Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 79


$$\frac{1}{8}\tan^8 \theta +c.$$

Work Step by Step

Let $u=\tan \theta$, then $du=\sec^2\theta d\theta$. We have $$\int \sec^2\theta \tan^7 \theta d\theta= \int u^7 du=\frac{u^8}{8}+c\\= \frac{1}{8}\tan^8 \theta +c.$$
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