Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.4 - Trigonometric Substitutions - Exercises 8.4 - Page 467: 39

Answer

$${\sec ^{ - 1}}\left| x \right| + C $$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} \cr & {\text{Using Trigonometric Substitutions}} \cr & {\text{We set}}{\text{, }}x = \sec \theta,{\text{ }}dx = \sec \theta \tan \theta d\theta,{\text{ }}0 \leqslant \theta \leqslant \frac{\pi }{2} \cr & {\text{With these substitutions}}{\text{, we have}} \cr & \int {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} = \int {\frac{{\sec \theta \tan \theta d\theta }}{{\left( {\sec \theta } \right)\sqrt {{{\left( {\sec \theta } \right)}^2} - 1} }}} \cr & = \int {\frac{{\sec \theta \tan \theta d\theta }}{{\sec \theta \sqrt {{{\sec }^2}\theta - 1} }}} \cr & {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr & = \int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\sec }^2}\theta \sqrt {{{\tan }^2}\theta } }}} \cr & {\text{simplifying}} \cr & = \int {\frac{{\sec \theta \tan \theta d\theta }}{{\sec \theta \tan \theta }}} \cr & = \int {\theta d\theta } \cr & {\text{integrate}} \cr & = \theta + C \cr & {\text{write in terms of }}x,{\text{ }}\sec \theta = x\,\,\,\, \Rightarrow \,\,\,\theta = {\sec ^{ - 1}}\left| x \right| \cr & = {\sec ^{ - 1}}\left| x \right| + C \cr} $$
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.