Quick evaluation of definite integral $\int_0^\pi \frac{\sin 5x}{\sin x}dx$ 
Find$$\int_0^{\pi}\frac{\sin 5x}{\sin x}dx$$

I can solve it by involving polynomials in sine and cosine as shown in the links below, but it’s huge (doing double angle formulas twice; I noticed that using polynomials in cosine is better because the integral spits out sines which are 0 between the limits) so I want a faster method, if it exists. Please don’t use contour integration:)
The only thing I noticed is that the integrand is symmetric about the midpoint in the given interval, i.e.$$\frac{\sin 5x}{\sin x}= \frac{\sin 5(\pi-x)}{\sin(\pi- x)}.$$
Determine the indefinite integral $\int \frac{\sin x}{\sin 5x}dx$
Integral of $\int \frac{\sin(3x)}{\sin(5x)} \, dx$
Expressing $\frac {\sin(5x)}{\sin(x)}$ in powers of $\cos(x)$ using complex numbers

We can generalized for positive integer k:
$\displaystyle \frac{\sin((2k-1)\,x)}{\sin(x)}
= 1 + 2\cos(2x) + 2\cos(4x) \;+\;...\;+\; 2\cos((2k-2)\,x)
$
Prove by induction:
Base case, $k=1$, is obviously true: $\displaystyle \frac{\sin x}{\sin x} = 1$
Assume formula is true for $n=k$, prove it is true for $n=k+1$
$\begin{align} \displaystyle 
\frac{\sin((2k-1)\,x)}{\sin(x)} + 2\cos((2k)\,x)
&=\frac{\sin((2k-1)\,x) + 2\cos((2k)\,x)×\sin(x)}{\sin(x)} \\
&=\frac{\sin((2k-1)\,x) + \sin((2k+1)\,x) - \sin((2k-1)\,x)}{\sin(x)} \\
&= \frac{\sin((2(k+1)-1)\,x)}{\sin(x)}
\end{align}$
QED
$\displaystyle ⇒ \int_0^\pi \frac{\sin((2k-1)\,x)}{\sin(x)}\;dx = \pi$

More direct way for proof, express as telescoping series.
$\sin((2k-1)\,x)$
$= \sin(x) + [\sin(3x)-\sin(x)] + [\sin(5x)-\sin(3x)] 
\;+\; ... \;+\; [\sin((2k-1)\,x)  - \sin((2k-3)\,x)]$
$= \sin(x) + 2\cos(2x)\sin(x) + 2\cos(4x)\sin(x)
\;+\; ... \;+\; 2\cos((2k-2)\,x)\sin(x)
$
Divide both side by $\sin(x)$, we have the required proof.

Proof using complex numbers, $z = e^{\,i\,x}$
$\begin{align} \displaystyle 
\frac{\sin((2k-1)\,x)}{\sin x} &= 
\frac{z^{2k-1}-1/z^{2k-1}}{z-1/z} \\
&= \left(\frac{z^{4k-2}-1}{z^2-1}\right)\bigg/z^{2k-2} \\
&= \frac{1+z^2+z^4 \;+\;...\;+\;z^{4k-4}}{z^{2k-2}}\\
&= 1+(z^2+1/z^2)+(z^4+1/z^4) \;+\;...\;+\;(z^{2k-2}+1/z^{2k-2}) \\
&= 1 + \;\;2\cos(2x)\;\, + \;\;2\cos(4x)\;\, \;+\;...\;+\;\;2\cos((2k-2)\,x)
\end{align}$
