Integral $\int_0^{\pi/2} \theta^2 \log ^4(2\cos \theta) d\theta =\frac{33\pi^7}{4480}+\frac{3\pi}{2}\zeta^2(3)$ $$
I=\int_0^{\pi/2}  \theta^2 \log ^4(2\cos \theta) d\theta =\frac{33\pi^7}{4480}+\frac{3\pi}{2}\zeta^2(3).
$$
Note $\zeta(3)$ is given by
$$
\zeta(3)=\sum_{n=1}^\infty \frac{1}{n^3}.
$$
I have a previous post related to this except the logarithm power is squared and not to the 4th power.  If you are interested in seeing this result go here: Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$..  However, I am wondering how to calculate the result shown above.  Thanks.

From Table of Integrals, Series, and Products Seventh Edition by I.S. Gradshteyn and I.M. Ryzhik equation $3.631\ (9)$ we have

$$
\int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)}
$$

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}
