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: 42


$\frac{1}{4} (\tan 2\theta)^2+c$

Work Step by Step

Since $u=\tan 2\theta $, then $du=2\sec^2 2\theta d\theta $. Now, we have $$ \int \sec^2(2\theta) \tan(2\theta) d\theta=\frac{1}{2}\int udu=\frac{1}{4} u^2+c\\ =\frac{1}{4} (\tan 2\theta)^2+c. $$
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