Using a derivative to find fourier coefficients The previous question was:

Find the fourier coefficients of $f(x)=x^2+1$

To which I found $a_0=\frac{\pi^2}{3}+1$ and $a_n=\frac{2}{n^2}(-1)^n$ (I am unsure on $b_n$ I get $b_n=(\frac{-2}{n\pi}-\frac{2\pi}{n}+\frac{4}{n^3\pi})(-1)^n$ but this is more method than correcting that)
I want to complete the next part:

By considering the derivative of $f$, write down the fourier series of $g(x)=x $ without explicitly calculating the coefficients

So as $\frac{1}{2}f'(x)=g(x)$ I get (again ignoring $b_n$ for now):
\begin{align}
g(x) & =\frac{1}{2}\int_{-\pi}^\pi \left(\frac{\pi^2}{3} + 1 + \sum_{n=1}^\infty \frac{2}{n^2} (-1)^n\cos(nx)\,dx\right) \\[10pt]
& =\frac{1}{2} \left[\frac{\pi^2x}{3} + x + \sum_{n=1}^\infty \frac{2}{n^3}(-1)^n\cos(nx)\right]_{-\pi}^\pi =\frac{2\pi^3}{6} + \frac{\pi}{2} + \sum_{n=1}^\infty \frac{1}{n^3}
\end{align}
Is this correct?

Firstly: $f(x)$ is an even function, so you expect all $b_n = 0$.
Now for $g(x)$. You are right to note that $g(x) = \frac{1}{2} f'(x)$, but why are you integrating? Futhermore: the last equation: LHS is a function of $x$ and RHS is just a constant.
To solve the problem write $f(x)$ as a sum of cosines (so in in it's Fourier form), take the derivative and then divide by $2$ to get $g(x)$. From there you can get the coefficients without much hustle.
