Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 15

Answer

$$-\frac{1}{60} \cos^{10}6\theta +\frac{1}{72} \cos^{12}6\theta+C$$

Work Step by Step

\begin{aligned} \int \cos ^{9} 6 \theta \sin ^{3} 6 \theta d \theta &=\int \cos ^{9} 6 \theta \sin ^{2} 6 \theta \sin 6 \theta d \theta \\ &=\int \cos ^{9} 6 \theta\left(1-\cos ^{2} 6 \theta\right) \sin 6 \theta d \theta \\ &=\int \cos ^{9} 6 \theta \sin 6 \theta-\cos ^{11} 6 \theta \sin 6 \theta d \theta \\ &=\int \cos ^{9} 6 \theta \sin 6 \theta d \theta-\int \cos ^{11} 6 \theta \sin 6 \theta d \theta \\ &= -\frac{1}{60} \cos^{10}6\theta +\frac{1}{72} \cos^{12}6\theta+C \end{aligned}
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