University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 42



Work Step by Step

$$A=\int\sin2x\cos4xdx$$ Recall the identity $$\sin a\cos b=\frac{\sin(a+b)+\sin(a-b)}{2}$$ Therefore, $$A=\int\frac{\sin6x+\sin(-2x)}{2}$$ Since $\sin(-a)=-\sin a$, $\sin(-2x)=-\sin2x$ $$A=\int\frac{\sin6x-\sin2x}{2}$$ $$A=\frac{1}{2}(\int \sin6x-\int\sin2x)$$ $$A=\frac{1}{2}\Big(-\frac{\cos6x}{6}+\frac{\cos2x}{2}\Big)+C$$ $$A=\frac{\cos2x}{4}-\frac{\cos6x}{12}+C$$
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