Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 252: 28


$$\frac{1}{2}\theta^2+\tan\theta +C$$

Work Step by Step

$\int(\theta+\sec^2\theta)d\theta$ Given $$\int{x}dx=\frac{1}{2}x^2+C$$ and$$\int\sec^2{x}dx=\tan{x}+C$$ Thus, $\int(\theta+\sec^2\theta)d\theta$ $=\frac{1}{2}\theta^2+\tan\theta +C$
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