How to evaluate $\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$ Find the value of
$$I=\displaystyle\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$$
We have the information that
$$J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{\pi^2}{8}\ln^2(2)-\dfrac{\pi^4}{192}$$
 A: This is not quite a complete answer but goes a good way towards showing that the idea of @kalpeshmpopat is not so far off the mark - if we want to answer the question that was orginally asked.
First, numerical investigation indicates that the correct integral is
$$I=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)dx=\dfrac{(\pi\ln{2})^2}{8}-\dfrac{\pi^4}{192}.$$
Now, as @kalpeshmpopat points out, a simple substitution, together with the facts that $\cos(\frac{\pi}{2}-x)=\sin(x)$ and vice-versa, shows that
$$\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)dx=\int_0^{\pi/2}(\frac{\pi}{2}-x)\ln(\sin x)\ln(\cos x)dx.$$
Thus, if we add these two together we get
$$\displaystyle\int_0^{\pi/2} \frac{\pi}{2} \ln(\sin x)\ln(\cos x)dx=2I.$$
All that remains to show is that
$$\displaystyle\int_0^{\pi/2} \frac{\pi}{2} \ln(\sin x)\ln(\cos x)dx =
\frac{1}{96} \pi ^2 \left(6 \log ^2(4)-\pi^2\right),$$
which Mathematica can do.  It's getting late, but my guess on this last integral would be to expand $\ln(\cos(x))$ into a power series (which is easy, since we know $\ln(1+y)$) and try to integrate $x^n \ln(\sin(x))$.
A: I'm still struggling with this integral, but I guess the following result may have a chance to be helpful:
\begin{align*}
\int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx
&= \frac{11 \pi^5}{1440} + \frac{\pi^3}{24} \log^2 2 + \frac{\pi}{2}\zeta(3) \log 2 \tag{1} \\
&\approx 4.2671523609840988652 \cdots.
\end{align*}
To prove this, let us consider the following identity
$$ \int_{0}^{\frac{\pi}{2}} \cos^{z}x \cos wx \, dx = \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}}.$$
You can find the proof of this identity at here. Thus it follows that
$$ \int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx = - \left. \frac{\partial^4}{\partial z^2 \partial w^2} \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}} \right|_{(z, w) = (0, 0)}. $$
Performing a bunch of calculations, we obtain $(1)$. Similar idea shows that
$$ \int_{0}^{\frac{\pi}{2}} \log^2 \cos x \, dx = \left. \frac{\partial^2}{\partial z^2} \frac{\pi}{2^{z+1}} \binom{z}{\frac{z+w}{2}} \right|_{(z, w) = (0, 0)} = \frac{\pi^3}{24} + \frac{\pi}{2}\log 2. \tag{2} $$

Indeed, starting from the identity
$$ \log^2 \left( \frac{\sin 2x}{2} \right) = \log^2 \cos x + \log^2 \sin x + 2\log \cos x \log \sin x, $$
I obtained
\begin{align*}I
&= -\frac{7}{8}\int_{0}^{\frac{\pi}{2}} x^2 \log^2 \cos x \, dx
   + \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} x \log^2 \cos x \, dx
   - \frac{3\pi^2}{32} \int_{0}^{\frac{\pi}{2}} \log^2 \cos x \, dx \\
&\quad -\frac{\log 2}{8}\int_{0}^{\pi} x^2 \log \sin x \, dx
   + \frac{\pi^3}{48} \log^2 2 \\
&\approx 0.077821979372293864338\cdots.
\end{align*}
From the identity
$$ \log \sin x = -\log 2 - \sum_{n=1}^{\infty} \frac{\cos 2nx}{n}, $$
we obtain
$$\int_{0}^{\pi} x^2 \log \sin x \, dx = -\frac{\pi}{2} \zeta (3) - \frac{\pi^3}{3} \log 2.  \tag{3}$$
Putting $(1)$, $(2)$ and $(3)$ together, I was able to derive
\begin{align*}I
&= -\frac{61 \pi^5}{5760} - \frac{3\pi}{8} \zeta (3) \log 2 -\frac{\pi^3}{48} \log^2 2
   + \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} x \log^2 \cos x \, dx.
\end{align*}
I'm not sure if this formula will be helpful, since the last remaining integral seems to defy my techniques.
A: This is yet another partial answer, and a verification of some other claims.
Using $(4)$ and $(8)$ from this answer, we get
$$
\int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x=\frac\pi2\log(2)^2-\frac{\pi^3}{48}\tag{1}
$$
Here is a way to extend kalpeshmpopat's suggestion about substituting $x\mapsto\frac\pi2-x$. Note that $g(x)=f(\sin(x))f(\cos(x))$ is even as a function of $x-\frac\pi4$; that is, $g(\frac\pi2-x)=g(x)$. Thus, if we multiply by an odd function of $x-\frac\pi4$, the integral over $[0,\frac\pi2]$ will be $0$. 
Therefore,
$$
\int_0^{\pi/2}\left(\frac\pi4-x\right)\log(\sin(x))\log(\cos(x))\,\mathrm{d}x=0\tag{2}
$$
Using $(1)$ and $(2)$, we get
$$
\begin{align}
\int_0^{\pi/2}x\log(\sin(x))\log(\cos(x))\,\mathrm{d}x
&=\frac\pi4\int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&=\frac\pi4\left(\frac\pi2\log(2)^2-\frac{\pi^3}{48}\right)\\
&=\frac{\pi^2}{8}\log(2)^2-\frac{\pi^4}{192}\tag{3}
\end{align}
$$
We also have
$$
\int_0^{\pi/2}\left(\frac\pi4-x\right)^3\log(\sin(x))\log(\cos(x))\,\mathrm{d}x=0\tag{4}
$$
Which, along with $(1)$ and $(3)$, implies that
$$
\begin{align}
\int_0^{\pi/2}x^3\log(\sin(x))\log(\cos(x))\,\mathrm{d}x
&=\frac{3\pi}{4}\int_0^{\pi/2}x^2\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&-\frac{3\pi^2}{16}\int_0^{\pi/2}x\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&+\frac{\pi^3}{64}\int_0^{\pi/2}\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&=\frac{3\pi}{4}\int_0^{\pi/2}x^2\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\
&-\frac{\pi^4}{64}\log(2)^2+\frac{\pi^6}{1536}\tag{5}
\end{align}
$$
Equation $(5)$ supports math110's claim that if we know $I_2$, we know $I_3$.
A: Related problem: (I), (II). The contribution of this post is to evaluate the integral

