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

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Theorem

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:

\(\ds \int \sin m x \cos m x \rd x\) \(=\) \(\ds \frac {\sin^2 m x} {2 m} + C\) Primitive of $\sin m x \cos m x$
\(\ds \leadsto \ \ \) \(\ds \int_0^\pi \sin m x \cos m x \rd x\) \(=\) \(\ds \intlimits {\frac {\sin^2 m x} {2 m} } 0 \pi\)
\(\ds \) \(=\) \(\ds \frac {\sin^2 m \pi} {2 m} - \frac {\sin^2 0} {2 m}\)
\(\ds \) \(=\) \(\ds 0\) Sine of Multiple of Pi

So in this case, $\ds \int_0^\pi \sin m x \cos n x \rd x = 0$ for $m + m$ even.


Let $m \ne n$.

\(\ds \int \sin m x \cos n x \rd x\) \(=\) \(\ds \frac {-\cos \paren {m - n} x} {2 \paren {m - n} } - \frac {\cos \paren {m + n} x} {2 \paren {m + n} } + C\) Primitive of $\sin m x \cos n x$
\(\ds \leadsto \ \ \) \(\ds \int_0^\pi \sin m x \cos n x \rd x\) \(=\) \(\ds \intlimits {\frac {-\cos \paren {m - n} x} {2 \paren {m - n} } - \frac {\cos \paren {m + n} x} {2 \paren {m + n} } } 0 \pi\)
\(\ds \) \(=\) \(\ds \paren {\frac {- \map \cos {\paren {m - n} \pi} } {2 \paren {m - n} } - \frac {\map \cos {\paren {m + n} \pi} } {2 \paren {m + n} } }\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \paren {\frac {-\cos 0} {2 \paren {m - n} } - \frac {\cos 0} {2 \paren {m + n} )} }\)
\(\ds \) \(=\) \(\ds \frac {- \map \cos {\paren {m - n} \pi} } {2 \paren {m - n} } - \frac {\map \cos {\paren {m + n} \pi} } {2 \paren {m + n} } + \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} }\) Cosine of Zero is One and simplifying
\(\ds \) \(=\) \(\ds \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {\paren {-1}^{m - n} } {2 \paren {m - n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m + n} }\) Cosine of Multiple of Pi and rearranging
\(\ds \) \(=\) \(\ds \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m - n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m + n} }\) $m + n$ and $m - n$ have the same parity

$\Box$


When $m + n$ is an even integer, we have:

\(\ds \) \(\) \(\ds \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m - n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m + n} }\)
\(\ds \) \(=\) \(\ds \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac 1 {2 \paren {m - n} } - \frac 1 {2 \paren {m + n} }\)
\(\ds \) \(=\) \(\ds 0\)

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:

\(\ds \) \(\) \(\ds \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m - n} } - \frac {\paren {-1}^{m + n} } {2 \paren {m + n} }\)
\(\ds \) \(=\) \(\ds \frac 1 {2 \paren {m - n} } + \frac 1 {2 \paren {m + n} } - \frac {-1} {2 \paren {m - n} } - \frac {-1} {2 \paren {m + n} }\)
\(\ds \) \(=\) \(\ds \frac 1 {m - n} + \frac 1 {m + n}\) simplification
\(\ds \) \(=\) \(\ds \frac {\paren {m + n} + \paren {m - n} } {\paren {m + n} \paren {m - n} }\)
\(\ds \) \(=\) \(\ds \frac {2 m} {\paren {m + n} \paren {m - n} }\)
\(\ds \) \(=\) \(\ds \frac {2 m} {m^2 - n^2}\) Difference of Two Squares

$\blacksquare$


Sources

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