Prove the orthogonality relations for sine and cosine using complex exponentials

Prove the integral formula,
$\int_0^{2 \pi} e^{inx} e^{-imx} \, dx = \begin{cases} 0 & \text{if } m \neq n, \\ 2 \pi & \text{if } m = n. \end{cases}$

for integers $m$ and $n$ .
Prove the following orthogonality relations of sine and cosine using the relation in part (a), where $m$ and $n$ are integers with $m^2 \neq n^2$ .
$\int_0^{2 \pi} \sin (nx) \cos (mx) \, dx = \int_0^{2 \pi} \sin (nx) \sin (mx) \, dx = \int_0^{2 \pi} \cos (nx) \cos (mx) \, dx = 0,$

$\int_0^{2 \pi} \sin^2 (nx) \, dx = \int_0^{2 \pi} \cos^2 (nx) \, dx = \pi \qquad \text{if } n \neq 0.$

Proof. First, if $n = m$ then we have
$\int_0^{2 \pi} e^{inx} e^{-imx} \, dx = \int_0^{2 \pi} e^{ix(n-m)} \, dx = \int_0^{2 \pi} dx = 2 \pi.$

If $n \neq m$ then we have

$\begin{align*} \int_0^{2 \pi} e^{inx} e^{-imx} \, dx &= \int_0^{2 \pi} e^{ix(n-m)} \, dx \\[9pt] &= \int_0^{2\pi} e^{ikx} \, dx &(\text{for some } k \neq 0) \\[9pt] &= \frac{1}{k} \left( e^{ikx} \Bigr \rvert_0^{2\pi} \right) \\[9pt] &= \frac{1}{k} \left( e^{2 \pi i k} - e^0 \right) \\[9pt] &= \frac{1}{k} (1-1) = 0. \qquad \blacksquare \end{align*}$
Proof. These are all direct computations using part (a). Here they are,
$\begin{align*} \int_0^{2 \pi} \sin (nx) \cos (mx) \, dx &= \int_0^{2 \pi} \left( \frac{e^{inx} - e^{-inx}}{2i} \right) \left( \frac{e^{imx} + e^{-imx}}{2} \right) \, dx \\[9pt] &= \frac{1}{4i} \int_0^{2 \pi} \big( e^{ix(n+m)} - e^{ix(m-n)} + e^{ix(n-m)} - e^{ix(-n-m)} \big) \, dx \\[9pt] &= 0. \end{align*}$

(The final line follows by part (a) and since $m^2 \neq n^2$ by hypothesis which implies $n \neq m$ , $-n \neq m$ and $-n \neq -m$ .) Next,

$\begin{align*} \int_0^{2\pi} \sin (nx) \sin (mx) \, dx &= \int_0^{2 \pi} \left( \frac{e^{inx} - e^{-inx}}{2i} \right) \left( \frac{e^{imx} - e^{-imx}}{2i} \right) \, dx \\[9pt] &= -\frac{1}{4} \int_0^{2 \pi} \big( e^{ix(n+m)} - e^{ix(n-m)} - e^{ix(m-n)} + e^{ix(-n-m)} \big) \, dx \\[9pt] &= 0. \end{align*}$

The third formula,

$\begin{align*} \int_0^{2 \pi} \cos (nx) \cos(mx) \, dx &= \int_0^{2 \pi} \left( \frac{e^{inx} + e^{-inx}}{2} \right) \left( \frac{e^{imx} + e^{-imx}}{2} \right) \, dx \\[9pt] &= \frac{1}{4} \int_0^{2 \pi} \big( e^{ix(n+m)} + e^{ix(n-m)} + e^{ix(m-n)} + e^{ix(-n-m)} \big) \, dx \\[9pt] &= 0. \end{align*}$

For the next one we use the identities for $\cos^2 \theta$ and $\sin^2 \theta$ derived in this exercise (Section 9.10, Exercise #4(b)).

$\begin{align*} \int_0^{2 \pi} \sin^2 (nx) \, dx &= \int_0^{2 \pi} \frac{1}{2}(1 - \cos(2nx)) \, dx \\[9pt] &= \pi - \int_0^{2 \pi} \cos (2nx) \, dx \\ &= \pi. \end{align*}$

Finally,

$\begin{align*} \int_0^{2 \pi} \cos^2 (nx) \, dx &= \frac{1}{2} \int_0^{2 \pi} (1 - \sin^2 nx)) \, dx \\[9pt] &= 2 \pi - \pi = \pi. \qquad \blacksquare \end{align*}$