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: 26



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^2 \theta (\sec \theta \tan \theta)d \theta }{(tan^5 \theta)} =\int \dfrac{\cos \theta }{\sin^4 \theta } d \theta $ Let us substitute $u=\sin \theta \implies du =\cos \theta d \theta$ we have $=\dfrac{1}{u^4} du$ Now, integrate . $\dfrac{-u^3}{3}+C=\dfrac{-1}{3\sin^3 \theta}+C=-\dfrac{1}{3}\csc^3 \theta +C$ and plug in $\tan \theta=\sin \theta /\cos \theta =\sqrt {x^2-1} \implies \csc \theta=\dfrac{1}{\sin \theta} =\dfrac{x}{\sqrt {x^2 -1}}$: $-\dfrac{1}{3}\csc^3 \theta +C=-\dfrac{1}{3}(\dfrac{x}{\sqrt{x^2-1}})^3+C$
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