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: 61


$$\int\sin\theta\cos\theta\cos3\theta d\theta=\frac{1}{4}\cos\theta-\frac{1}{20}\cos5\theta+C$$

Work Step by Step

$$A=\int\sin\theta\cos\theta\cos3\theta d\theta=\frac{1}{2}\int(2\sin\theta\cos\theta)\cos3\theta d\theta$$ $$A=\frac{1}{2}\int\sin2\theta\cos3\theta d\theta$$ Use the identity: $$\sin mx\cos nx=\frac{1}{2}[\sin(m-n)x+\sin(m+n)x]$$ we have $$A=\frac{1}{4}\int\Big(\sin(-\theta)+\sin5\theta\Big)d\theta$$ $$A=\frac{1}{4}\int(-\sin\theta+\sin5\theta)d\theta$$ $$A=\frac{1}{4}(\cos\theta-\frac{1}{5}\cos5\theta)+C$$ $$A=\frac{1}{4}\cos\theta-\frac{1}{20}\cos5\theta+C$$
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