Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 68

Prove that $\displaystyle \int^\pi_{-\pi} \sin (mx) \sin (nx) dx = \left\{
\begin{array}{c}
0 & \text{if} & m \neq n\\
\pi & \text{if} & m = n
\end{array}\right.
$


where $m$ and $n$ are positive integers.
If we use the sum and difference angles formula for cosine, we get


$
\begin{equation}
\begin{aligned}
\cos (mx - nx) &= \cos (mx) \cos (nx) - \sin (mx) \sin (nx) \qquad \text{and}\\
\\
\cos (mx + nx) &= \cos (mx) \cos (nx) + \sin (mx) \sin (nx)
\end{aligned}
\end{equation}
$

Then, $\cos (mx - nx) - \cos(mx+nx) = \cos (mx) \cos (nx) - \sin (mx) \sin (nx) - \cos (mx) \cos (nx) + \sin (mx) \sin (nx) = 2 \sin (mx) \sin (nx)$

Therefore,
$\displaystyle \int^\pi_{-\pi} \sin (mx)\sin(nx)dx = \int^\pi_{-\pi}\left[ \frac{\cos(mx-nx)-\cos(mx+nx)}{2} \right]dx$

if $m \neq n$

$
\begin{equation}
\begin{aligned}
\int^\pi_{-\pi} \sin (mx)\sin(nx)dx &= \frac{1}{2} \left[ \frac{\sin(mx-nx)}{(m-n)} - \frac{\sin(mx+nx)}{(m+n)} \right]^\pi_{-\pi}\\
\\
\int^\pi_{-\pi} \sin (mx)\sin(nx)dx &= \frac{1}{2} \left(\left[ \frac{\sin(m\pi - n \pi)}{(m-n)} - \frac{\sin(m\pi+n\pi)}{(m+n)} \right] - \left[ \frac{\sin(m(-\pi) - n(-\pi))}{(m-n)} - \frac{\sin(m(-\pi) + n (-\pi))}{(m+n)} \right] \right)\\
\\
&= 0
\end{aligned}
\end{equation}
$


if $m =n$,

$
\begin{equation}
\begin{aligned}
\int^\pi_{-\pi} \sin (mx) \sin(nx) dx &= \int^\pi_{-\pi} \left[ \frac{\cos (mx - mx) - \cos (mx + mx)}{2} \right] dx\\
\\
&= \frac{1}{2} \int^\pi_{-\pi} [\cos(0) - \cos(2mx)] dx\\
\\
&= \frac{1}{2} \int^\pi_{-\pi} [ 1 - \cos(mx)] dx\\
\\
&= \frac{1}{2} \left[ x - \frac{\sin(2mx)}{2m} \right]^\pi_{-\pi}\\
\\
&= \frac{1}{2} \left( \left[ \pi - \frac{\sin(2m\pi)}{m} \right] - \left[ (-\pi) - \frac{\sin(2m(-\pi))}{2m} \right] \right)\\
\\
&= \frac{1}{2} [ \pi + \pi]\\
\\
&= \frac{2\pi}{2} \\
\\
&= \pi
\end{aligned}
\end{equation}
$