<s>[INST] Ways to prove $ \int_0^\pi \frac{\sin^2 nx}{\sin^2 x} dx= n\pi$ In how many ways can we prove the following theorem?
$$I(n):= \int_0^\pi \frac{\sin^2 nx}{\sin^2 x} dx= n\pi$$
where $n$ is a nonnegative integer.
The proof I found is by considering $I(n+1)-I(n)$, which can be reduced to
$$
g(n):= \int_0^\pi \frac{\sin(2 n x) \cos x}{\sin x}dx
$$
I then showed that $g(n)=g(n+1)$, with $g(n) = ng(1) = n\pi$. This completes the proof.
I was wondering if there is a more direct way to prove it. By 'direct' I mean without deriving auxiliary recursions.
 [/INST] I can post my solution , from my document. It can be found on page 83. If one truly want to shoot this problem with a canon one can use the following generalization
$$
\int _{0}^{\pi }\! \left( {\frac {\sin \left( nx \right) }{\sin
 \left( x \right) }} \right)^{m}{dx}=\pi \sum _{l=0}^{\large\left\lfloor {\frac{m\left(n-1\right)}{2n}} \right\rfloor }\left( -1 \right) ^{l}{m\choose l}{\dfrac{m}{2}\left( n+1\right) -ln-1\choose m-1}\tag{10}
$$
Which is proved by Graham Hesketh, see equation 10  here. Set $m=1$ use the properties of the floor function and one is is done.
Another way to prove $(n)$ can be found in the same answer

Proof

Lemma: Let $k \in \mathbb{Z}$ then
$$\begin{align*}
\int_0^{\pi}\frac{\sin 2kx}{\sin x}\mathrm{d}x & = 0 \tag{1} \\ \int_0^{\pi}\frac{\sin (2k-1)x}{\sin x}\mathrm{d}x & = \pi \tag{2} \end{align*}$$

Proof:  We first define the following function $\displaystyle I_n = \int_0^\pi \frac{\sin 2n x}{\sin x} \mathrm{d}x$. Note that we now have $I_0 = 0$ and $I_1=\pi$. since $\sin 0 = 0$ and $\sin 2x = 2\cos x \sin x$. We have the following relation for all $n$
$$
I_n-I_{n-2}
=\int_0^{\pi}\frac{\sin{nx}-\sin{(n-2)x}}{\sin x}\mathrm{d}x
=2\int_0^{\pi}\cos(n-1)x\mathrm{d}x
=2\left[\frac{\sin{(n-1)x}}{n-1}\right]_0^{\pi}
=0
$$
For $|n|\geq 3$. This means $I_{2k}=I_{2k-2}=\cdots=I_{2}=\pi$ and simmilarly $I_{2k+1} = I_{2k-1}=\cdots I_1=0$. Which is what we wanted to show.

Proposition: Let $k \in \mathbb{Z}$ then
$$
\ell_2(k) 
= \int_0^\pi \left( \frac{\sin kx}{\sin x} \right)^2\mathrm{d}x
= \int_0^\pi \frac{1 - \cos kx}{1 - \cos x}\,\mathrm{d}x
= |k|\pi \tag{3}
$$

Proof 1: Here we will follow in the footsteps of the other answer and use the lemma above. We define $\displaystyle J_n=\int_0^{\pi}\left(\frac{\sin nx}{\sin x}\right)^2\mathrm{d}x$. We can note that $J_n - J_{n-1}$ is constant eg
$$ 
\begin{align*}
    J_k - J_{k-1}
& = -\frac{1}{2}\int_0^{\pi}\frac{\cos{2kx}-\cos{(2n-2)x}}{(\sin x)^2}
  = -\frac{1}{2}\int_0^{\pi}\frac{-2\sin\left(\frac{4k-2}{2}x\right) 
    \sin\left(\frac{2x}{2}\right)}{(\sin x)^2} \\
& = \int_0^{\pi}\frac{\sin{(2k-1)x}}{\sin x}
  = I_{2k-1}=\pi.
\end{align*}
$$
Then we can write
$$J_n=J_{n-1}+\pi=J_{n-2}+2\pi=\cdots=J_1+(n-1)\pi=n\pi.$$
This completes the proof.

Proof 2: Here is another way to solve this problem. I got the idea from chat, but the tecnique is much older. We want to prove that $(I_{n+1} + I_{n-2}) / 2 = I_n$. Eg that $I_n$ is the average of the next and previous term
Some calculations show that
\begin{align*}
     \frac{I_{n+1}+I_{n-1}}{2}
 & = \frac{1}{2}\int_0^\pi \frac{1 - \cos(n+1)x }{1 - \cos x} + \frac{1-\cos(n-1)x}{1-\cos x} \,\mathrm{d}x \\
 & = \frac{1}{2}\int_0^\pi \frac{2 -\bigl[ \cos(n+1)x + \cos(n-1)x\bigr]}{1-\cos x} \,\mathrm{d}x \\
 & =  \int_0^\pi \frac{1 - \cos nx \cos x}{1-\cos x} \,\mathrm{d}x \\
 & =  \int_0^\pi \frac{(1-\cos nx) +  (1-\cos x) \cos nx}{1-\cos x} \,\mathrm{d}x \\
 & =  \int_0^\pi \frac{1 - \cos nx}{1-\cos x} \,\mathrm{d}x  
   = I_n
\end{align*}
Several things was used here. Like $\cos (n+1)x + \cos(n-1)x = 2 \cos nx \cos x$ and $ \int_0^\pi \cos n x\,\mathrm{d}x = 0 \ \forall \ n \in\mathbb{Z} \backslash \{0\}$. Now we have shown that
$$
 I_n = \frac{I_{n+1}+I_{n-1}}{2}
 \ \Rightarrow \
 2I_n = I_{n+1}+I_{n-1}
 \ \Rightarrow \
 I_n - I_{n-1} = I_{n+1} - I_n
$$
Which is just an arithmetric sequence because the difference between two terms is constant (same as the previous answer). Hence
$I_n = I_0 + (n-0)d = nd$, where $d$ is the difference between two terms. We have $d = I_1 - I_0 = \pi - 0 = \pi$. This completes the second proof. $\qquad\square$
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