University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 26

Answer

$$\int\tan^7\frac{x}{2}\sec^2\frac{x}{2}dx=\frac{1}{4}\tan^8\frac{x}{2}+C$$

Work Step by Step

$$A=\int\tan^7\frac{x}{2}\sec^2\frac{x}{2}dx$$ We set $u=\tan\frac{x}{2}$ Then $$du=\frac{1}{2}\sec^2\frac{x}{2}dx$$ That means $$\sec^2\frac{x}{2}dx=2du$$ Therefore, $$A=2\int u^7du=\frac{2u^8}{8}+C=\frac{u^8}{4}+C$$ $$A=\frac{1}{4}\tan^8\frac{x}{2}+C$$
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.