University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 41



Work Step by Step

Consider $x=\sec \theta \implies dx=\sec \theta \tan \theta d \theta $ Now, the given integral can be written as: $\int\dfrac{\sec^2 \theta \tan \theta}{ \sqrt{\tan^2 \theta}} d\theta= \int \sec^2 \theta$ or, $= \tan d \theta+C$ or, $=\sqrt{x^2-1}+C$
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