Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 32


$$\frac{1 }{7}.$$

Work Step by Step

Since $(\cot7y )'=-7\csc^2 7y $, we have $$\int_{\pi/28}^{\pi / 14} \csc^2 7y dy=-\frac{1}{7}\cot7y |_{\pi/28}^{\pi / 14} \\ =-\frac{1}{7}(\cot \frac{\pi}{2}-\cot \frac{\pi}{4})=\frac{1 }{7}.$$
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