If $I_n=\int_{-\pi}^{\pi} \frac{\sin nx}{(1+\pi^x)(\sin x)}dx$, then prove the following 
$I_n=I_{n+2}$


$ \sum_{m=1}^{10} I_{2m+1} =10 \pi$


$\sum _{m=1}^{10} I_2m =0$

My simplification of the given integral is
$$I_n =\int_0^{\pi} \frac{\sin nx}{\sin x} dx$$
Where $n$ is an integer
I believe the expression needs further simplification before the above statements can be proven, but I am unable to achieve that. How should I proceed?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\left.\vphantom{\Large A}I_{n}\,
\right\vert_{\,n\ \in\ \mathbb{N}_{\,\geq 1}}\ & \equiv
\bbox[5px,#ffd]{\int_{0}^{\pi}{\sin\pars{nx} \over \sin\pars{x}}\dd x} =
\int_{-\pi/2}^{\pi/2}{\sin\pars{nx + n\pi/2} \over \cos\pars{x}}\dd x
\\[5mm] & =
\int_{0}^{\pi/2}{\sin\pars{nx + n\pi/2} + \sin\pars{-nx + n\pi/2} \over \cos\pars{x}}\dd x
\\[5mm] & =
2\sin\pars{n\pi \over 2}\
\underbrace{\int_{0}^{\pi/2}{\cos\pars{nx} \over \cos\pars{x}}\dd x}
_{\ds{\cal J_{n}}}
\\ & \underline{\mbox{Note that}\ \color{red}{I_{n} = 0}\ \mbox{whenever}\ n\
\mbox{is}\ \color{red}{even}}
\\[2mm] & \mbox{and}\
\left.2\sin\pars{n\pi \over 2}\,\right\vert_{\,n\ \color{red}{odd}} = 2\pars{-1}^{\pars{n - 1}/2}.
\label{1}\tag{1}
\end{align}

With $\ds{n \equiv 2k + 1}$ where $\ds{k \in \mathbb{N}_{\,\geq\ 0}\,\,}$:
\begin{align}
\left.\vphantom{\Large A}{\cal J}_{n}
\,\right\vert_{\ds{\,n\ odd}} & =
{\cal J}_{2k + 1}\
\equiv
\bbox[5px,#ffd]{\int_{0}^{\pi/2}
{\cos\pars{\bracks{2k + 1}x} \over
\cos\pars{x}}\,\dd x}
\\[5mm] & =
\Re\int_{0}^{\pi/2}{\expo{\ic\pars{2k + 1}x} - \expo{\ic\pars{2k + 1}\pi/2} \over \cos\pars{x}}
\,\,\dd x
\\[5mm] & =
\left.\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}\
{z^{2k + 1} - \pars{-1}^{k}\,\ic \over
\pars{z + 1/z}/2}\,{\dd z \over \ic z}
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] & =
\left.-2\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\
{\pars{-1}^{k}\,\ic - z^{2k + 1} \over
1 + z^{2}}\,\dd z
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] & =
2\ \underbrace{\Im\int_{1}^{0}\
{\pars{-1}^{k}\,\ic - \ic^{2k + 1}\,\,y^{2k + 1} \over
1 - y^{2}}\,\ic\,\dd y}
_{\ds{= \color{red}{0}\ \mbox{because the integrand}\ \in\ \mathbb{R}}}
\\[2mm] &\
+ 2\,\Im\int_{0}^{1}
{\pars{-1}^{k}\,\ic - x^{2k + 1} \over
1 + x^{2}}\,\dd x
\\[5mm] = &\
2\pars{-1}^{k}\int_{0}^{1}
{\dd x \over 1 + x^{2}} 
\bbx{{\pi \over 2}\pars{-1}^{k}} \\ &
\end{align}

Finally, see (\ref{1}),
$$
\mbox{}\\
\left.\vphantom{\Large A}I_{n}\,
\right\vert_{\,n\ \in\ \color{red}{\mathbb{Z}}}\ =
\bbx{\left\{\begin{array}{lcl}
\ds{-I_{-n}} & \mbox{if} & \ds{n \leq 0}
\\[2mm]
\ds{0} & \mbox{if} & \ds{n > 0\ \mbox{is}\ even}
\\[2mm]
\ds{\pi} & \mbox{if} & \ds{n > 0\ \mbox{is}\ odd}
\end{array}\right.} \\
$$

The three statements at the beginning of the OP question are $\underline{correct}$ but the first one when $\ds{n = -1}$.
