Exercise 8.5.2

Question

Using trigonometric identities when necessary, verify the following integrals.
\begin{enumerate}[label=(\alph*)] 
\item For all $n\in \mathbf{N}$,
$$\int_{-\pi}^\pi \cos(nx) dx = 0 \text{\quad and \quad} \int_{-\pi}^\pi \sin(nx)dx = 0$$
\item For all $n\in \mathbf{N}$,
$$\int_{-\pi}^\pi \cos^2(nx) dx = \pi \text{\quad and \quad} \int_{-\pi}^\pi \sin^2(nx)dx = \pi$$
\item For all $m,n \in \mathbf{N}$,
$$\int_{-\pi}^\pi \cos(mx) \sin (nx) dx = 0$$
For $m \neq n$,
$$\int_{-\pi}^\pi \cos(mx) \cos(nx) dx = 0 \text{\quad and \quad} \int_{-\pi}^{\pi} \sin(mx) \sin(nx) dx = 0$$
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item Setting $u = nx$,
$$\int_{-\pi}^\pi \cos(nx) dx = \frac{1}{n} \int_{-n \pi} ^{n\pi} \cos(u) du = \frac{1}{n}\left(\sin(n\pi) - \sin(-n\pi)\right) = 0$$
$$\int_{-\pi}^\pi \sin(nx) dx = \frac{1}{n} \int_{-n \pi} ^{n\pi} \sin(u) du = \frac{-1}{n}\left(\cos(n\pi) - \cos(-n\pi)\right) = 0$$
\item Using part (a),
$$\int_{-\pi}^\pi \cos^2(nx) dx = \frac{1}{2}\int_{-\pi}^\pi 1 + \cos (2nx) dx = \pi$$
$$2 \pi = \int_{-\pi}^\pi 1 dx = \int_{-\pi}^{\pi} \sin^2(nx) + \cos^2 (nx)dx \implies \int_{-\pi}^\pi \sin^2 (nx) dx = \pi$$

\item  If $m = n$:
\newcommand{\intpi}{\int_{-\pi}^\pi}
$$\intpi \cos(nx) \sin(nx) dx = \intpi \sin(2nx) dx = 0$$
If $m \neq n$:
$$\begin{aligned}
\intpi \cos(nx) \sin(mx) dx &= \frac{\sin(n\pi) \sin(m\pi) - \sin(-n\pi)\sin(-m\pi)}{m} \
    &\qquad\qquad - \frac{n}{m}\intpi \cos(nx) \sin(mx) dx \
&= - \frac{n}{m}\intpi \cos(nx) \sin(mx) dx \
&= -\frac{n}{m} \left( \frac{-\cos(n\pi) \cos(m\pi) + \cos(-n\pi)\cos(-m \pi)}{n}\right) \
    &\qquad\qquad + \frac{n^2}{m^2}\intpi \sin(nx) \cos(mx) dx  \
&= \frac{n^2}{m^2}\intpi \sin(nx) \cos(mx) dx
\end{aligned}
    $$
Since $n^2 / m^2 \neq 1$, this implies
$$\intpi \sin(nx) \cos(mx) dx = 0$$
A similar process shows
$$\intpi \cos(mx) \cos(nx) dx = \frac{n^2}{m^2} \intpi \cos(mx) \cos(nx) = 0$$
and
$$\intpi \sin(mx) \sin(nx) dx = \frac{n^2}{m^2} \intpi \sin(mx) \sin(nx) = 0$$
 \end{enumerate}

Exercise 8.5.2

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Exercise 8.5.2

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