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: 46


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

Work Step by Step

Consider $u=x^{3/2} \implies dx=\dfrac{2 du}{3\sqrt u}$ Now, the given integral can be written as: $\dfrac{2}{3}\dfrac{1}{\sqrt{1-u^2}} du=\dfrac{2}{3} \sin^{-1} u+C$ or, $=\dfrac{2}{3} \sin^{-1} (x^{3/2})+C$
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