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: 34


$\dfrac{-1}{7} (\dfrac{\sqrt{1-r^2}}{r})^7 +C$

Work Step by Step

Let us consider $r=\sin \theta \implies dr = \cos \theta d\theta$ Now, the given integral can be written as: $\int\dfrac{(1-\sin^2 \theta)^{5/2} \cos \theta d\theta}{\sin^8 \theta} =-\dfrac{1}{7} \cot^7 \theta+C $ or, $=\dfrac{-1}{7} (\dfrac{\sqrt{1-r^2}}{r})^7 +C$
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