compute the following integral in closed form : $\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2}(x)+8\cos^{2} x}dx$ Evaluate 
$I=\int_0^{\frac{π}{2}}\frac{x}{(1+\sqrt 2)\sin^{2} (x)+8\cos^{2} x}dx$
How can I starte in this hard integral , at first use $y=\frac{π}{2}$ but no result so I this use : 
$y=\tan \frac{x}{2}$ then $dx=2\frac{1}{1+y^2}dy$ 
$x=2\arctan y$ 
$\cos x=\frac{1-y^2}{1+y^2}$ $&$ $\sin x=2\frac{y}{1+y^2}$ 
So : 
$8\cos^{2} x+(1+\sqrt 2)sin^{2} x=\frac{8(1-y^2)^2+4(1+\sqrt 2)y^2}{(1+y^2)^2}$
Now I get $arctan$ integral 
$I=2\int_0^{\infty}\frac{(1+y^2)\arctan y}{8(1-y^2)^2+4(1+\sqrt 2)y^2}dy$ 
But I don't know how to complete this work! 
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\pi/2}\frac{x}{\pars{1 + \root{2}}
\sin^{2}\pars{x} + 8\cos^{2}\pars{x}}\,\dd x:\ {\LARGE ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\pi/2}{x \over
\pars{1 + \root{2}}
\sin^{2}\pars{x} + 8\cos^{2}\pars{x}}\,\dd x}
\\[5mm] & =
\int_{0}^{\pi/2}{x \over
\pars{1 + \root{2}}
\bracks{1 - \cos\pars{2x}}/2 + 8\bracks{1 + \cos\pars{2x}}/2}\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{\pi/2}{2x \over
9 + \root{2}
 + \pars{7 - \root{2}} \cos\pars{2x}}\,2\,\dd x
\\[5mm] & =
{7 + \root{2} \over 94}
\int_{0}^{\pi}{x \over
a + \cos\pars{x}}\,\dd x
\end{align}
where $\ds{a \equiv {65 + 16\root{2} \over 47} > 1}$.

Then,
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\pi}{x \over
a + \cos\pars{x}}\,\dd x}
\\[5mm] = &\
\left.\Re\int_{x\ =\ 0}^{x\ =\ \pi}{-\ic\ln\pars{z} \over
a + \pars{z + 1/z}/2}\,{\dd z \over \ic z}
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left. -2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi}{\ln\pars{z} \over
z^{2} + 2az + 1}\,\dd z
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\tag{1}\label{1}
\end{align}
Roots of $\ds{z^{2} + 2az + 1 = 0}$ are given by
$\ds{r_{\pm} \equiv -a \pm \root{a^{2} - 1}}$ where $\ds{r_{-} < -1}$ and $\ds{-1 < r_{+} < 0}$.


\eqref{1} can be handle with the
  "Polylogarithm Machinery".

