$f(x)= \begin{cases} \sin(2x),0\le x\le \frac{\pi}{2}\\0,\frac{\pi}{2}\le x\le\pi\end{cases}$
Prove that f(x)= $\frac{1}{\pi}+\frac{1}{2}\sin(2x)-\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{\cos(4nx)}{(2n-1)(2n+1)}$
I figured $a_n$ , $b_n$ , and $a_0$ would be found using:
$a_n$= $\frac{1}{L}\int_{-L}^{L}f(x)\cos(\frac{n\pi x}{L})dx$ = $\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(2x)\cos(2nx)dx$ = $\frac{2}{\pi}(\frac{\cos(\pi n)+1}{2-2n^2})$
$b_n$ = $\frac{1}{L}\int_{-L}^{L}f(x)\sin(\frac{n\pi x}{L})dx$ = $\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(2x)\sin(2nx)dx$ = $\frac{2}{\pi}(\frac{2\sin(\pi n)}{2-2n^2})$ = 0 since "n" is an integer.
$a_0$ = $\frac{1}{2L}\int_{-L}^{L}f(x)dx$ = $\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(2x)dx$ = $\frac{1}{\pi}$
Then, putting that all together into a Fourier Series we get:
$$f(x) = \frac{1}{2\pi}+\sum_{n=1}^{\infty}[(\frac{2}{\pi})(\frac{\cos(\pi n)+1}{2-2n^2})\cos(2nx)]$$
which reduces to $$f(x) = \frac{1}{2\pi}-\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{\cos(4nx)}{(2n-1)(2n+1)}$$ (which I found thanks to Winther's comment below!)
But I am still unsure of where the $\frac{1}{2}\sin(2x)$ term comes from. Any help is much appreciated!