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
 A: Here's a fast and convenient method that doesn't rely on remembering any trig formulas and can be easily used to calculate similar integrals.
It involves complex numbers, but don't worry, I won't use any contour integration.
Using $\sin \alpha = [e^{i \alpha}-e^{-i\alpha}]/(2i)$, I rewrite your original integral as
$$
\int_0^\pi \frac{\sin 5x}{\sin x} \, dx = \int_0^\pi \frac{e^{5ix}-e^{-5ix}}{e^{ix}-e^{-ix}} \, dx = 
\int_0^\pi \frac{e^{4ix}-e^{-6ix}}{1-e^{-2ix}} \, dx.
$$
Now I'll use the formula $(1-y)^{-1}=1 + y + y^2 + y^3 + y^4 + ... $, also known as the sum of an infinite geometric series, to rewrite the integral as
$$
\int_0^\pi \left[e^{4ix}-e^{-6ix}\right](1+e^{-2ix}+e^{-4ix} + e^{-6ix} + ...)  \, dx.
$$
We've got an infinite sum inside the round brackets, but the thing is that when we open the brackets, we get integrals of the form
$\int_0^\pi e^{i k x}\, dx$, where $k$ is an even integer, and all such integrals are zero except for $k=0$. The $k=0$ term comes from multiplying $e^{4ix}$, present inside the square brackets, by $e^{-4ix}$, present inside the round brackets. The integral of this term is $\int_0^\pi dx = \pi$.
Hence the original integral is $\pi$.

UPDATE: Here's why $\int_0^\pi e^{i k x}\, dx=0$ for any non-zero even integer $k$:
$$
\int e^{ikx} \,d x = \frac{e^{ikx}}{ik} + const,
$$
$$
\int_0^\pi e^{ikx} \,d x = \left. \frac{e^{ikx}}{ik} \right|_{x=2 \pi}-\left. \frac{e^{ikx}}{ik} \right|_{x=0} = \frac{1}{ik} - \frac{1}{ik} = 0.
$$
A shorter method to prove the same thing is to use $e^{ikx}= \cos (ikx) + i \sin (ikx)$ and observe that $\int_0^\pi \cos (ikx) \, dx=0$ and $\int_0^\pi \sin (ikx) \,dx=0$ for any non-zero even integer $k$.
A: 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, $z = e^{\,i\,x}$
$\begin{align} \displaystyle 
\frac{\sin((2k-1)\,x)}{\sin x} &= 
\frac{z^{2k-1}-1/z^{2k-1}}{z-1/z} \\
&= \left(\frac{z^{4k-2}-1}{z^2-1}\right)\bigg/z^{2k-2} \\
&= \frac{1+z^2+z^4 \;+\;...\;+\;z^{4k-4}}{z^{2k-2}}\\
&= 1+(z^2+1/z^2)+(z^4+1/z^4) \;+\;...\;+\;(z^{2k-2}+1/z^{2k-2}) \\
&= 1 + \;\;2\cos(2x)\;\, + \;\;2\cos(4x)\;\, \;+\;...\;+\;\;2\cos((2k-2)\,x)
\end{align}$
A: Update: By using
$$ 2\sin(A)\cos(B)=\sin(A+B)+\sin(A-B) $$
twice, one has
\begin{eqnarray}
\int_0^\pi\frac{\sin(5x)}{\sin(x)}dx&=&\int_0^\pi\frac{2\sin(5x)\cos(x)}{2\sin (x)\cos(x)}dx\\
&=&\int_0^\pi\frac{2\sin(6x)+\sin(4x)}{\sin(2x)}dx\\
&=&\int_0^\pi\frac{\sin(6x)}{\sin (2x)}dx+2\int_0^\pi\cos(2x)dx\\
&=&\int_0^\pi\frac{2\sin(6x)\cos(2x)}{2\sin (2x)\cos(2x)}dx\\
&=&\int_0^\pi\frac{\sin(8x)+\sin(4x)}{\sin (4x)}dx\\
&=&\int_0^\pi(2\cos(4x)+1)dx\\
&=&\pi.
\end{eqnarray}
A: Using the identity
$$
\begin{aligned}
\frac{\sin 5 x}{\sin x}-1 &=\frac{\sin 5 x-\sin x}{\sin x} \\
&=\frac{2 \sin 2 x \cos 3 x}{\sin x} \\
&=4 \cos x \cos 3 x \\
&=2(\cos 4 x+\cos 2 x),
\end{aligned}
$$
we have $$\int_{0}^{\pi}\left(\frac{\sin 5 x}{\sin x}-1\right) d x=2 \int_{0}^{\pi}(\cos 4 x+\cos 2 x) d x=0$$
and hence $$\int_{0}^{\pi} \frac{\sin 5 x}{\sin x} d x=\pi$$
In general, for any non-negative integer $n$,$$
\begin{aligned}
\frac{\sin (2 n+1) x}{\sin x}-\frac{\sin (2 n-1) x}{\sin x}=\frac{2 \sin x \cos 2 n x}{\sin x}=2 \cos 2 n x,
\end{aligned}
$$
we have
$$
\boxed{\int_{0}^{\pi} \frac{\sin (2 n+1) x}{\sin x}=\int_{0}^{\pi} \frac{\sin (2 n-1) x}{\sin x} d x=\cdots= \int_{0}^{\pi} 1 d x=\pi}
$$
A: With $2\sin a \cos b = \sin (b+a)-\sin (b-a)$
$$2\sin x\ (\cos 2x + \cos 4x)
= \sin {5x}-\sin x $$
Then
$$\begin{align}
\int_0^\pi \frac{\sin{5x}}{\sin x}dx 
=&\int_0^\pi(1+2\cos 2x + 2\cos4x)\ dx= \int_0^\pi dx=\pi
\end{align}$$
