Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 19


$$ \int \sin^2\theta \cos \theta d\theta =\frac{1}{3}\sin^3\theta +c $$

Work Step by Step

Since $ u=\sin\theta $, then $ du=\cos \theta d\theta $ and hence, $$ \int \sin^2\theta \cos \theta d\theta=\int u^2 d u= \frac{1}{3}u^3 +c\\ =\frac{1}{3}\sin^3\theta +c. $$
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