Integral over 2 pi of Sine of m x by Cosine of n x

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $m, n \in \Z$ be integers.

Let $\alpha \in \R$ be a real number.


Then:

$\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \cos n x \rd x = 0$


Proof

Let $m \ne n$.

\(\displaystyle \int \sin m x \cos n x \rd x\) \(=\) \(\displaystyle \frac {-\map \cos {m - n} x} {2 \paren {m - n} } - \frac {\map \cos {m + n} x} {2 \paren {m + n} } + C\) Primitive of $\sin m x \cos n x$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \cos n x \rd x\) \(=\) \(\displaystyle \intlimits {\frac {-\map \cos {m - n} x} {2 \paren {m - n} } - \frac {\map \cos {m + n} x} {2 \paren {m + n} } } \alpha {\alpha + 2 \pi}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\frac {-\map \cos {\paren {m - n} \paren {\alpha + 2 \pi} } } {2 \paren {m - n} } - \frac {\map \cos {\paren {m + n} \paren {\alpha + 2 \pi} } } {2 \paren {m + n} } }\)
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \paren {\frac {-\map \cos {m - n} \alpha} {2 \paren {m - n} } - \frac {\map \cos {m + n} \alpha} {2 \paren {m + n} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\frac {-\map \cos {m - n} \alpha} {2 \paren {m - n} } - \frac {\map \cos {m + n} \alpha} {2 \paren {m + n} } }\) Corollary to Cosine of Angle plus Full Angle
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \paren {\frac {-\map \cos {m - n} \alpha} {2 \paren {m - n} } - \frac {\map \cos {m + n} \alpha} {2 \paren {m + n} } }\)
\(\displaystyle \) \(=\) \(\displaystyle 0\) after simplification


When $m = n$ we have:

\(\displaystyle \int \sin m x \cos m x \rd x\) \(=\) \(\displaystyle \frac {\sin^2 m x} {2 m} + C\) Primitive of $\sin m x \cos m x$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \cos m x \rd x\) \(=\) \(\displaystyle \intlimits {\frac {\sin^2 m x} {2 m} } \alpha {\alpha + 2 \pi}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\map {\sin^2} {m \paren {\alpha + 2 \pi} } } {2 m} - \frac {\sin^2 m \alpha} {2 m}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sin^2 m \alpha} {2 m} - \frac {\sin^2 m \alpha} {2 m}\) Corollary to Cosine of Angle plus Full Angle
\(\displaystyle \) \(=\) \(\displaystyle 0\)

$\blacksquare$


Sources