{u} = -\sum_{k=0}^\infty \frac{u^k}{k+1}$$ and integrate term-wise: $$\int_0^1 u^k \ln(u) \mathrm{d} u \stackrel{u=\exp(-t)}{=} \int_0^\infty t \exp(-t(k+1)) \mathrm{d} t =  -\frac{1}{(k+1)^2}$$ which yields the result immediately.
A: Rewrite the integral as
$$
\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx=\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{\sqrt{1-\sin^2 x}}}{\sin x}\cdot\cos x\ dx.
$$
Set $t=\sin x\ \color{red}{\Rightarrow}\ dt=\cos x\ dx$, then we obtain
\begin{align}
\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx&=\frac12\int_0^1\frac{\ln t\ \ln(1-t^2)}{t}\ dt\\
&=-\frac12\int_0^1\ln t\sum_{n=1}^\infty\frac{t^{2n}}{nt}\ dt\tag1\\
&=-\frac12\sum_{n=1}^\infty\frac{1}{n}\int_0^1t^{2n-1}\ln t\ dt\\
&=\frac12\sum_{n=1}^\infty\frac{1}{n}\cdot\frac{1}{(2n)^2}\tag2\\
&=\large\color{blue}{\frac{\zeta(3)}{8}}.
\end{align}

Notes :
$[1]\ \ $Use Maclaurin series for natural logarithm: $\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\ $ for $|x|<1$.
$[2]\ \ $$\displaystyle\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}\ $ for $  n=0,1,2,\cdots$
A: $$\begin{align*}
I &= \int_0^{\frac\pi2} \frac{\log(\cos(x)) \log(\sin(x))}{\tan(x)} \, dx \\[1ex]
&= \int_0^\infty \frac{\log\left(\frac1{\sqrt{x^2+1}}\right) \log\left(\frac x{\sqrt{x^2+1}}\right)}{x} \, \frac{dx}{1+x^2} \tag{1} \\[1ex]
&= \frac14 \int_0^\infty \frac{\log^2(x^2+1)-\log(x^2)\log(x^2+1)}{x} \, \frac{dx}{1+x^2} \\[1ex]
&= \frac18 \int_0^\infty \frac{\log^2(x+1)-\log(x)\log(x+1)}{x(1+x)} \, dx \tag{2} \\[1ex]
&= \frac18 \int_0^1 \left(\frac{\log^2(x+1)}{x} - \frac{\log(x)\log(x+1)}x\right) \, dx \tag{3} \\[1ex]
&= \frac14 \int_0^1 \frac{\log(x)\log(x+1)}x \, dx - \frac12 \int_0^1 \frac{\log(x+1)\log(x)}{x+1} \, dx \tag{4} \\[1ex]
&= -\frac14 \sum_{n=1}^\infty \frac{(-1)^n}n \int_0^1 x^{n-1} \log(x) \, dx + \frac12 \sum_{n=1}^\infty (-1)^n H_n \int_0^1 x^n \log(x) \, dx \tag{5} \\[1ex]
&= -\frac14 \sum_{n=1}^\infty \frac{(-1)^n}{n^3} + \frac12 \sum_{n=1}^\infty \frac{(-1)^n H_n}{(n+1)^2} \tag{4} \\[1ex]
&= \boxed{\frac{\zeta(3)}8} \tag{6}
\end{align*}$$


*

*$(1)$ : substitute $x\mapsto\arctan(x)$

*$(2)$ : substitute $x\mapsto\sqrt x$

*$(3)$ : split up the integral at $x=1$; substitute $x\mapsto\frac1x$ in the integral over $[1,\infty)$; partial fractions

*$(4)$ : integrate by parts

*$(5)$ : recall the power series $-\log(1-x)=\sum\limits_{n\ge1}\frac{x^n}n$ and $-\frac{\log(1-x)}{1-x}=\sum\limits_{n\ge1} H_n x^n$, where $H_n$ is the $n^{\rm th}$ harmonic number

*$(6)$ : see e.g. section $7$ for a contour-integration method to evaluating the Euler sum; I'm sure there are more efficient means

