University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 41

Answer

$\sqrt{x^2-1}+C$

Work Step by Step

Consider $x=\sec \theta \implies dx=\sec \theta \tan \theta d \theta $ Now, the given integral can be written as: $\int\dfrac{\sec^2 \theta \tan \theta}{ \sqrt{\tan^2 \theta}} d\theta= \int \sec^2 \theta$ or, $= \tan d \theta+C$ or, $=\sqrt{x^2-1}+C$
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.