Answer
$$\tan1.$$
Work Step by Step
Let $u=\cos \theta $, then $du=-\sin\theta d\theta$ and $u:1\to 0$. Now we have
$$\int_{0}^{\pi / 2} \sec ^{2}(\cos \theta) \sin \theta d \theta=-\int_1^0\sec^2u du\\
=\int_0^1\sec^2u du=\tan u|_0^1=\tan1.$$
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 - Chapter Review Exercises - Page 278: 69
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