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 276: 52


$$\frac{1}{9} (\sin \theta)^{9}+C$$

Work Step by Step

Given $$\int \sin ^{8} \theta \cos \theta d \theta$$ Let $$ u=\sin \theta \ \ \ \ \Rightarrow \ \ \ du = \cos \theta d \theta$$ Then \begin{aligned} \int \sin ^{8} \theta \cos \theta d \theta &=\int u^{8} d u \\ &=\frac{1}{9} u^{9}+C\\ &=\frac{1}{9} (\sin \theta)^{9}+C \end{aligned}
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