Boundary Values & Fourier Series · Challenge problem

The half-range cosine series of $\sin x$ on $[0,\pi]$ is $\dfrac2\pi - \dfrac4\pi\sum_{m\ge1}\dfrac{\cos 2mx}{4m^2 - 1}$. Use it to find $\displaystyle\sum_{m=1}^{\infty}\dfrac{(-1)^m}{4m^2 - 1}$.