Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 426: 33

Answer

$\ln \left| {\tan \theta } \right| + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{{{\sec }^2}\theta }}{{\tan \theta }}} d\theta \cr & {\text{Let }}u = \tan \theta ,{\text{ then }}du = {\sec ^2}\theta d\theta ,{\text{ then substituting}} \cr & \int {\frac{{{{\sec }^2}\theta }}{{\tan \theta }}} d\theta = \int {\frac{{du}}{u}} \cr & {\text{Integrate }} \cr & = \ln \left| u \right| + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}u = \tan \theta \cr & = \ln \left| {\tan \theta } \right| + C \cr} $$
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