Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises: 51

Answer

$$\tan \left( {\ln \theta } \right) + \frac{{{{\tan }^3}\left( {\ln \theta } \right)}}{3} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{{\sec }^4}\ln \left( \theta \right)}}{\theta }d\theta } \cr & {\text{substitute }}u = \ln \theta ,{\text{ }}du = \frac{1}{\theta }d\theta \cr & = \int {{{\sec }^4}u} du \cr & {\text{split off }}{\sec ^4}u \cr & = \int {{{\sec }^2}u} {\sec ^2}udu \cr & {\text{pythagorean identity}} \cr & = \int {\left( {1 + {{\tan }^2}u} \right)} {\sec ^2}udu \cr & = \int {\left( {{{\sec }^2}u + {{\tan }^2}u{{\sec }^2}u} \right)} du \cr & {\text{evaluate}} \cr & = \tan u + \frac{{{{\tan }^3}u}}{3} + C \cr & {\text{replace }}u = \ln \theta \cr & = \tan \left( {\ln \theta } \right) + \frac{{{{\tan }^3}\left( {\ln \theta } \right)}}{3} + C \cr} $$
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.