University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 46


$\dfrac{3}{5} \sin 5\theta +C $

Work Step by Step

Given: $\int (3 \cos 5\theta) d\theta$ Thus, $\int (3 \cos 5\theta) d\theta=\dfrac{3}{5} \sin 5\theta +C $ and $\dfrac{d}{d \theta}(\dfrac{3}{5} \sin 5\theta +C )=3 \cos 5\theta$
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