Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 279: 89


$$(\sin x)^3 \cos x. $$

Work Step by Step

By making use of Theorem 1 (FTC II), we have $$ G'(x)=\frac{d}{d x} \int_{-2}^{\sin x} t^3 dt =\left(\frac{d}{d u} \int_{-2}^{u} t^3 dt\right)\frac{du}{dx}=(\sin x)^3 (\cos x)\\ =(\sin x)^3 \cos x. $$
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