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: 22


$$\int\sin2x\cos3xdx=\frac{\cos x}{2}-\frac{\cos5x}{10}+C$$

Work Step by Step

$$A=\int\sin2x\cos3xdx$$ 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 $a=2$ and $b=3$: $$A=-\frac{\cos5x}{2\times5}-\frac{\cos(-x)}{2\times(-1)}+C$$ We have $\cos(-x)=\cos x$ $$A=-\frac{\cos5x}{10}-\frac{\cos x}{(-2)}+C$$ $$A=\frac{\cos x}{2}-\frac{\cos5x}{10}+C$$
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