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: 9


$\dfrac{1}{2} \ln|\dfrac{2x}{7} +\dfrac{\sqrt {4x^2-49}}{7}|+C$

Work Step by Step

Let us consider $t =\dfrac{7}{2} \sec \theta $ and $dt=\dfrac{7}{2} \sec\theta \tan \theta d \theta $ Now, the given integral becomes: $\int \dfrac{\dfrac{7}{2} \sec\theta \tan \theta d \theta}{\sqrt{4(\dfrac{49}{4} \sec^2 \theta-49)}} = \dfrac{1}{2} \int \sec \theta d \theta$ Now, integrate and plug in $\sec \theta =(2x/7)$: $ \dfrac{1}{2} \ln |\sec\theta + \tan \theta|+C=\dfrac{1}{2} \ln|\dfrac{2x}{7} +\dfrac{\sqrt {4x^2-49}}{7}|+C$
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