Answer
$=-\frac{15(1-x^2)^{4/3}}{8}+C$
Work Step by Step
$let$ $1-x^2=u$
$du=-2x$ $dx$
$\int u^{1/3}$ $du$
$=-\frac{5}{2}\int (1-x^2)^{1/3}$$-2x$ $dx$
$=-\frac{5}{2}(\frac{3(1-x^2)^{4/3}}{4})+C$
$=-\frac{15(1-x^2)^{4/3}}{8}+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 4 - Integration - 4.5 Exercises - Page 301: 15
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