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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.28$