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: 19



Work Step by Step

Let us consider $w =2 \sin \theta $ and $dw=2 \cos \theta d\theta $ and $2 \cos \theta =\sqrt{4-w^2}$ Now, the given integral becomes: $\int \dfrac{(16\cos \theta)}{4 \sin^2 \theta} d \theta = 2 \int \csc^2 \theta d \theta$ Integrate and plug in $\cot \theta =\cos\theta/\sin \theta=\dfrac{\sqrt{4-w^2}}{w}$ Thus, $-2\cot \theta+C=-\dfrac{2\sqrt{4-w^2}}{w}+C$
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