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: 39


$sec^{-1} |x|+C$

Work Step by Step

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