<s>[INST] interesting square of log sin integral I ran across this challenging log sin integral and am wondering what may be a good approach.
$$
\int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}(2\cos(x))dx=\frac{11{{\pi}^{5}}}{1440}
$$
This looks like it may be able to be connected to the digamma or incomplete beta function somehow. 
I tried using the identity 
$$
\cos(x)=\frac{e^{ix}+e^{-ix}}{2},
$$ 
but that square threw a wrench in my plans.
Maybe write it as 
$$
(x\ln(2\cos(x)))^{2}=\ln^{2}(2)x^{2}+2\ln(2)x^{2}\ln(\cos(x))+x^{2}\ln^{2}(\cos(x))
$$ and expand. I doubt if that does any good though.
Does anyone know of a clever way to approach this?  It looks like a fun one if I knew a good starting point.  
Is there an identity that goes with 
$$
(2\cos(t))^{a}.
$$
If 
$$
t^{2}(2\cos(t))^{a}
$$
were differentiated twice w.r.t a, then we would have
$$
(2\cos(t))^{a}t^{2}\ln^{2}(2\cos(t))
$$
Maybe the trig form of Beta which would give 
$$
2^{a}\cdot B(1/2, \frac{a+1}{2}).
$$
Then, differentiate this twice and it would somehow relate to the digamma?
$$
\frac{1}{2}B(1/2,\frac{a+1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{a+1}{2})}{2\Gamma(\frac{a+2}{2})}.
$$
Differentiate twice and obtain some sort of Psi relation.
Cheers Everyone.
 [/INST] As I remarked in a comment, I have spent a while trying to derive this identity by means of contour integration. Though I have not succeeded at that, (I take great pleasure in striking that out; see my other answer below!) I have found some related identities in my search which surprised me quite a bit, and I think they're worth sharing. For example, using the integral you asked about, I have shown that
$$
\begin{equation}
\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{5}{4}\cdot \frac{\pi^4}{90} = \frac{5}{4} \zeta(4). \tag{1}
\end{equation}
$$
Here
$$
H_n = \sum_{k = 1}^n \frac{1}{k}
$$
is the $n$th harmonic number. Here are a couple more examples. I will use the first to derive $(1)$.


*

*With the method from this answer, one can show that
$$
\int_0^{\pi/2} x^2\log^2(2\cos{x})\,dx = \frac{1}{5}\left(\frac{\pi}{2}\right)^5 + \pi \int_0^\infty y \log^2(1-e^{-2y})\,dy,
$$
and it then follows from the result of your question that
$$
\pi \int_0^\infty y \log^2(1- e^{-2y})\,dy = \frac{11}{45}\left(\frac{\pi}{2}\right)^5 - \frac{1}{5}\left(\frac{\pi}{2}\right)^5 = \frac{\pi}{8} \cdot \frac{\pi^4}{90}.
$$

*I also shared this example in the other answer, but for completeness, let me mention that
$$
\int_0^{\pi/2} x^2\log^2(2\cos{x})\,dx = \frac{1}{30}\left(\frac{\pi}{2}\right)^5 +  \frac{1}{6}\int_0^{\pi/2} \log^4(2\cos{x})\,dx,
$$
hence
$$
\int_0^{\pi/2} \log^4(2\cos{x})\,dx = \frac{19}{15}\left(\frac{\pi}{2}\right)^5.
$$


More details about the derivations of 1. and 2. are contained in the answer I referenced above (near the bottom).

Let me now turn to deriving $(1)$. Taking $x = - e^{-2y}$ in the series expansion
$$
\log{(1+x)} = \sum_{n = 1}^\infty \frac{(-1)^{n+1}x^n}{n}
$$
and inserting the result into the integral evaluated in 1. gives
$$
\begin{align}
\frac{1}{8}\cdot\frac{\pi^4}{90} &= \int_0^\infty y \log^2(1 - e^{-2y})\,dy \\
& = \int_0^\infty y \left(\sum_{n = 1}^\infty \frac{e^{-2ny}}{n}\right)^2\,dy \\
& = \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{1}{nm} \int_0^\infty y e^{-2(n + m)y}\,dy \\
& = \frac{1}{4}\sum_{m = 1}^\infty\sum_{n = 1}^\infty \frac{1}{nm(n+m)^2}. \tag{2}
\end{align}
$$
Now put $r = m+n$ and $s=n$ in order to write
$$
\begin{align}
\sum_{m = 1}^\infty\sum_{n = 1}^\infty \frac{1}{nm(n+m)^2} & = \sum_{r = 2}^\infty \frac{1}{r^2}\sum_{s = 1}^{r-1} \frac{1}{s(r-s)} = 2 \sum_{r=2}^\infty \frac{1}{r^3} \sum_{s=1}^{r-1}\frac{1}{s}, \tag{3}
\end{align}
$$
with the last equation a consequence of the identity $\frac{1}{s(r-s)} = \frac{1}{r}\left(\frac{1}{s}+ \frac{1}{r-s}\right)$. Insert $(3)$ into $(2)$ and multiply both sides by $2$ to get
$$
\frac{1}{4} \cdot \frac{\pi^4}{90} = \sum_{r=2}^\infty \frac{1}{r^3} \sum_{s=1}^{r-1}\frac{1}{s} = \sum_{r=2}^\infty \frac{H_{r-1}}{r^3}.
$$
Since $H_r - H_{r-1} = 1/r$, the identity $(1)$ now follows from the equations
$$
\frac{\pi^4}{90} = \sum_{r=1}^\infty \frac{1}{r^4} = \sum_{r = 1}^\infty \frac{H_r}{r^3} - \sum_{r=2}^\infty \frac{H_{r-1}}{r^3} = \sum_{r = 1}^\infty \frac{H_r}{r^3} - \frac{1}{4}\cdot\frac{\pi^4}{90}.
$$
</s>