University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 85


$$\int^{\pi/3}_{0}\sec^2\theta d\theta=\sqrt3$$

Work Step by Step

$$A=\int^{\pi/3}_{0}\sec^2\theta d\theta$$ $$A=\tan\theta\Big]^{\pi/3}_{0}$$ $$A=\tan\frac{\pi}{3}-\tan0$$ $$A=\sqrt3-0$$ $$A=\sqrt3$$
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