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: 31


$$\frac{-1+\sqrt 2}{5}.$$

Work Step by Step

Since $(\csc 5x )'=-5\csc 5x\cot 5x $, we have $$\int_{\pi/20}^{\pi / 10} \csc 5x\cot 5x d x=-\frac{1}{5}\csc 5x |_{\pi/20}^{\pi / 10}\\ =-\frac{1}{5}(\csc \frac{\pi}{2}-\csc \frac{\pi}{4})=\frac{-1+\sqrt 2}{5}.$$
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