Answer
$$\frac{1 }{7}.$$
Work Step by Step
Since $(\cot7y )'=-7\csc^2 7y $, we have
$$\int_{\pi/28}^{\pi / 14} \csc^2 7y dy=-\frac{1}{7}\cot7y |_{\pi/28}^{\pi / 14} \\
=-\frac{1}{7}(\cot \frac{\pi}{2}-\cot \frac{\pi}{4})=\frac{1 }{7}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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