Determine the limit of a series, involving trigonometric functions: $\sum \frac{\sin(nx)}{n^3}$ and $\frac{\cos(nx)}{n^2}$ I have $$\sum^\infty_{n=1} \frac{\sin(nx)}{n^3}.$$
I did prove convergence:
$0<\theta<1$
$$\left|\frac{\sin((n+1)x)n^3}{(n+1)^3\sin(nx)}\right|< \left|\frac{n^3}{(n+1)^3}\right|<\theta$$
Now I want to determine the limit. I found a similar proof but I need help understanding it; it goes like this. :
$$ F(x):=\sum^\infty_{n=1} \frac{\cos(nx)}{n^2}$$
As for this series we have uniform convergence. The series of derivatives: $$-\sum^\infty_{n=1} \frac{\sin(nx)}{n}$$ converges for every $\delta >0$ on the interval $[\delta, 2\pi-\delta]$ uniform against $\frac{x-\pi}{2}$
so, for every $x \in]0,2\pi[$ : $\displaystyle F'(x) = \frac{x-\pi}{2}$$\displaystyle F(x) = \left(\frac{x-\pi}{2}\right)^2+c,c\in \mathbb{R}$.
To determine the constant we calculate:
$$ \int^{2\pi}_0F(x)dx=\int^{2\pi}_0\left(\frac{x-\pi}{2}\right)^2dx+\int^{2\pi}_0cdx=\frac{\pi^3}{6}+2\pi c$$
(Question: Why can we do this do get the constant?)
Because $\int^{2\pi}_0cos(nx)dx= 0 \forall n≥1$ we have:
$$\int^{2\pi}_0F(x)dx = \sum^\infty_{n=1}\int^{2\pi}_0\frac{\cos(nx)}{n^2}=0,$$ so $c = -\frac{\pi^2}{12}$. (Question: How does he get to that term $\frac{\pi^2}{12}$?) With that we have proven, that
$$\sum^\infty_{n=1} \frac{\cos(nx)}{n^2}=\left(\frac{x-\pi}{2}\right)^2-\frac{\pi^2}{12}$$
If you can explain one of the questions about this proof, or if you know how to calculate the limit in my situation above, it would be cool if you leave a quick post here, thanks!

$$\sum^\infty_{n=1}\frac{sin(nx)}{n^3}= ?$$  
We know that $$\sum^\infty_{n=1}\frac{sin(nx)}{n}= \frac{x-\pi}{2}$$.
$$(\frac{sin(nx)}{n^3})''= (\frac{cos(nx)}{n^2})'=-\frac{sin(nx)}{n}$$
$$-\sum^\infty_{n=1}\frac{sin(nx)}{n}= -(\frac{x-\pi}{2})$$
$$ - \int \frac{x-\pi}{2}= -\int \frac{x}{2}-\frac{\pi}{2} = \frac{x^2}{4}+\frac{\pi x}{2}+c$$
$$\int\frac{x^2}{4}+\frac{\pi x}{2}+c=\frac{\pi x^2}{4}+\frac{x^3}{12}+xc$$
We can also determine the constant. We know that $-f'(0) = 0 + c$.
$$-\sum^\infty_{n=1} \frac{cos(n0)}{n^2}= - \sum^\infty_{n=1} \frac{1}{n^2} = -\frac{\pi^2} {6} = 0+c$$
$$-\int\frac{\pi^2} {6} = -\frac{x\pi^2} {6}$$ So 
$$\sum^\infty_{n=1}\frac{sin(nx)}{n^3}= \frac{\pi x^2}{4}+\frac{x^3}{12}+(-\frac{x\pi^2}{6}) =  \frac{x\pi^2}{6}-\frac{\pi x^2}{4}+\frac{x^3}{12}$$
