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: 11


$(\sqrt {y^2-49}-7 \sec^{-1} (y/7))+C$

Work Step by Step

Let us consider $y = \sec \theta $ and $dy=7\sec\theta \tan \theta d \theta $ Now, the given integral becomes: $\int \dfrac{\sqrt{(49\sec^2 \theta-49)}}{7 \sec \theta} = 7 \int (\sec^2 \theta-1) d \theta$ Now, integrate and plug in $\sec \theta =(y/7)$: $ 7( \tan \theta -\theta)+C=(\sqrt {y^2-49}-7 \sec^{-1} (y/7))+C$
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