Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - Review - Exercises - Page 577: 7



Work Step by Step

Given $$\int_{0}^{\pi/2}\sin^3\theta\cos^2\theta d \theta$$ Since \begin{align*} \int_{0}^{\pi/2}\sin^3\theta\cos^2\theta d \theta&=\int_{0}^{\pi/2}\sin^2\theta\cos^2\theta \sin\theta d \theta\\ &=\int_{0}^{\pi/2}(1-\cos^2\theta)\cos^2\theta \sin\theta d \theta\\ &=\int_{0}^{\pi/2}\cos^2\theta \sin\theta d \theta-\int_{0}^{\pi/2}\cos^4\theta \sin\theta d \theta \\ &=-\frac{1}{3}\cos^3\theta + \frac{1}{5}\cos^5 \theta \bigg|_{0}^{\pi/2}\\ &=\frac{1}{3} - \frac{1}{5}\\ &=\frac{2}{15} \end{align*}
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