What is the Sum of a Series Involving Sine Squared for $0 < x < \pi$?

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The problem regarding the order of $\operatorname{SL}_n(\Bbb F_p)$ was intended to be posted as a University POTW, but I accidentally posted it as a Graduate POTW! My apologies. Here is another problem.

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For $0 < x < \pi$, find the sum of the series

$$\sum_{n = 1}^\infty \frac{\sin^2(nx)}{n^2}$$

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Congratulations to Opalg for his correct solution, which can be read below.

Spoiler

I found this by looking at the Fourier cosine series for the function $f(x) = \frac12x(\pi-x)$ on the interval $[0,\pi]$. The Fourier cosine series is the series \(\displaystyle \tfrac12a_0 + \sum_{k=1}^\infty a_k\cos(kx)\), where \(\displaystyle a_k = \frac2\pi\int_0^\pi f(x)\cos(kx)\,dx.\)

When $k=0$, \(\displaystyle a_0 = \frac2\pi\int_0^\pi \frac12x(\pi-x)\,dx = \frac2\pi\Bigl[\frac14x^2\pi - \frac16x^3\Bigr]_0^\pi = \frac16\pi^2.\)

For $k>0$, integrate by parts twice to get $$\begin{aligned}a_k &= \frac1\pi\int_0^\pi x(\pi - x)\cos(kx)\,dx \\ &= \frac1\pi\Bigl[x(\pi-x)\frac{\sin(kx)}k\Bigr]_0^\pi - \frac1{k\pi}\int_0^\pi(\pi-2x)\sin(k\pi)\,dx \\ &= -\frac1{k\pi}\Bigl[(\pi-2x)\frac{-\cos(kx)}k\Bigr]_0^\pi + \frac1{k^2\pi}\int_0^\pi2\cos(kx)\,dx \\ &= \frac{-\pi(-1)^k -\pi}{k^2\pi} = \begin{cases}0 &\text{if $k$ is odd,} \\ -\frac2{k^2}&\text{if $k$ is even.} \end{cases} \end{aligned}$$

The function $f$ is continuous, and vanishes at both ends of the interval, so it is equal to the sum of its Fourier cosine series. Therefore $$\frac12x(\pi-x) = \frac{\pi^2}{12} - \sum_{n=1}^\infty \frac2{(2n)^2}\cos(2nx) = \frac{\pi^2}{12} - \sum_{n=1}^\infty \frac1{2n^2}(1 - 2\sin^2(nx)) = \sum_{n=1}^\infty\frac{\sin^2(nx)}{n^2},$$ using the well-known fact that \(\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6.\)

So the conclusion from this backwards calculation is that \(\displaystyle \sum_{n=1}^\infty\frac{\sin^2(nx)}{n^2} = \frac12x(\pi-x).\)