Find the sums $S_1$ and $S_2$ $$ S_1=\sum_{k=1}^\infty\frac{\cos^2 kx}{k^2}\\ S_2=\sum_{k=1}^\infty\frac{\sin^2 kx}{k^2} $$ using the following expansion: $$ I_{[a,b]}(x)=\begin{cases} 1,\ a\leqslant x\leqslant b\\ 0\ \text{otherwise} \end{cases};\ \ \ \ [a,b]\subset[-\pi,\pi]\\ I_{[a,b]}(x)=\frac{b-a}{2\pi}+\frac{1}{\pi}\sum_{n=1}^\infty\left(\frac{2}{n}\sin\frac{n(b-a)}{2}\cos\frac{n(b+a-2x)}{2}\right) $$
I found another solution to this problem (not using Fourier series of Indicator function), and here is my answer: $$ \begin{aligned} &S_1=\frac{\pi^2}{6}+\frac{x^2}{2}-\frac{\pi x}{2}\\ &S_2=-\frac{x^2}{2}+\frac{\pi x}{2} \end{aligned} $$ I did it by finding the sum of $\sum_{k=1}^\infty\frac{\cos 2kx}{k^2}$ as a subtask. But now I have to somehow apply that Indicator function. I've been told that the solution should be straightforward. However, I haven't succeeded in finding it so far.
See my answer at Calculate $\sum \limits_{n = 1}^{\infty} \frac{\cos 2n}{n^2}$. You can regard your series as sums of the squares of coefficients of Fourier series. In your expansion, choose $b+a=0$ so that the cosine becomes $\cos nx$, and use Parseval’s theorem to replace the sum of the squares of the coefficients by the square of the integral of the (squared) indicator function. This gives you $S_2$, and then $S_1=\frac{\pi^2}6-S_2$ since $\cos^2+\sin^2=1$. (Here $\frac{b-a}2$ and $n$ in the expansion play the role of $x$ and $k$, respectively, in the series.)