Question

# Derive equations$\int_{0}^{T} \sin n \omega_{T} t \sin m \omega_{T} t d t=\left\{\begin{array}{ll}{0} & {m \neq n} \\ {T / 2} & {m=n}\end{array}\right.$,$\int_{0}^{T} \cos n \omega_{T} t \cos m \omega_{T} t d t=\left\{\begin{array}{ll}{0} & {m \neq n} \\ {T / 2} & {m=n}\end{array}\right.,$and $\int_{0}^{T} \cos n \omega_{T} t \sin m \omega_{T} t d t=0$ and hence verify the equations for the Fourier coefficient given by equations $a_{0}=\frac{2}{T} \int_{0}^{T} F(t) d t,$ $a_{n}=\frac{2}{T} \int_{0}^{T} F(t) \cos n \omega_{T} t d t \quad n=1,2, \ldots,$ and $b_{n}=\frac{2}{T} \int_{0}^{T} F(t) \sin n \omega_{T} t d t \quad n=1,2, \ldots.$

Solution

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We begin with the first given integral for the product of sines. We'll divide it into two cases, for $m \neq n$ and for $m = n$. For all of the integrals the frequency is defined as $\omega_T = \dfrac{2 \pi}{T}$.

For $m \neq n$ we have:

\begin{align*} \int_0^T \sin(n \omega_T t) \sin(m \omega_T t) dt &= \left[ \dfrac{\sin((n - m)\omega_T t)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m)\omega_T t)}{2 (n + m) \omega_T} \right]_0^T \\ &= \left[ \begin{array} {ll} \dfrac{\sin((n - m)\omega_T T)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m)\omega_T T)}{2 (n + m) \omega_T} \\ - \dfrac{\sin((n - m)\omega_T \cdot 0)}{2 (n - m) \omega_T} + \dfrac{\sin((n + m)\omega_T \cdot 0)}{2 (n + m) \omega_T} \end{array} \right] \\ &= \dfrac{\sin((n - m)\omega_T T)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m)\omega_T T)}{2 (n + m) \omega_T} - 0 \\ &= \dfrac{\sin((n - m)\dfrac{2 \pi}{T} T)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m)\dfrac{2 \pi}{T} T)}{2 (n + m) \omega_T} \\ &= \dfrac{\sin((n - m) 2 \pi)}{2 (n - m) \omega_T} - \dfrac{\sin((n + m) 2 \pi)}{2 (n + m) \omega_T} \\ &= 0 - 0 \\ \int_0^T \sin(n \omega_T t) \sin(m \omega_T t) dt &= 0 \qquad m \neq n \end{align*}

$m$ and $n$ are both integers so their sum or difference will be an integer and any sine of a product of $2 \pi$ with an integer will be 0.