Integral over 2 pi of Sine of n x

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Theorem

Let $m \in \Z$ be an integer.


Then:

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


Proof

Let $m \ne n$.

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

$\Box$


Let $m = 0$.

\(\ds \int \sin 0 x \rd x\) \(=\) \(\ds \int 0 \rd x\) Sine of Zero is Zero
\(\ds \leadsto \ \ \) \(\ds \int_\alpha^{\alpha + 2 \pi} \sin 0 x \rd x\) \(=\) \(\ds \bigintlimits 0 \alpha {\alpha + 2 \pi}\)
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$


Sources

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