Is it true that $\int_{0}^{1}\sin(m \pi x)\cos(n \pi x) dx = 0$ only when $n+m $ is even? I would just like to verify this is correct:
$$\int_{0}^{1}\sin(m \pi x)\cos(n \pi x) dx = -\frac{1}{2}\bigg[\frac{(-1)^{n+m}}{(n+m)\pi} + \frac{(-1)^{n-m}}{(n-m)\pi} -\frac{1}{(n+m)\pi} - \frac{1}{(n-m)\pi} \bigg]$$
and so the above integral is equal to zero only when $n+m$ is even.
 A: Disclaimer: I assume that $m$ and $n$ are integers because of the usual notation and because the OP suggests studying their parity.

The integral is pretty straightforward when you start by using
$$\sin a \cos b = \frac{\sin(a+b) + \sin(a-b)}2.$$
This converts it to
$$\frac1{2\pi} \left(
\int_0^\pi \sin((m+n)y)\,\mathrm{d}y +
\int_0^\pi \sin((m-n)y)\,\mathrm{d}y
\right)$$
from which the result you stated follows unless $m = \pm n$.
Edit: With some sign corrections, that is:
$$\frac1{2\pi}\left(
\frac{1 - (-1)^{m+n}}{m+n} + \frac{1 - (-1)^{m-n}}{m-n}
\right).$$
In these cases ($m = \pm n$) the part of the formula which would divide by $m\pm n$ must be replaced by zero (because it's the result of integration of $\sin((m \mp n)y)$ in the case where that reduces to $\int 0\,\mathrm{d}y$).
This can become zero for several reasons.


*

*All of the terms are trivially zero: this happens when $m = n = 0$.

*Two of the terms are trivially zero and the other two cancel each other out: Let $m = \pm n$, the other terms are of the form:
$$\frac1{2\pi} \frac{1 - (-1)^{m\mp n}}{m \mp n}.$$
This can only be zero if $m \mp n$ is even, which is equivalent to $m+n$ being even, as you found. A generalization of such condition covers the case (1) as well.

*None of the terms are trivially zero but they still cancel each out: This is a possibility you left out. It can happen than
$$\frac{1-(-1)^{m+n}}{m+n} = -\frac{1-(-1)^{m-n}}{m-n}$$
for $m \not\equiv n \pmod 2$ if $m+n = -(m-n)$. That means, if $m = 0$. Indeed, in this case the sine term in the integral is constantly zero so the cosine can be anything.
So the overall result is zero if $m \equiv n \pmod 2$ (read: $m+n$ is even) or $m = 0$. The latter was missing from your analysis.
