Answer
$$\frac{1}{8}\tan^8 \theta +c.$$
Work Step by Step
Let $u=\tan \theta$, then $du=\sec^2\theta d\theta$. We have
$$\int \sec^2\theta \tan^7 \theta d\theta= \int u^7 du=\frac{u^8}{8}+c\\=
\frac{1}{8}\tan^8 \theta +c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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