Answer
$$\frac{\tan ^{5} x}{5}+C $$
Work Step by Step
Let $ u= \tan x\ \ \ \Rightarrow \ du=\sec^2xdx $, then
\begin{align*}
\int \tan ^{4} x \sec ^{2} x d x&=\int u ^{4} d u\\
&=\frac{u ^{5} }{5}+C \\
&=\frac{\tan ^{5} x}{5}+C
\end{align*}
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 8: Techniques of Integration - Practice Exercises - Page 518: 39
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