Answer
$\dfrac{2}{5} \sec \theta+C$
Work Step by Step
Here, we have $\int (\dfrac{2}{5} \sec \theta \tan \theta) d\theta= (\dfrac{2}{5})\int (\sec \theta \tan \theta)) d\theta$
or, $=\dfrac{2}{5} \sec \theta+C$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 239: 42
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