Theorem: Definite Integral from 0 to Pi of Sine of m x by Cosine of n x

Let $m, n \in \Z$ be integers.
Then:
:$\ds \int_0^\pi \sin m x \cos n x \rd x = \begin{cases} 0 & : m + n \text { even} \\ \dfrac {2 m} {m^2 - n^2} & : m + n \text { odd} \end{cases}$

Proof:
First we address the special case where $m = n$.
In this case $m + n = m + m = 2 m$ is even.
We have:
{{begin-eqn}}
{{eqn | l = \int \sin m x \cos m x \rd x
      | r = \frac {\sin^2 m x} {2 m} + C
      | c = Primitive of $\sin m x \cos m x$
}}
{{eqn | ll= \leadsto
      | l = \int_0^\pi \sin m x \cos m x \rd x
      | r = \intlimits {\frac {\sin^2 m x} {2 m} } 0 \pi
      | c = 
}}
{{eqn | r = \frac {\sin^2 m \pi} {2 m} - \frac {\sin^2 0} {2 m}
      | c = 
}}
{{eqn | r = 0
      | c = Sine of Multiple of Pi
}}
{{end-eqn}}
So in this case, $\ds \int_0^\pi \sin m x \cos n x \rd x = 0$ for $m + m$ even.
Let $m \ne n$.
{{begin-eqn}}
{{eqn | l = \int \sin m x \cos n x \rd x
      | r = \frac {-\cos \paren {m - n} x} {2 \paren {m - n} } - \frac {\cos \paren {m + n} x} {2 \paren {m + n} } + C
      | c = Primitive of $\sin m x \cos n x$
}}
{{eqn | ll= \leadsto
      | l = \int_0^\pi \sin m x \cos n x \rd x
      | r = \intlimits {\frac {-\cos \paren {m - n} x} {2 \paren {m - n} } - \frac {\cos \paren {m + n} x} {2 \paren {m + n} } } 0 \pi
      | c = 
}}
{{eqn | r = \paren {\frac {- \map \cos {\paren {m - n} \pi} } {2 \paren {m - n} } - \frac {\map \cos {\paren {m + n} \pi} } {2 \paren {m + n} } }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \paren {\frac {-\cos 0} {2 \paren {m - n} } - \frac {\cos 0} {2 \paren {m + n} )} }
      | c = 
}}
{{eqn | r = \frac {- \map \cos {\paren {m - n} \pi} } {2 \paren {m - n} } - \frac {\map \cos {\paren {m + n} \pi} } {2 \paren {m + n} } + \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} }
      | c = Cosine of Zero is One and simplifying
}}
{{eqn | r = \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {\paren {-1}^{m - n} } {2 \paren {m - n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m + n} }
      | c = Cosine of Multiple of Pi and rearranging
}}
{{eqn | r = \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m - n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m + n} }
      | c = $m + n$ and $m - n$ have the same parity
}}
{{end-eqn}}
{{qed|lemma}}
When $m + n$ is an even integer, we have:
{{begin-eqn}}
{{eqn | o = 
      | r = \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m - n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m + n} }
      | c = 
}}
{{eqn | r = \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac 1 {2 \paren {m - n} } - \frac 1 {2 \paren {m + n} }
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
This shows that in all cases $\ds \int_0^\pi \sin m x \cos n x \rd x = 0$ for $m + m$ even.
When $m + n$ is an odd integer, we have:
{{begin-eqn}}
{{eqn | o = 
      | r = \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m - n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m + n} }
      | c = 
}}
{{eqn | r = \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {-1} {2 \paren {m - n} } - \frac {-1} {2 \paren {m + n} }
      | c = 
}}
{{eqn | r = \frac 1 {m - n} + \frac 1 {m + n}
      | c = simplification
}}
{{eqn | r = \frac {\paren {m + n} + \paren {m - n} } {\paren {m + n} \paren {m - n} }
      | c = 
}}
{{eqn | r = \frac {2 m} {\paren {m + n} \paren {m - n} }
      | c = 
}}
{{eqn | r = \frac {2 m} {m^2 - n^2}
      | c = Difference of Two Squares
}}
{{end-eqn}}
{{qed}}