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: 69



Work Step by Step

Let $u=\cos \theta $, then $du=-\sin\theta d\theta$ and $u:1\to 0$. Now we have $$\int_{0}^{\pi / 2} \sec ^{2}(\cos \theta) \sin \theta d \theta=-\int_1^0\sec^2u du\\ =\int_0^1\sec^2u du=\tan u|_0^1=\tan1.$$
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