Compute $\int_0^{\pi/2}\frac{\sin 2013x }{\sin x} \ dx\space$ How would you approach
$$\int_0^{\pi/2}\frac{\sin 2013x }{\sin x} \ dx\space?$$
The way I see here involves Dirichlet kernel. I wonder what else can we do, maybe some easy/elementary approaching ways. Thanks !
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\pi/2}{\sin\pars{2013 x} \over \sin\pars{x}}\,\dd x:\ {\large ?}}$

We'll consider $\ds{\pars{~\mbox{with}\ \color{#c00000}{\large n=\ {\tt 2013}}~}}$:
  \begin{align}
&\color{#c00000}{\int_{0}^{\pi/2}{\sin\pars{nx} \over \sin\pars{x}}\,\dd x}=
\half\int_{-\pi/2}^{\pi/2}{\sin\pars{nx} \over \sin\pars{x}}\,\dd x
=
\half\int_{0}^{\pi}{-\cos\pars{nx} \over -\cos\pars{x}}\,\dd x
=
{1 \over 4}\int_{-\pi}^{\pi}{\cos\pars{nx} \over \cos\pars{x}}\,\dd x
\\[3mm]&={1 \over 4}\Re
\oint_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
{z^{n} \over \pars{z^{2} + 1}/\pars{2z}}\,{\dd z \over \ic z}
\end{align}

$$
\color{#c00000}{\int_{0}^{\pi/2}{\sin\pars{nx} \over \sin\pars{x}}\,\dd x}
=\half\Im
\oint_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}\verts{{\rm Arg}\pars{z}}\ <\ \pi}}
{z^{n} \over z^{2} + 1}\,\dd z\tag{1}
$$

The only contribution to the integration arises from two 'small' semicircles around
  $\ds{\pm\ic}$:
  \begin{align}
&\color{#c00000}{\int_{0}^{\pi/2}{\sin\pars{nx} \over \sin\pars{x}}\,\dd x}
\\[3mm]&=\half\Im\lim_{\epsilon \to 0^{+}}\bracks{%
-\int_{2\pi}^{\pi}{\pars{\ic + \epsilon\expo{\ic\theta}}^{n} \over
\pars{\ic + \epsilon\expo{\ic\theta}}^{2} + 1}\,
\epsilon\expo{\ic\theta}\ic\,\dd\theta
-\int_{\pi}^{0}{\pars{-\ic + \epsilon\expo{\ic\theta}}^{n} \over
\pars{-\ic + \epsilon\expo{\ic\theta}}^{2} + 1}\,
\epsilon\expo{\ic\theta}\ic\,\dd\theta}
\\[3mm]&=\half\pars{{\pi \over 2} + {\pi \over 2}}.\qquad\qquad
\mbox{Note that}\quad \pars{\pm \ic}^{2013} = \pm\ic.
\end{align}

$$
\color{#00f}{\Large\int_{0}^{\pi/2}{\sin\pars{2013 x} \over \sin\pars{x}}\,\dd x
={\pi \over 2}}
$$
A: Let $I=\displaystyle\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx$
As $\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx,$
$\displaystyle I=\int_0^{\frac\pi2} \frac{\sin (2n+1)(\frac\pi2-x)}{\sin (\frac\pi2-x)} dx$
$\displaystyle =\int_0^{\frac\pi2} \frac{\sin \{n\pi+\frac\pi2-(2n+1)x\}}{\cos x} dx$
$\displaystyle =\int_0^{\frac\pi2} \frac{\cos (2n+1)x}{\cos x} dx$ if $n$ is even.
$\displaystyle =-\int_0^{\frac\pi2} \frac{\cos (2n+1)x}{\cos x} dx$ if $n$ is odd.
If  $n$ is odd, 
$\displaystyle 2I=\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx-\int_0^{\frac\pi2} \frac{\cos (2n+1)x}{\cos x} dx$
$\displaystyle =\int_0^{\frac\pi2} \frac{\sin (2n)x}{\sin x\cos x} dx$
$\displaystyle =2\int_0^{\frac\pi2} \frac{\sin (2n)x}{\sin 2x} dx$
$\displaystyle =2\frac12\int_0^{\pi} \frac{\sin ny}{\sin y} dy$
$\displaystyle =2\int_0^{\frac{\pi}2} \frac{\sin ny}{\sin y} dy$ as $\displaystyle\frac{\sin ny}{\sin y}$ is an even function.
So, $\displaystyle\int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx=I=\int_0^{\frac{\pi}2} \frac{\sin nx}{\sin x} dx$ if $n$ is odd.
Similarly, $\displaystyle \int_0^{\frac\pi2} \frac{\sin (2n+1)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin (n+1)x}{\sin x} dx$ if $n$ is even.
If we put, $2n+1=2013, n=1006$ which is even.
S0,  $\displaystyle \int_0^{\frac\pi2} \frac{\sin (2013)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin (1007)x}{\sin x} dx$
Now, if we put $2n+1=1007,n=503$ which is odd.
So, $\displaystyle \int_0^{\frac{\pi}2} \frac{\sin (1007)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin (503)x}{\sin x} dx$
Now, if $2n+1=503,n=251$
The reduction of $n$ will follow :$2013,1007,503,251,125,63,31,15,7,3,1$
So,$\displaystyle \int_0^{\frac\pi2} \frac{\sin (2013)x}{\sin x} dx=\int_0^{\frac{\pi}2} \frac{\sin x}{\sin x} dx=\frac{\pi}2$
A: $\sin(m+2)x-\sin mx=2\sin x\cos(m+1)x$
So, $\displaystyle\frac{\sin(m+2)x}{\sin x}-\frac{\sin mx}{\sin x}=2\cos (m+1)x$
$\displaystyle\int_0^{\frac{\pi}2}\frac{\sin(m+2)x}{\sin x}dx-\int_0^{\frac{\pi}2}\frac{\sin mx}{\sin x}dx=2\int_0^{\frac{\pi}2}\cos (m+1)xdx=\frac2{m+1}\sin(m+1)\frac{\pi}2$
Putting $m=2n-1$
$\displaystyle\int_0^{\frac{\pi}2}\frac{\sin(2n+1)x}{\sin x}dx-\int_0^{\frac{\pi}2}\frac{\sin (2n-1)x}{\sin x}dx=\frac2{m+1}\sin(2n-1+1)\frac{\pi}2=0$
So, $$\int_0^{\frac{\pi}2}\frac{\sin(2n+1)x}{\sin x}dx=\int_0^{\frac{\pi}2}\frac{\sin (2n-1)x}{\sin x}dx$$ and so on up to $\displaystyle\int_0^{\frac{\pi}2}\frac{\sin x}{\sin x}dx=\frac{\pi}2$
