calculus - 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?

Answer

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*}