Find $\int_0^{2\pi}g(x)\space dx$ where $g(x)$ is the sum of the series $$\sum^\infty_{n=1}\frac{\cos(2nx)}{n^3}$$
Let $g(x)$ denote the sum of the series. 
Find $$\int_0^{\frac{\pi}{2}}g(x)\space dx.$$
I have already shown that the series converges uniformly on $\mathbb{R}$. 
I used this theorem:
$$\int^b_af=\int^b_a\sum^\infty_{n=1}f_n=\sum^\infty_{n=1}\int^b_af_n$$
My method was too use the theorem to swap the integral and sum around, in order to find the pointwise on $x\in [0, 2\pi]$. Then using the same theorem find the pointwise, and integrate it.
I found the pointwise limit $f$ to be $0$, as $\lim\limits_{n \to \infty} \frac{1}{n^3}\to 0$. 
Then using the pointwise limit to find the integral:$$\int^{\frac{\pi}{2}}_00\space dx=0$$
I think I've gone wrong here somewhere, either by using the wrong method or the wrong application of the correct method. I would be grateful if someone could point some potential mistakes for me.

You have done well! Now
$$
\int_0^{2\pi}\frac{\cos(2nx)}{n^3}=
\Bigl[\frac{\sin(2nx)}{2n^4}\Bigr]_0^{2\pi}=0
$$
Now use that
$$
\int_0^{2\pi}\sum_{n>0}f_n=\sum_{n>0}\int_0^{2\pi}f_n
$$
Nothing else is needed.
