Definite Integral from 0 to Pi of Cosine 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 \cos m x \cos n x \rd x = \begin{cases}

0 & : m \ne n \\ \dfrac \pi 2 & : m = n \end{cases}$


That is:

$\ds \int_0^\pi \cos m x \cos n x \rd x = \dfrac \pi 2 \delta_{m n}$

where $\delta_{m n}$ is the Kronecker delta.


Proof

Let $m \ne n$.

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

$\Box$


When $m = n$ we have:

\(\ds \int \cos m x \cos m x \rd x\) \(=\) \(\ds \int \cos^2 m x \rd x\)
\(\ds \) \(=\) \(\ds \frac x 2 + \frac {\sin 2 m x} {4 m} + C\) Primitive of $\cos^2 m x$
\(\ds \leadsto \ \ \) \(\ds \int_0^\pi \cos m x \cos m x \rd x\) \(=\) \(\ds \intlimits {\frac x 2 + \frac {\sin 2 m x} {4 m} } 0 \pi\)
\(\ds \) \(=\) \(\ds \paren {\frac \pi 2 + \frac {\map \sin {2 m \pi} } {4 m} } - \paren {\frac 0 2 + \frac {\sin 0} {4 m} }\)
\(\ds \) \(=\) \(\ds \frac \pi 2 + \frac {\map \sin {2 m \pi} } {4 m}\) Sine of Zero is Zero
\(\ds \) \(=\) \(\ds \frac \pi 2\) Sine of Multiple of Pi

$\blacksquare$


Sources

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