Apparent inconsistency between integral table and integration using trigonometric identity 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?
 A: If $a=0$, then $\cos(ax)$ should be simplified before finding its antiderivative.  This will keep you from dividing by zero, and it will make the answer come out right.  (E.g., if $a=(m-m)\pi/L$ or $a=(0+0)\pi/L$.)
A: The integration is wrong if $n=m$ or if $n=-m$, because it is false that $\int \cos(0\pi x)\,dx = \frac{\sin(0\pi x)}{0}+C$. So the very use of the formula assumes that $|n|\neq|m|$.
But if $|n|\neq|m|$, then your formula does say that the integral should be $0$, the same thing you get after the substitution. What makes you say that it "does not jive"?
A: What Arturo says is correct that the answer you get is not quite right on the diagonal ( m=n) and the anti diagonal (m=-n). In a certain sense it is almost right. What do I mean? If we take a limit of the
$$ lim_{n-m\rightarrow 0}
\frac{L}{\pi} \left [\frac{\sin \left ( \pi (n - m) \right )}{n - m} + \frac{\sin \left ( \pi (n+m) \right )}{n + m}\right ] =$$
$$\frac{L}{\pi}[\pi+\frac{sin(2\pi n)}{2n}]$$.
Now if $n$ is a nonzero integer, you recover the answer $L$. If $n=0$, then when we take the limit of the new expression as $n$ approaches zero, we get the answer $2L$. 
Now a technical note. I took this limit in a particular way. To make this well defined, what one must show is that the two dimensional limits, $$lim_{(x,y)\rightarrow (n,n)}\frac{sin(x-y)}{x-y}=1$$. Also note that the case where $m=-n$ is similar.
