Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 24


$$\int \sec 3\theta \tan 3\theta d\theta =\frac{1}{3} \sec u+c=\frac{1}{3}\sec 3\theta +c $$

Work Step by Step

Let $ u=3\theta $, then $ du=3d\theta $ and since we have $(\sec x)'=\sec x\tan x $ then we get $$\int \sec 3\theta \tan 3\theta d\theta =\frac{1}{3}\int \sec u\tan u du=\frac{1}{3} \sec u+c=\frac{1}{3}\sec 3\theta +c $$
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