Fourier coefficients of $f(x)$ with symmetry $f(x)=f(\pi-x)$ 
Let $f(x) =\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos(nx)+b_n\sin(nx))$ and $f(x) = f(\pi-x)$. Considering this symmetry, prove that the Fourier coefficients satisfy $a_{2m+1}= b_{2m}= 0$.

My attempt so far:
Considering the form of the Fourier series of $f(x)$, the period is $2\pi$. Thus, we can write:
$$
a_{n} = \frac{1}{\pi} \int_{x_0}^{x_0+2\pi} f(x) \cos({n x}) \,dx \\
= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos({n x}) \,dx \\
=\frac{1}{\pi} \int_{-\pi}^{0} f(x) \cos({n x}) \,dx + \frac{1}{\pi} \int_{0}^{\pi} f(x) \cos({n x}) \,dx
$$
Now working with just the first integral and making the substitution $x=\pi - t$
$$
I_1=\frac{1}{\pi} \int_{t = 2\pi}^{\pi} f(\pi-t) \cos({n (\pi-t)}) \,(-dt) = \frac{1}{\pi} \int_{\pi}^{2\pi} f(t) [\cos(n\pi)\cos(nt)-\sin(n\pi)\sin(nt)] \,dt \\
=\frac{1}{\pi} (-1)^n\int_{\pi}^{2\pi} f(t) \cos(nt) \,dt
$$
But I'm sure where to go from here/if this works. I was expecting to get something of the form:
$$
a_n = \frac{1}{\pi} (1+(-1)^n)\int_{\alpha}^{\beta} f(x) \cos(nx) \,dx
$$
To then show that only $a_{2m}$ are non-zero etc. but I can't seem to make the bounds work to do this. Is the along the right track?
 A: If you really want to split up the integral like that, you may want to leave the formula as
$$\begin{align*}a_n &= \frac{1}{\pi}\int_{x_0}^{x_0 +2\pi} f(x)\cos(nx)\,dx \\ &= \frac{1}{\pi}\int_{x_0}^{x_0+\pi} f(x)\cos(nx)\,dx + \frac{1}{\pi}\int_{x_0+\pi}^{x_0+2\pi}f(x)\cos(nx)\,dx \\ &:= I_1 + I_2.\end{align*}$$
Then as you've done, we can pick the first integral and write $$\begin{align*}I_1 &= \frac{1}{\pi}\int_{x_0}^{x_0+\pi} f(x)\cos(nx)\,dx \\ &= \frac{1}{\pi}\int_{\pi-x_0}^{-x_0}f(\pi-t)\cos(n(\pi-t))\,(-dt) \\ &= \frac{1}{\pi}(-1)^n\int_{-x_0}^{\pi-x_0}f(t)\cos(nt)\,dt\end{align*}$$
Then compare the limits on both integrals.  They both have length $\pi$, so for them to be equal, we'd need the lower limits to equal:
$$-x_0 = x_0 + \pi \implies x_0 = -\pi/2.$$
That is, by setting $x_0=-\pi/2$ originally, we have $$a_n = \frac{1}{\pi}((-1)^n+1)\int_{\pi/2}^{3\pi/2} f(x)\cos(nx)\,dx$$
which is the form you were looking for.

That said, we don't actually need to split the integral in two, and by using the whole integral, we don't need to care about which $x_0$ we choose.  Using the same substitution but with the whole integral, we'd have
$$a_n = \frac{1}{\pi}\int_{0}^{2\pi} f(x)\cos(nx)\,dx = \frac{1}{\pi}(-1)^n\int_{-\pi}^{\pi} f(x)\cos(nx)\,dx = (-1)^na_n \\ \implies \\(1 - (-1)^n)a_n = 0$$
which gives $2a_{2m+1} = 0.$
