How would I find the Fourier series of $\cos\left(\, 5x/2\,\right) $ on $\left[-\pi,\pi\right]$?
Progress
$$A_0={1\over 2\pi}\int_{-\pi}^\pi \cos(5x/2)dx={2\over 5\pi}$$
$$A_n = {1\over \pi} \int_{-\pi}^\pi \cos(5x/2) \cos(nx)dx={20 \cos(n\pi)\over 25-4n^2}$$
$$B_n = {1\over \pi} \int_{-\pi}^\pi \cos(5x/2)\sin(n x)dx=0$$
What will the final form of the answer look like?
After your recent edit, your result $A_0 = 2/(5\pi)$ is correct. Also, you are correct that all of the $B_n$'s are zero because the integrand of $$B_n = \frac{1}{\pi}\int_{-\pi}^{\pi} \cos(5x/2)\sin(nx) dx$$ is an even function times an odd function, hence odd. So that leaves the $A_n$ integral for $n \neq 0$. We have $$A_n = \frac{1}{\pi}\int_{-\pi}^{\pi} \cos(5x/2)\cos(nx) dx$$ We may use the trig identity $$\cos(a)\cos(b) = \frac{1}{2}\left(\cos(a+b) + \cos(a-b)\right)$$ and the fact that $$\frac{1}{\pi}\int_{-\pi}^{\pi} \cos(\omega x) dx = \frac{2\sin(\omega\pi)}{\omega\pi}$$ to obtain $$A_n = \frac{\sin((5/2 + n)\pi)}{(5/2 + n)\pi} + \frac{\sin((5/2 - n)\pi)}{(5/2 - n)\pi}$$ Note that for every integer $n$, $$\sin((5/2 + n)\pi) = \sin((5/2 - n)\pi)$$ and that $$\sin((5/2 + n)\pi) = \begin{cases} 1 & \text{if }n\text{ is even} \\ -1 & \text{if }n\text{ is odd} \\ \end{cases}$$ In other words, $\sin((5/2 + n)\pi) = (-1)^n$. This allows us to simplify the expression as follows: $$A_n = \frac{1}{\pi}\left( \frac{(-1)^n}{(5/2 + n)} + \frac{(-1)^n}{5/2 - n}\right) = \frac{1}{\pi}\left(\frac{5(-1)^n}{25/4 - n^2} \right) = \frac{1}{\pi}\left(\frac{20(-1)^n}{25 - 4n^2} \right)$$ which is about as simplified as one might hope for. This nearly matches your $$\frac{20\cos(n\pi)}{25-4n^2}$$ since $\cos(n\pi) = (-1)^n$. One of us is off by a factor of $\pi$.
The final answer is $$A_0 + \sum_{n=1}^{\infty}A_n \cos(nx) + \sum_{n=1}^{\infty}B_n\sin(nx) = \frac{2}{5\pi} + \frac{1}{\pi}\sum_{n=1}^{\infty}\left(\frac{20(-1)^n}{25 - 4n^2} \right) \cos(nx)$$