Please verify my solution to this exercise
$f(x)=\left\{\begin{array}{rcl} \pi & \mbox{ si } & -\pi \leq x \leq \pi/2 \\ 0 & \mbox{ si } & \pi/2 < x < \pi \end{array}\right.$
$$a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx=\frac{1}{2\pi}\int_{-\pi/2}^{\pi}\pi dx=\frac{1}{2}\int_{-\pi}^{\pi/2}dx=\frac{1}{2}((\pi/2)-(-\pi))=\frac{3}{4}\pi$$
$$a_n= \frac{1}{\pi}\int_{-\pi}^{\pi/2}\pi \cos(\frac{n\pi x}{\pi})dx=\int_{-\pi}^{\pi/2}\cos nx = \frac{1}{n}(\sin n\frac{\pi}{2} -\sin(-n\pi))=\frac{1}{n}$$
$$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi/2}\pi\sin(\frac{n\pi x}{\pi})dx=\int_{-\pi}^{\pi/2}=-\frac{1}{n}(\cos (n\frac{\pi}{2})-\cos(-n\pi))=\frac{1}{n}$$
$F(x)=\frac{3}{4}\pi+\sum_{n=1}^{\infty}\frac{1}{n}(\cos nx+\sin nx)$
Observe that
$$\frac1n\left(\sin\frac{n\pi}2-\sin(-n\pi)\right)=\frac1n\sin\frac{n\pi}2=\begin{cases}0,&n=0\pmod2\\{}\\\frac1n,&n=1\pmod4\\{}\\-\frac1n,&n=3\pmod4\end{cases}$$
and also
$$-\frac{1}{n}\left(\cos\frac{n\pi}2-\cos(n\pi)\right)=\begin{cases}-\frac1n(0-(-1))=-\frac1n,&n=1\pmod2\\{}\\-\frac1n(-1-1)=\frac2n,&n=2\pmod4\\{}\\-\frac1n(1-1)=0,&n=0\pmod4\end{cases}$$
Thus, you can see things aren't quite as you wrote them