I am trying to follow a proof from my textbook for the equation which gives the value of the $a_r$ coefficients of the Fourier series:
$$\int^{x_0 + L}_{x_0} f(x) \cos (\frac{2 \pi r x}{L}) dx = \frac{a_r}{2}L$$
The approach of the proof is to multiply $f(x)$ (using the general form for a Fourier series) by $\cos(\frac{2 \pi p x}{L})$ and consider the case where $r=p$. The resulting expression is:
$$ \int^{x_0 + L}_{x_0} f(x) \cos (\frac{2 \pi r x}{L}) dx = \frac{a_0}{2} \int^{x_0 + L}_{x_0} \cos(\frac{2 \pi p x}{L}) dx \\ + \sum^{\infty}_{r = 1} a_r \int^{x_0 + L}_{x_0} \cos(\frac{2 \pi r x}{L}) \cos(\frac{2 \pi p x}{L}) dx \\ + \sum^{\infty}_{r = 1} b_r \int^{x_0 + L}_{x_0} \sin(\frac{2 \pi r x}{L}) \cos(\frac{2 \pi p x}{L}) dx $$
Using mutual orthogonality rules for sine and cosine and considering the case $r = p$, the first and third terms reduce to 0 and the second term reduces to $\frac{a_r}{2} L $ as in the original expression. I understand why the second and third terms reduce the way that they do, but I can't understand why the first term disappears.
The rules in question are:
$$ \int^{x_0 + L}_{x_0} \sin(\frac{2 \pi r x}{L}) \cos(\frac{2 \pi p x}{L}) dx = 0 \text{ for all } r, p$$
$$ \int^{x_0 + L}_{x_0} \cos(\frac{2 \pi r x}{L}) \cos(\frac{2 \pi p x}{L}) dx = 0 \begin{cases} L & \text{ for } r = p = 0, \\ \frac{1}{2} L & \text{ for r = p > 0, } \\ 0 & \text{ for } r \neq p \end{cases}$$
$$ \int^{x_0 + L}_{x_0} \sin(\frac{2 \pi r x}{L}) \sin(\frac{2 \pi p x}{L}) dx = 0 \begin{cases} 0 & \text{ for } r = p = 0, \\ \frac{1}{2} L & \text{ for r = p > 0, } \\ 0 & \text{ for } r \neq p \end{cases}$$