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: 30


$$\frac{2\sqrt 3}{3}-1.$$

Work Step by Step

Since $(\sec \theta )'=\sec \theta \tan \theta $, we have $$\int_{0}^{\pi / 6} \sec \theta \tan \theta d \theta=\sec \theta |_{0}^{\pi / 6} \\ =\sec {\pi / 6}-\sec {0} =\frac{2\sqrt 3}{3}-1.$$
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