Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.3 Exercises - Page 530: 29


$\frac{\sec^7x}{7} - \frac{\sec^5x}{5}+C$

Work Step by Step

Find the indefinite integral $\int sec^5xtan^3xdx$ $\int sec^4xtan^2x secxtanxdx$ $\int sec^4x(sec^2x -1) secx tanx dx$ $ \int sec^6x secxtanxdx - \int sec^4xsecxtanxdx$ Use u-substitution, let $u=secx$, $du= secxtanxdx$ $\int u^6du - \int u^4du$, Integrate $ \frac{u^7}{7} - \frac{u^5}{5} +C$ $\frac{\sec^7x}{7} - \frac{\sec^5x}{5}+C$
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.