According to my textbook: $\int_{-L}^{L} \cos\frac{n \pi x}{L} \cos\frac{m \pi x}{L} dx = \begin{cases} 0 & \mbox{if } n \neq m \\ L & \mbox{if } n = m \neq 0 \\ 2L& \mbox{if } n = m = 0 \end{cases} $
According to the trig identity given on this cheat sheet:
$ \cos{\alpha}\cos{\beta} = \frac{1}{2}\left [ \cos \left (\alpha -\beta \right ) + \cos \left(\alpha +\beta \right ) \right ] $
Substituting this trig identity in and integrating from $-L \mbox{ to } L$ gives:
$\int_{-L}^{L} \cos\frac{n \pi x}{L} \cos\frac{m \pi x}{L} dx = \frac{L}{\pi} \left [\frac{\sin \left ( \pi (n - m) \right )}{n - m} + \frac{\sin \left ( \pi (n+m) \right )}{n + m}\right ] $
Evaluating the right side at $n = m$ gives a zero denominator, making the whole expression undefined. Evaluating the right hand side at $n \neq m$ gives $0$ because the sine function is always $0$ for all integer multiples of $\pi$ as can be clearly seen with the unit circle. None of these results jive with the first equation.
Could you explain what mistakes I am making with my thinking?