Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises: 20

Answer

$\frac{1}{4}\sec^{4}\theta +C$

Work Step by Step

$\int \tan\theta \sec^{4}\theta d\theta$ $=\int \tan\theta \sec\theta \sec^{3}\theta d\theta$ Use u-substitution. Let $u=\sec\theta$ $\frac{du}{dx}= \tan\theta \sec\theta$ $dx=\frac{1}{\tan\theta \sec\theta}du$ $=\int u^{3}du$ $=\frac{1}{4}u^{4}+C$ $=\frac{1}{4}\sec^{4}\theta +C$
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