Show that: $$2x\cos x-\sin x=4\sum_{n=2}^\infty \frac{(-1)^n}{n^2-1}\sin(nx)$$
Supposedly, this can be proved by using Fourier series, by choosing the right function but I have been thus far unable to do so. Any hints will be welcome, thanks in advance!
By using the Fourier sine series \begin{align} f(x) &= \sum_{n=1}^{\infty} B_{n} \, \sin(nx) \hspace{15mm} x \in [0,\pi] \\ B_{n} &= \frac{2}{\pi} \, \int_{0}^{\pi} f(x) \, \sin(nx) \, dx \end{align} the function $f(x) = a x \cos(x) + b \sin(x)$ can be seen to be represented by \begin{align} a x \cos(x) + b \sin(x) = \frac{2b - a}{2} + 2a \, \sum_{n=2}^{\infty} \frac{(-1)^{n} \, n}{n^{2} -1} \, \sin(nx). \end{align} This is obtained by making use of \begin{align} B_{1} &= \frac{2}{\pi} \, \int_{0}^{\pi} (ax \cos(x) + b \sin(x)) \, \sin(x) \, dx = \frac{2b-a}{2} \end{align} and calculating the general case for $n \geq 2$. Now choosing $a$ and $b$ as $a=2$ and $b=1$ then the reduction provides \begin{align} 2 x \cos(x) + \sin(x) = 4 \, \sum_{n=1}^{\infty} \frac{(-1)^{n} \, n}{n^{2}-1} \, \sin(n x). \end{align}