Answer
$\frac{\theta^3}{3}+\tan{\theta}+C$
Work Step by Step
$\int(\theta^2+\sec^2{\theta})d\theta=\int\theta^2d\theta+\int\sec^2{\theta}d\theta$
$=\frac{\theta^3}{3}+\tan{\theta}+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 4 - Integration - 4.1 Exercises - Page 251: 28
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