$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left( #1 \right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{I \equiv
{1 \over \pi}\int_{0}^{\pi}\theta^{2}\ln^{2}\pars{2\cos\pars{\theta \over 2}}
\,\dd\theta = {11\pi^{4} \over 180} = {11 \over 2}\,\zeta\pars{4}}$
\begin{align}
I&={1 \over 2\pi}\int_{-\pi}^{\pi}\theta^{2}
\ln^{2}\pars{2\root{1 + \cos\pars{\theta} \over 2}}\,\dd\theta
\\[3mm]&={1 \over 2\pi}
\int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
\bracks{-\ln^{2}\pars{z}}\ln^{2}\pars{\root{2}\root{1 + {z^{2} + 1 \over 2z}}}\,
{\dd z \over \ic z}
\\[3mm]&={\ic \over 2\pi}
\int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
\ln^{2}\pars{z}\ln^{2}\pars{z + 1 \over z^{1/2}}\,{\dd z \over z}
\\[3mm]&={\ic \over 2\pi}\lim_{\mu \to -1 \atop \nu \to 0}
\partiald[2]{}{\mu}\partiald[2]{}{\nu}
\int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
z^{\mu}\pars{z + 1 \over z^{1/2}}^{\nu}\,\dd z
\\[3mm]&={\ic \over 2\pi}\lim_{\mu \to -1 \atop \nu \to 0}
\partiald[2]{}{\mu}\partiald[2]{}{\nu}
\int_{\verts{z} = 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}z^{\mu - \nu/2}\pars{z + 1}^{\nu}\,\dd z
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\pars{1}
\end{align}
The integration in $\pars{1}$ is given by:
\begin{align}
&\int_{\verts{z} = 1}
z^{\mu - \nu/2}\pars{z + 1}^{\nu}\,\dd z
\\[3mm]&=-\int_{-1}^{0}\pars{-x}^{\mu - \nu/2}\expo{\ic\pi\pars{\mu - \nu/2}}
\pars{x + 1}^{\nu}\,\dd x
-\int_{0}^{-1}\pars{-x}^{\mu - \nu/2}\expo{-\ic\pi\pars{\mu - \nu/2}}
\pars{x + 1}^{\nu}\,\dd x
\\[3mm]&=-\expo{\ic\pi\pars{\mu - \nu/2}}\int_{0}^{1}x^{\mu - \nu/2}
\pars{-x + 1}^{\nu}\,\dd x
+\expo{-\ic\pi\pars{\mu - \nu/2}}\int_{0}^{1}x^{\mu - \nu/2}
\pars{-x + 1}^{\nu}\,\dd x
\\[3mm]&=2\ic\sin\pars{\pi\bracks{{\nu \over 2} - \mu}}
{\rm B}\pars{\mu - {\nu \over 2} + 1,\nu + 1}\tag{2}
\end{align}
where $\ds{{\rm B}\pars{x,y} = \int_{0}^{1}t^{x - 1}\pars{1 - t}^{y - 1}\,\dd t}$,
$\ds{\pars{~\mbox{with}\ \Re\pars{x} > 0,\ \Re\pars{y} > 0~}}$ is the
Beta Function.
