How to evaluate the following integral $\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx$? How to evaluate the following integral
$$\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx$$
It seems that it evaluates to$$\frac{1}{4}-\frac{\pi^2}{48}$$
Is this true? How would I prove it?
 A: Alternatively, using the general trigonometric form of beta function from equation $(14)$ Wolfram MathWorld - Beta Function we have
$$\int_0^{\pi/2}\sin^nx\cos^mx\,dx=\frac{1}{2}\text{B}\left(\frac{n+1}{2},\frac{m+1}{2}\right)$$
Differentiating with respect to $m$ and $n$ once, then putting $m=1$ and $n=1$ we directly obtain the desired result
$$\begin{align}\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx&=\frac{\text{B}\left(1,1\right)}{8}\bigg[\left(\psi_0(1)-\psi_0(2)\right)^2-\psi_1(2)\bigg]\\&=\frac{1}{4}-\frac{\pi^2}{48}\end{align}$$
Here I use equation $(26)$ from Wolfram MathWorld - Beta Function and also equation $(8)$ & equation $(15)$ from Wolfram MathWorld - Polygamma Function.
A: Find this
$$I=\int_{0}^{\frac{\pi}{2}}\sin{x}\cos{x}\ln{(\cos{x})}\ln{(\sin{x})}dx$$
Solution
Since 
$$\sin(2x) = 2\sin(x)\cos(x)$$
then
$$I=\dfrac{1}{8}\int_{0}^{\frac{\pi}{2}}\ln{(\sin^2{x})}
\ln{(\cos^2{x})}\sin{(2x)}dx$$
Let $\cos{(2x)}=y$, and since 
$$\cos(2x) = 2\cos^2x - 1 = 1 - 2\sin^2x$$
we get
$$I=\dfrac{1}{16}\int_{-1}^{1}\ln{\left(\dfrac{1-y}{2}\right)}
\ln{\left(\dfrac{1+y}{2}\right)}dy$$
Let $\dfrac{1-y}{2}=z$, then we have
\begin{align*}I&=\dfrac{1}{8}\int_{0}^{1}\ln{z}\ln{(1-z)}dz=\dfrac{-1}{8}\sum_{n=1}^{\infty}\dfrac{1}{n}
\int_{0}^{1}z^n\ln{z}dz\\
&=\dfrac{1}{8}\sum_{n=1}^{\infty}
\dfrac{1}{n(n+1)^2}=\dfrac{1}{8}\sum_{n=1}^{\infty}
\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)-\dfrac{1}{8}\sum_{n=1}^{\infty}
\dfrac{1}{(n+1)^2}\\
&=\dfrac{1}{4}-\dfrac{\pi^2}{48}
\end{align*}
A: Hint:-
$\ln \cos x=z \implies \tan x\ dx=-dz$
$\therefore\displaystyle\int\sin{x}\cos{x}\ln{(\cos{x})}\ln{(\sin{x})}dx\\=\dfrac{1}{2}\displaystyle\int\sin{2x}\ln{(\cos{x})}\ln{(\sin{x})}dx\\=\displaystyle\int\left(\dfrac{\tan x}{\sec^2 x}\right)\ln{(\cos{x})}\ln{(\sin{x})}dx\\=-\displaystyle\int\left(\dfrac{z\ln{(\sqrt{1-e^{2z}})}}{e^{2z}}\right)dz$
$e^{-2z}=t \implies e^{-2z}dz=-\dfrac{1}{2}dt, e^{-2z}=t\implies{z}=-\dfrac{1}{2}\ln t $
$\therefore \dfrac{1}{4}\displaystyle\int{\ln t\ln{\left(e^z\sqrt{t-1}\right)}}dt\\=\dfrac{1}{8}\displaystyle\int{\ln t\ln{(t-1)}}dt+\dfrac{1}{4}\displaystyle\int{(\ln t)^2}dt$
$\displaystyle\int{\ln t\ln{(t-1)}}dt=\ln (t-1)\displaystyle\int\ln t\ dt-\color{green}{\displaystyle\int\left(\dfrac{t\ln t-t}{t-1}\right)\ dt}$
$\color{green}{\therefore\displaystyle\int\left(\dfrac{t\ln t-t}{t-1}\right)\ dt=\displaystyle\int\left(\dfrac{t\ln t-\ln t+\ln t-t}{t-1}\right)\ dt=\displaystyle\int \ln t \ dt+\color{red}{\displaystyle\int\dfrac{\ln t}{t-1}dt}-\displaystyle\int\dfrac{t}{t-1}dt}$
$\color{red}{t-1=y \implies dt = dy\implies\displaystyle\int\dfrac{\ln t}{t-1}dt=\displaystyle\int\dfrac{\ln (y+1)}{y}dy=-\operatorname{Li}_2(-y)+C}$
${\color{blue}{\displaystyle\int{(\ln t)^2}dt=t\left((\ln t)^2 -2\ln t+2\right)+C'}}$
