University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.5 - Integral Tables and Computer Algebra Systems - Exercises - Page 451: 21

Answer

$$\int\sin3x\cos2xdx=-\frac{\cos5x}{10}-\frac{\cos x}{2}+C$$

Work Step by Step

$$A=\int\sin3x\cos2xdx$$ Use Formula 69a, which states that $$\int\sin ax\cos bxdx=-\frac{\cos(a+b)x}{2(a+b)}-\frac{\cos(a-b)x}{2(a-b)}+C$$ with $a=3$ and $b=2$: $$A=-\frac{\cos5x}{2\times5}-\frac{\cos x}{2\times1}+C$$ $$A=-\frac{\cos5x}{10}-\frac{\cos x}{2}+C$$
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