With $\pars{1}$ and $\pars{2}$, $\ds{I}$ is reduced to:
$$
I=-\,{1 \over \pi}\lim_{\mu \to -1 \atop \nu \to 0}
\partiald[2]{}{\mu}\partiald[2]{}{\nu}\bracks{%
\sin\pars{\pi\bracks{{\nu \over 2} - \mu}}
{\rm B}\pars{\mu - {\nu \over 2} + 1,\nu + 1}}
$$
Since $\ds{{\rm B}\pars{x,y}=
{\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$
( $\ds{\Gamma\pars{z}}$ is the
GammaFunction ):
\begin{align}
I&=-\,{1 \over \pi}\
\lim_{\mu \to -1 \atop \nu \to 0}
\partiald[2]{}{\mu}\partiald[2]{}{\nu}\bracks{%
\sin\pars{\pi\bracks{{\nu \over 2} - \mu}}
{\Gamma\pars{\mu - \nu/2 + 1}\Gamma\pars{\nu + 1}
\over \Gamma\pars{\mu + \nu/2 + 2}}}
\\[3mm]&=-\
\overbrace{\lim_{\mu \to -1 \atop \nu \to 0}
\partiald[2]{}{\mu}\partiald[2]{}{\nu}\bracks{%
{\Gamma\pars{\nu + 1}
\over \Gamma\pars{\nu/2 - \mu}\Gamma\pars{\mu + \nu/2 + 2}}}}
^{\ds{=\ -\,{11\pi^{4} \over 180}}}
\end{align}
where we used the identity
$\ds{\Gamma\pars{z}\Gamma\pars{1 - z} = {\pi \over \sin\pars{\pi z}}}$
Then,
$$
I \equiv
\color{#00f}{\large%
{1 \over \pi}\int_{0}^{\pi}\theta^{2}\ln^{2}\pars{2\cos\pars{\theta \over 2}} \,\dd\theta}
= \color{#00f}{\large{11\pi^{4} \over 180} = {11 \over 2}\,\zeta\pars{4}}
$$
Best Answer
From Table of Integrals, Series, and Products Seventh Edition by I.S. Gradshteyn and I.M. Ryzhik equation $3.631\ (9)$ we have
Proof
Integrating $(1+z)^p z^q$, for $p,q\ge0$, in the $z=u+iv$ plane around the contour bounded by the $u$-axis from $-1$ to $1$ and the upper semicircle of unit radius yields $$ \int_{-1}^1(1+z)^p z^q\ dz=-i\int_0^\pi\left(1+e^{it}\right)^p e^{i(q+1)t}\ dt, $$ since $(1+z)^p z^q$ is holomorphic within and continuous on and within the given contour. The imaginary part of the RHS is $$ \Im\left[-i\int_0^\pi\left(1+e^{it}\right)^p e^{i(q+1)t}\ dt\right]=-\int_0^\pi\left(2\cos\frac t2\right)^p \cos bt\ dt, $$ where $b=q+\frac12p+1$. The LHS integral is equal to $$ \int_{0}^1(1+u)^p u^q\ du+e^{i\pi q}\int_{0}^1(1-u)^p u^q\ du $$ of which the imaginary part is \begin{align} \operatorname{B}\left(p+1,q+1\right)\sin\pi q&=-\frac{\Gamma\left(p+1\right)\Gamma\left(q+1\right)}{\Gamma\left(p+q+2\right)}\sin\pi q\\ &=-\frac{\Gamma\left(p+1\right)\Gamma\left(b-\frac12p\right)}{\Gamma\left(b+\frac12p+1\right)}\sin\pi \left(b-\frac12p\right)\\ &=-\frac{\pi\Gamma\left(p+1\right)}{\Gamma\left(1+\frac12p+b\right)\Gamma\left(1+\frac12p-b\right)}. \end{align} Final step is setting $t=2x$, $p=n-1$, and $a=2b$. $\qquad\color{blue}{\mathbb{Q.E.D.}}$
Thus \begin{align} \int_0^{\Large\frac\pi2}\theta^2\ln^4(2\cos \theta)\ d\theta&=\lim_{n\to5}\lim_{a\to0}\frac{\partial^6}{\partial n^4\partial a^2}\left[\int_0^{\Large\frac\pi2}(2\cos \theta)^{n-1}\cos a\theta\ d\theta\right]\\ &=\lim_{n\to5}\lim_{a\to0}\frac{\partial^6}{\partial n^4\partial a^2}\left[\frac{\pi}{2 n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)}\right]\\ &=\large\color{blue}{\frac{33\pi^7}{4480}+\frac{3\pi}{2}\zeta^2(3)}. \end{align}