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: 23

Answer

$4 \sqrt 3-\dfrac{4 \pi}{3}$

Work Step by Step

Let us consider $x = \sin \theta $ and $dx=\cos \theta d \theta $ Now, the given integral becomes: $\int \dfrac{4 \sin^2 \theta \cos \theta d \theta }{(1- \sin^2 \theta)^{3/2}} = 4\int_0^{\pi/3} (\sec^2 \theta -1) d \theta$ Now, integrate and use $x = \sin \theta $: $[4(\tan \theta -\theta)+C]_0^{\pi/3}=4 \sqrt 3-\dfrac{4 \pi}{3}$
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