University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 66

Answer

$\theta+\tan \theta +C$

Work Step by Step

Given: $\int 2 d\theta + \int (\tan^2 \theta )d\theta$ Thus, $\int 2 d\theta + \int (\tan^2 \theta )d\theta= 2 \theta + \int (\sec^2 \theta )d\theta -\int 1 d\theta=\theta+\tan \theta +C$
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