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: 20


$\dfrac{-\sqrt {9-w^2}}{w} -\sin^{-1} (w/3)+C$

Work Step by Step

Let us consider $w =3 \sin \theta $ and $dw=3 \cos \theta d\theta $ Now, the given integral becomes: $\int \dfrac{\sqrt{9-9 \sin^2 \theta}}{9 \sin^2 \theta} (3 \cos \theta) d \theta = \int (\csc^2 \theta -1)d \theta$ Now, integrate and plug in $w =3 \sin \theta $: $- \cot \theta -\theta+C=\dfrac{-\sqrt {9-w^2}}{w} -\sin^{-1} (w/3)+C$
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