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: 27



Work Step by Step

Let us consider $x = \sin \theta $ and $dx= \cos \theta d \theta $ and $\cos \theta=\sqrt {1-x^2} \implies \cos^3 \theta=(\sqrt {1-x^2})^3$ Now, the given integral becomes: $\int\dfrac{(\cos^3 \theta)(\cos \theta)}{\sin ^6 \theta} =\int \dfrac{\cos^4 \theta}{\sin ^6 \theta} =\int csc^2 \theta (\cot^4 \theta) d \theta $ Now, integrate and plug in $\dfrac{\cos \theta }{\sin \theta }=\cot \theta =\dfrac{\sqrt{1-x^2}}{x}$ $-\dfrac{\cot^5 \theta}{5} +C=-\dfrac{1}{5}(\dfrac{\sqrt{1-x^2}}{x})^5+C$
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