Preliminary results

Using Euler's formula,

$$e^{ix} = \cos x + i \sin x$$

we can write

$$e^{imx} e^{inx} = ( \cos mx + i \sin mx ) ( \cos nx + i \sin nx )$$

$$= \cos mx \cos nx + i \sin nx \cos mx + i \sin mx \cos nx - \sin nx \sin mx$$

from which,

$$\begin{split} I := \int_0^\pi e^{imx} e^{inx} \d x = & \in... ...l( \sin mx \cos nx + \cos mx \sin nx \biggr) \d x \end{split}$$

We now evaluate $I$,

$$I = \int_0^\pi e^{i(m+n)x} \d x$$

$$= \frac{1}{i(m+n)} \biggl[ e^{i(m+n)x} \biggr]_0^\pi$$

$$= \frac{-i}{m+n} \biggl( (-1)^{m+n} - 1 \biggr)$$

Equating real and imaginary parts gives

$$\int_0^\pi \cos mx \cos nx \d x = \int_0^\pi \sin mx \sin nx \d x$$ (1)

We now consider

$$e^{-imx} e^{inx} = ( \cos mx - i \sin mx ) ( \cos nx + i \sin nx )$$

$$= \cos mx \cos nx + i \cos mx \sin nx - i \sin mx \cos nx + \sin mx \sin nx$$

and so

$$\begin{split} J := \int_0^\pi e^{-imx} e^{inx} \d x &= \int... ...l( \cos mx \sin nx - \sin mx \cos nx \biggr) \d x \end{split}$$

Evaluate $J$,

$$J = \int_0^\pi e^{i(n-m)x} \d x$$

$$= \frac{1}{i(n-m)} \biggl[ e^{i(n-m)x} \biggr]_0^\pi$$

$$= \frac{i}{m-n} \biggl( (-1)^{n-m} - 1 \biggr)$$

and compare real and imaginary parts:

$$\int_0^\pi \cos mx \cos nx \d x + \int_0^\pi \sin mx \sin nx \d x = 0$$ (2)

$$\int_0^\pi \cos mx \sin nx \d x = \frac{1}{m-n}\biggl((-1)^{n-m} - 1\biggr) + \int_0^\pi \sin mx \cos nx \d x$$ (3)

Clearly, (2) and (3) are only valid for $n\neq m$ since the derivation involves expressions containing $\frac{1}{m-n}$.