Compute the Fourier Series for $f(x)=\cos(\frac{x}{2})$ for the domain $(-\pi<x≤\pi)$ and $f(x)=f(x+2\pi)$

Question

I know that $b_{n}$ is an even function, due to the function being an even function, however I am struggling to compute $a_{n}$.

Answer 1

1. $$a_0=\frac{1}{\pi} \int_{-\pi}^{\pi}\cos(\frac{x}{2})dx$$ $$=\frac{2}{\pi} \int_{-\pi}^{\pi}\cos(\frac{x}{2})d(\frac{x}{2})$$ $$=\frac{2}{\pi} \left[\sin(\frac{x}{2})\right]_{-\pi}^{\pi}$$ $$=\frac{4}{\pi}$$

2. $$a_n=\frac{1}{\pi} \int_{-\pi}^{\pi}\cos(\frac{x}{2})\cdot \cos nx dx$$ $$=\frac{1}{2\pi} \int_{-\pi}^{\pi}2\cos(\frac{x}{2})\cdot \cos nx dx$$ $$=\frac{1}{2\pi} \int_{-\pi}^{\pi}\left[\cos\frac{(2n+1)x}{2} + \cos\frac{(2n-1)x}{2} \right] dx$$ $$=\frac{1}{2\pi} \cdot \frac{2}{2n+1} \left[\sin\frac{(2n+1)x}{2} + \sin\frac{(2n-1)x}{2} \right]_{-\pi}^{\pi}$$ $$=\frac{1}{\pi} \cdot \frac{1}{2n+1} \left[2\sin\frac{(2n+1)\pi}{2} + 2\sin\frac{(2n-1)\pi}{2} \right]$$

I think you can complete the last line and that $b_n=0$ since $f(x)$ is even.