Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 7



Work Step by Step

Given $$\int \frac{d x}{x \sqrt{x^{2}-9}} $$ Let $$x=3\sec \theta \to dx= 3\sec \theta \tan \theta d \theta $$ Then \begin{aligned} \int \frac{d x}{x \sqrt{x^{2}-9}} &=\int \frac{3 \sec \theta \tan \theta d \theta}{3 \sec \theta \sqrt{(3 \sec \theta)^{2}-9}} \\ &=\int \frac{3 \sec \theta \cdot \tan \theta d \theta}{3 \sec \theta \sqrt{9 \sec ^{2} \theta-9}} \\ &=\int \frac{\tan \theta d \theta}{3 \sqrt{\sec ^{2} \theta-1}} \\ &=\int \frac{\tan \theta d \theta}{3 \tan ^{2} \theta-1} \\ &=\int \frac{\tan \theta d \theta}{3 \tan ^{2} \theta} \\ &=\int \frac{\tan \theta d \theta}{3 \tan \theta} \\ &=\frac{1}{3} \int d \theta \\ &=\frac{1}{3} \theta+C\\ &=\frac{1}{3}\sec^{-1}\left(\frac{x}{3}\right)+C \end{aligned}
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.