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

## 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 $m \ne n$.

 $\displaystyle \int \sin m x \cos n x \rd x$ $=$ $\displaystyle \frac {-\cos \left({m - n}\right) x} {2 \left({m - n}\right)} - \frac {\cos \left({m + n}\right) x} {2 \left({m + n}\right)} + C$ Primitive of $\sin m x \cos n x$ $\displaystyle \leadsto \ \$ $\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \cos n x \rd x$ $=$ $\displaystyle \left[{\frac {-\cos \left({m - n}\right) x} {2 \left({m - n}\right)} - \frac {\cos \left({m + n}\right) x} {2 \left({m + n}\right)} }\right]_\alpha^{\alpha + 2 \pi}$ $\displaystyle$ $=$ $\displaystyle \left({\frac {-\cos \left({\left({m - n}\right) \left({\alpha + 2 \pi}\right)}\right)} {2 \left({m - n}\right)} - \frac {\cos \left({\left({m + n}\right) \left({\alpha + 2 \pi}\right)}\right)} {2 \left({m + n}\right)} }\right)$ $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \left({\frac {-\cos \left({m - n}\right) \alpha} {2 \left({m - n}\right)} - \frac {\cos \left({m + n}\right) \alpha} {2 \left({m + n}\right)} }\right)$ $\displaystyle$ $=$ $\displaystyle \left({\frac {-\cos \left({m - n}\right) \alpha} {2 \left({m - n}\right)} - \frac {\cos \left({m + n}\right) \alpha} {2 \left({m + n}\right)} }\right)$ Corollary to Cosine of Angle plus Full Angle $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle \left({\frac {-\cos \left({m - n}\right) \alpha} {2 \left({m - n}\right)} - \frac {\cos \left({m + n}\right) \alpha} {2 \left({m + n}\right)} }\right)$ $\displaystyle$ $=$ $\displaystyle 0$ 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 \ \$ $\displaystyle \int_\alpha^{\alpha + 2 \pi} \sin m x \cos m x \rd x$ $=$ $\displaystyle \left[{\frac {\sin^2 m x} {2 m} }\right]_\alpha^{\alpha + 2 \pi}$ $\displaystyle$ $=$ $\displaystyle \frac {\sin^2 \left({m \left({\alpha + 2 \pi}\right)}\right)} {2 m} - \frac {\sin^2 m \alpha} {2 m}$ $\displaystyle$ $=$ $\displaystyle \frac {\sin^2 m \alpha} {2 m} - \frac {\sin^2 m \alpha} {2 m}$ Corollary to Cosine of Angle plus Full Angle $\displaystyle$ $=$ $\displaystyle 0$

$\blacksquare$