Fourier series of an even function

Question

Given the figure above i have to show that the Fourier series of the functions is: $$f(t)=\frac{1}{2}+\frac{6}{\pi}(cos\omega_0t -\frac{1}{3}cos3\omega_0t+\frac{1}{5}cos5\omega_0t-\frac{1}{7}cos7\omega_0t+...)$$

I know that the function is even so bn=0, so : $$a_n=\frac{4}{T}\int_{0}^{T/2} f(2) cosn\omega_0dt+\frac{4}{T}\int_{T/4}^{T/2} f(-1) cosn\omega_0dt$$ $$=\frac{4}{T}[\frac{sinn\omega_0t}{n\omega_0}]_{0}^{T/4}+\frac{4}{T}[\frac{-sinn\omega_0t}{n\omega_0}]_{T/4}^{T/2}$$

$$=\frac{4}{T}[\frac{sinn\omega_0\frac{T}{4}}{n\omega_0}-\frac{sin(0)}{n\omega_0}]+\frac{4}{T}[-\frac{sinn\pi}{n\omega_0}+\frac{sinn\frac{\pi}{2}}{n\omega_0}]$$

I don't know how to continue from here.

Answer 1

I am going to assume that this is all on the standard period $[-\pi,\pi]$

$a_n = $$\frac 1{\pi}\int_{-\pi}^{\pi} f(t) \cos nt \;dt\\ \frac 1{\pi}[\int_{-\pi}^{-\frac {\pi}{2}} f(t) \cos nt \;dt+\int_{-\frac{\pi}{2}}^{\frac {\pi}{2}} f(t) \cos nt \;dt+\int_{\frac{\pi}{2}}^{\pi} f(t) \cos ntT \;dt]\\ \frac 1{\pi}[\int_{-\pi}^{-\frac {\pi}{2}} - \cos nt \;dt+\int_{-\frac{\pi}{2}}^{\frac {\pi}{2}} 2 \cos nt \;dt+\int_{\frac{\pi}{2}}^{\pi} - \cos nt \;dt]\\ \frac {1}{n\pi}[-\sin nt|_{-\pi}^{-\frac{\pi}{2}}+ 2\sin nt|_{-\frac{\pi}{2}}^{\frac \pi2}-\sin nt|_{\frac{\pi}{2}}^{\pi}]$

$\sin nt = 0$ for all $n$ a $t = -\pi, \pi$

If $n$ is even $\sin nt = 0$ evaluated at $\frac {pi}{2}$

If $n\equiv 1\pmod {4}\\ \frac {1}{n\pi}[-\sin nt|_{-\pi}^{-\frac{\pi}{2}}+ 2\sin nt|_{-\frac{\pi}{2}}^{\frac \pi2}-\sin nt|_{\frac{\pi}{2}}^{\pi}] = \frac 1{n\pi}[1+2+0-(0 -2 -1)] = \frac 6{n\pi}$

If $n\equiv -1\pmod {4}\\ \frac {1}{n\pi}[-\sin nt|_{-\pi}^{-\frac{\pi}{2}}+ 2\sin nt|_{-\frac{\pi}{2}}^{\frac \pi2}-\sin nt|_{\frac{\pi}{2}}^{\pi}] =- \frac 6{n\pi}$

$a_0$ is simple enough to find.