Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 484: 8

Answer

$$\int_{0}^{2\pi}sin^{2}\frac{\theta}{3}d\theta=\pi+\frac{3\sqrt{3}}{8}$$

Work Step by Step

$$\int_{0}^{2\pi}sin^{2}\frac{\theta}{3}d\theta=\frac{1}{2}\int_{0}^{2\pi}(1-cos\frac{2\theta}{3})d\theta$$$$=\frac{1}{2}\left [ \theta -\frac{3}{2}sin\frac{2\theta }{3}\right ]_{0}^{2\pi}=\pi+\frac{3\sqrt{3}}{8}$$
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