Calculus (3rd Edition)

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

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 60


$\frac{3(1+\sqrt 2)}{5}.$

Work Step by Step

We have \begin{align*} \int_{\pi/2}^{\pi} \sin\left(\frac{5\theta -\pi}{6}\right) d \theta &=\frac{6}{5}\int_{\pi/2}^{\pi}\frac{5}{6} \sin\left(\frac{5\theta -\pi}{6}\right) d \theta \\ &=-\frac{6}{5 }\cos\left(\frac{5\theta -\pi}{6}\right)|_{\pi/2}^{\pi}\\ &=-\frac{6}{5 }\cos\left(\frac{4\pi}{6}\right)+\frac{6}{5 }\cos\left(\frac{3\pi}{12}\right)\\ &=\frac{3(1+\sqrt 2)}{5}. \end{align*}
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