University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 435: 56


$$\int^{\pi/2}_{-\pi/2}\cos x\cos 7xdx=0$$

Work Step by Step

$$A=\int^{\pi/2}_{-\pi/2}\cos x\cos 7xdx$$ Apply the identity of the product of 2 cosine functions: $$\cos mx\cos nx=\frac{1}{2}[\cos(m-n)x+\cos(m+n)x]$$ we have $$A=\frac{1}{2}\int^{\pi/2}_{-\pi/2}\Big[\cos(-6x)+\cos8x\Big]dx$$ Since $\cos(-x)=\cos x$, $$A=\frac{1}{2}\int^{\pi/2}_{-\pi/2}(\cos6x+\cos8x)dx$$ $$A=\frac{1}{2}\Big(\frac{1}{6}\sin6x+\frac{1}{8}\sin8x\Big)\Big]^{\pi/2}_{-\pi/2}$$ $$A=\Big(\frac{1}{12}\sin6x+\frac{1}{16}\sin8x\Big)\Big]^{\pi/2}_{-\pi/2}$$ $$A=\frac{1}{12}\sin3\pi+\frac{1}{16}\sin4\pi-\Big(\frac{1}{12}\sin(-3\pi)+\frac{1}{16}\sin(-4\pi)\Big)$$ $$A=0+0-(0+0)=0$$
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