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: 13


$\dfrac{\sqrt{x^2 -1}}{x}+C$

Work Step by Step

Let us consider $x = \sec \theta $ and $dx= \sec\theta \tan \theta d \theta $ Now, the given integral becomes: $\int \dfrac{ \sec \theta \tan \theta }{ \sec^2 \theta\sqrt{\sec^2 \theta -1}} d \theta = \int cos \theta d \theta$ Now, integrate and plug in $\sec \theta =x$: $ \sin \theta+C=\dfrac{\sqrt{x^2 -1}}{x}+C$
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