Thomas' Calculus 13th Edition

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


$\dfrac{2}{5} \sec \theta+C$

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, $=\dfrac{2}{5} \sec \theta+C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.