Evaluating by real methods $\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx$ $\def\Li{{\rm{Li}}}$I'm sure you guys can briefly get the result by some methods of complex analysis, but now
I'm only interested in real analysis methods of proving the result. What would you propose
for that?
\begin{align*}
\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx=&\,\frac{\pi^6 \sqrt{2}}{768}+\frac{5 \sqrt{2}\pi^4}{64}\Li_2\left(2\sqrt2-3\right)-\frac{15\sqrt{2}\pi^2}{16}\Li_4\left(2\sqrt2-3\right)\\
&+\,\frac{15\sqrt{2}}{8}\bigg[\Li_6\left(2\sqrt2-3\right)-\Li_6\left(3-2\sqrt2\right)\bigg]
\end{align*}
And a supplementary question  for another version, that is  
$$\int_0^{\pi/2} \frac{x^5}{1+\cos^2(x)}\ dx$$
again, by real analysis methods only.

Using double angle formula, the integrand can be rewritten as
\begin{equation}
I=\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx=\int_0^{\pi/2} \frac{2x^5}{3-\cos(2x)}\ dx
\end{equation}
Mapping the variable $2x\mapsto x$, we have
\begin{equation}
I=\frac{1}{32}\int_0^{\pi} \frac{x^5}{3-\cos x}\ dx
\end{equation}
Using identity (proof can be seen here)
\begin{equation}
1+2\sum_{n=1}^\infty \left(\frac{b}{a}\right)^n\cos(n x)=\frac{a^2-b^2}{a^2+b^2-2ab\cos x}\qquad,\qquad\mbox{for}\, |b|<a
\end{equation}
and the correspondence values $a=\dfrac{2+\sqrt{2}}{2}$ and $b=\dfrac{2-\sqrt{2}}{2}$, one may find
\begin{equation}
1+2\sum_{n=1}^\infty \left(3-2\sqrt{2}\right)^n\cos(n x)=\frac{2\sqrt{2}}{3-\cos x}
\end{equation}
Therefore
\begin{align}
I&=\frac{1}{64\sqrt{2}}\int_0^{\pi} \left[x^5+2\sum_{n=1}^\infty \left(3-2\sqrt{2}\right)^n x^5\cos(n x)\right]\ dx\\
&=\frac{1}{64\sqrt{2}} \left[\frac{\pi^6}{6}+2\sum_{n=1}^\infty \left(3-2\sqrt{2}\right)^n \int_0^{\pi} x^5 \cos(n x)\ dx\right]\\
\end{align}
The rest part can be done by multiple times integration by parts and using $\sin(n\pi)=0$ for $n\in\mathbb{Z}$. We will obtain
\begin{equation}
I=\frac{\pi^6 \sqrt{2}}{768}+\frac{\sqrt{2}}{64}\sum_{n=1}^\infty \left(3-2\sqrt{2}\right)^n \left[\frac{5\pi^4\cos(\pi n)}{n^2}-\frac{60\pi^2\cos(\pi n)}{n^4}+\frac{120\cos(\pi n)}{n^6}-\frac{120}{n^6}\right]\
\end{equation}
Using
\begin{equation}
\cos(n\pi)=\begin{cases}\,\,+1&,\,\,\mbox{if}\,\, n\,\,\mbox{is even}\\[12pt]
\,\,-1&,\,\,\mbox{if}\,\, n\,\,\mbox{is odd}\\
\end{cases}
\end{equation}
and the representation of polylogarithm function in term of its infinite series, we finally obtain
\begin{align*}
\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx=&\,\frac{\pi^6 \sqrt{2}}{768}+\frac{5 \sqrt{2}\pi^4}{64}\text{Li}_2\left(2\sqrt2-3\right)-\frac{15\sqrt{2}\pi^2}{16}\text{Li}_4\left(2\sqrt2-3\right)\\
&+\,\frac{15\sqrt{2}}{8}\bigg[\text{Li}_6\left(2\sqrt2-3\right)-\text{Li}_6\left(3-2\sqrt2\right)\bigg]
\end{align*}
The same approach can be applied for evaluating the second integral.
