interesting set of integrals When integrating functions $\dfrac {\sin x}{\sin x}$, as well as $\dfrac {\sin3x}{\sin x}$ , $\dfrac {\sin 5x}{\sin x}$, $\dfrac {\sin 7x}{\sin x}$ etc. on interval from $0$ to $\dfrac {\pi}{2}$, the answer turns out to be $\dfrac {\pi}{2}$. How can this be proven?
 A: Hint:
$$1+2 \cos{2 x} + 2 \cos{4 x} + \ldots+2 \cos{2 n x} = \frac{\sin{(2 n+1) x}}{\sin{x}} $$
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}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\pi/2}{%
\sin\pars{\bracks{2n + 1}x} \over \sin\pars{x}}
\,\dd x\,\right\vert_{\,n\ \in\ \mathbb{N}_{\,\geq\ 0}}} \\[5mm] = &\
\Im\int_{0}^{\pi/2}{\expo{\ic\pars{2n + 1}x}\,\,\, -\ 1 \over \sin\pars{x}}\,\dd x
\\[5mm] = &\
\left.\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\,\,
{z^{2n + 1} -1 \over \pars{z - 1/z}/\pars{2\ic}}
\,{\dd z \over \ic z}\,\right\vert_{\,z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left.2\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}\,\,
{1 - z^{2n + 1} \over 1 - z^{2}}
\,\dd z\,\right\vert_{\,z\ =\ \exp\pars{\ic x}}
\end{align}
Hereafter, I'll "close" the integration around a quarter circle in the complex plane first quadrant: Thetre isn't any pole inside such contour.
Namely,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{0}^{\pi/2}{%
\sin\pars{\bracks{2n + 1}x} \over \sin\pars{x}}
\,\dd x\,\right\vert_{\,n\ \in\ \mathbb{N}_{\,\geq\ 0}}}
\\[5mm] = &\
-2\,\Im\int_{1}^{0}\,\,
{1 - \ic^{2n + 1}\,\,y^{2n + 1}\,\, \over 1 - \ic^{2}\,y^{2}\,}
\,\ic\,\dd y\ -\
\pars{\substack{\mbox{An integral, along the}\ \ds{x}\mbox{-axis}
\\[1mm]
\mbox{segment}\ \ds{\pars{0,1}},\ \mbox{which}
\\[1mm] \ds{vanishes\ out}\ \mbox{because}
\\[1mm]\mbox{the integrand}\ \ds{\in \mathbb{R}.}}}
\\[5mm] = &\
2\int_{0}^{1}{\dd y \over 1 + y^{2}} = \bbx{\pi \over 2} \\ &
\end{align}
A: I recommend you to prove by induction with all odd integers since $$\int \frac{\sin3x}{\sin x}dx$$
will depend on $$\int \frac{\sin x}{\sin x}dx$$
using integration by parts.
A: $\sin((k+2)x)-\sin kx=2\cos (k+1)x \sin x$ Use this to reduce the problem.
