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: 25


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

Work Step by Step

Let us consider $x = \sec \theta $ and $dx= \sec \theta \tan \theta d \theta $ and $\tan \theta=\sin \theta /\cos \theta =\sqrt {x^2-1}$ Now, the given integral becomes: $\int \dfrac{\sec \theta \tan \theta d \theta }{(\tan^3 \theta)} d\theta = = \int \dfrac{\cos \theta}{\sin^2 \theta} d \theta$ Let us substitute $u=\sin \theta \implies du =\cos \theta d \theta$ we have $=\dfrac{1}{u^2} du$ Now, integrate: $\dfrac{-1}{u}+C=\dfrac{-1}{\sin \theta}+C=-\csc \theta +C$ and plug in $\tan \theta=\sin \theta /\cos \theta =\sqrt {x^2-1} \implies \csc \theta=\dfrac{1}{\sin \theta} =\dfrac{x}{\sqrt {x^2 -1}}$: Thus, $=-\dfrac{x}{\sqrt {x^2 -1}}+C$
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