University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Practice Exercises - Page 420: 2


$$\int e^t\cos(3e^t-2)dt=\frac{\sin(3e^t-2)}{3}+C$$

Work Step by Step

$$A=\int e^t\cos(3e^t-2)dt$$ Set $u=3e^t-2$, which means $$du=3e^tdt$$ $$e^tdt=\frac{1}{3}du$$ Therefore, $$A=\frac{1}{3}\int\cos udu$$ $$A=\frac{\sin u}{3}+C$$ $$A=\frac{\sin(3e^t-2)}{3}+C$$
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