$$ I = \int_0^{\pi/2}\ln(\sin x)\ln(\cos x)dx  $$

symbolically. Now to find $I$, we use the first change of variables $ t = \sin(x) $ which results in

$$ I = \frac{1}{2}\int_{0}^{1}\frac{\ln(t)\ln(1-t^2)}{\sqrt{1-t^2}}dt. $$

Following it by the change of variables $u=t^2$ gives

$$ I = \frac{1}{8}\int_{0}^{1}\frac{ \ln(u) \ln(1-u) }{ \sqrt{u} \sqrt{1-u} } du .$$

To evaluate the last integral, we consider the integral

$$ F = \frac{1}{8}\int_{0}^{1}u^{a-\frac{1}{2}} (1-u)^{b-\frac{1}{2}} du = \beta(a+1/2,b+1/2) ,$$

where $\beta(u,v)$ is the beta function.

$$ \implies I = D_{b}\,D_{a} \beta(a+1/2,b+1/2)|_{a=0,b=0}= \frac{\pi}{48} \, \left( 24\, \left( \ln  \left( 2 \right)\right)^{2} -{\pi }^{2} \right),$$

where $D_a=\frac{\partial }{\partial a}$ and $D_b=\frac{\partial }{\partial b}$.
A: Hint: Replace x by π/2- x
then simplify so we will get one term same as I
A: Presented below is a complete solution that evaluates the integral to the following close-form
\begin{align}
&\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ dx\\
=& -\frac\pi2 \text{Li}_4(\frac12)-\frac{3\pi}8\ln2\ \zeta(3)
+\frac{\pi^5}{320}+\frac{\pi^3}{16}\ln^22-\frac\pi{48}\ln^42\tag1
\end{align}
To derive (1), evaluate the integrals below sequentially, with $\int_0^{\pi/2}\ln^2(2\sin x)dx=\frac{\pi^3}{24}$ and $\int_0^{\pi/2}x^2 \ln^2(2\cos x) dx=\frac{11\pi^5}{1440}$
\begin{align}
&\int_0^{\pi/2}x^2
\overset{x\to\frac\pi2-x}{\ln^2(2\sin x)}\ dx 
 =\pi\int_0^{\pi/2}x
\ln^2(2\sin x)\ dx-\frac{\pi^5}{360}\\
\\
&\int_0^{\pi/2}x^2 
\ln^2(2\sin 2x)\overset{2x\to x}{dx} \\
 =& \ \frac18\int_0^{\pi/2}(\pi^2-2\pi x+2x^2 )
\ln^2(2\sin x)\ dx=\frac{13\pi^5}{2880}\\
\\
&\int_0^{\pi/2}x^2\ln(2\sin x)\ln(2\cos x)\ dx\\
=& \ \frac12 \int_0^{\pi/2}[x^2 
\ln^2(2\sin 2x)-x^2 \ln^2(2\sin x)-x^2 \ln^2(2\cos x)]\ dx\\
=& - \frac{\pi^5}{5760}-\frac\pi2 \int_0^{\pi/2}x
\ln^2(2\sin x)\ dx\
\end{align}
Writing out both sides of the last integral and utilizing below to yield the close-form (1)
\begin{align}
\int_0^{\pi/2}x\ln(2\sin x)dx=\frac7{16}\zeta(3),\>\>\>
\int_0^{\pi/2}x^2\ln(2\sin x)dx=\frac{3\pi}{16}\zeta(3)\\
\end{align}
$$\int_0^{\pi/2}x\ln^2(\sin x)dx 
=  \text{Li}_4(\frac12)
-\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42
$$
