Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 68

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Prove that $\displaystyle \int^\pi_{-\pi} \sin (mx) \sin (nx) dx = \left\{ \begin{array}{c} 0 & \text{if} & m \neq n\\ \pi & \text{if} & m = n \end{array}\right. $

where $m$ and $n$ are positive integers.

If we use the sum and difference angles formula for cosine, we get

$ \begin{equation} \begin{aligned} \cos (mx - nx) &= \cos (mx) \cos (nx) - \sin (mx) \sin (nx) \qquad \text{and}\\ \\ \cos (mx + nx) &= \cos (mx) \cos (nx) + \sin (mx) \sin (nx) \end{aligned} \end{equation} $

Then, $\cos (mx - nx) - \cos(mx+nx) = \cos (mx) \cos (nx) - \sin (mx) \sin (nx) - \cos (mx) \cos (nx) + \sin (mx) \sin (nx) = 2 \sin (mx) \sin (nx)$

Therefore,

$\displaystyle \int^\pi_{-\pi} \sin (mx)\sin(nx)dx = \int^\pi_{-\pi}\left[ \frac{\cos(mx-nx)-\cos(mx+nx)}{2} \right]dx$

if $m \neq n$

$ \begin{equation} \begin{aligned} \int^\pi_{-\pi} \sin (mx)\sin(nx)dx &= \frac{1}{2} \left[ \frac{\sin(mx-nx)}{(m-n)} - \frac{\sin(mx+nx)}{(m+n)} \right]^\pi_{-\pi}\\ \\ \int^\pi_{-\pi} \sin (mx)\sin(nx)dx &= \frac{1}{2} \left(\left[ \frac{\sin(m\pi - n \pi)}{(m-n)} - \frac{\sin(m\pi+n\pi)}{(m+n)} \right] - \left[ \frac{\sin(m(-\pi) - n(-\pi))}{(m-n)} - \frac{\sin(m(-\pi) + n (-\pi))}{(m+n)} \right] \right)\\ \\ &= 0 \end{aligned} \end{equation} $

if $m =n$,

$ \begin{equation} \begin{aligned} \int^\pi_{-\pi} \sin (mx) \sin(nx) dx &= \int^\pi_{-\pi} \left[ \frac{\cos (mx - mx) - \cos (mx + mx)}{2} \right] dx\\ \\ &= \frac{1}{2} \int^\pi_{-\pi} [\cos(0) - \cos(2mx)] dx\\ \\ &= \frac{1}{2} \int^\pi_{-\pi} [ 1 - \cos(mx)] dx\\ \\ &= \frac{1}{2} \left[ x - \frac{\sin(2mx)}{2m} \right]^\pi_{-\pi}\\ \\ &= \frac{1}{2} \left( \left[ \pi - \frac{\sin(2m\pi)}{m} \right] - \left[ (-\pi) - \frac{\sin(2m(-\pi))}{2m} \right] \right)\\ \\ &= \frac{1}{2} [ \pi + \pi]\\ \\ &= \frac{2\pi}{2} \\ \\ &= \pi \end{aligned} \end{equation} $