Find $ S=\frac{\cos 2x}{1\cdot 3}+\frac{\cos 4x}{3\cdot 5}+\frac{\cos 6x}{5\cdot 7}+\dots=\sum_{n=1}^\infty\frac{\cos (2nx)}{(2n-1)(2n+1)} $ 
Find a sum of the series:
  $$
S=\frac{\cos 2x}{1\cdot 3}+\frac{\cos 4x}{3\cdot 5}+\frac{\cos 6x}{5\cdot 7}+\dots=\sum_{n=1}^\infty\frac{\cos (2nx)}{(2n-1)(2n+1)}
$$

My attempt:
$$
\begin{aligned}
&z=\cos x+i\sin x\\
&S=\frac{1}{2}\text{Re}\sum_{n=1}^\infty\frac{z^{2n}}{2n-1}-\frac{1}{2}\text{Re}\sum_{n=1}^\infty\frac{z^{2n}}{2n+1}
\end{aligned}
$$
But calculating these sums seems a bit difficult to me. Perhaps there is a better approach to this problem?

Note that
$$1+z^2+z^4+\cdots = \frac{1}{1-z^2}.$$
Integrating both sides,
$$z+\frac{z^3}{3}+\frac{z^5}{5}+\cdots=\frac 12 (\log(1+z)-\log(1-z)) = \tanh^{-1}(z).$$
Observe that multiplying both sides by $z$ gives
$$\sum_{n=1}^\infty \frac{z^{2n}}{2n-1} = z\tanh^{-1}(z)$$
and multiplying the left hand side by $\frac 1z$ and subtracting the $1$ term gives
$$\sum_{n=1}^\infty \frac{z^{2n}}{2n+1} = -1 + \frac{\tanh^{-1}(z)}{z}.$$
Rewriting $\tanh^{-1}(z)$ as $\ln\left(\frac{1-z}{1+z}\right)$, remembering that $z$ is on the unit circle, we can draw vectors $1+z$ and $1-z$ in the complex plane. Doing some basic geometry, we can see that the angle between these is $\frac \pi 2$, and that the lengths of $1+z$ and $1-z$ are $2 \cos \left(\frac \theta 2 \right)$ and $2 \cos \left( \frac \pi 2 - \frac \theta 2 \right) = 2 \sin \left( \frac \theta 2 \right)$.
So $\frac{1-z}{1+z} = -i \cdot \tan \left(\frac x2 \right)$, and so the $\log$ of this is
$$-\frac{i \pi}{2} + \ln\left(\tan \left(\frac x2 \right)\right)$$
(since $\log(-i) = -\frac{i \pi}{2}$).
From here, everything is easily calculatable.
