Answer
$$
\int \sec^2x\tan xdx
=\frac{1}{2}\tan^2x +c
$$
Work Step by Step
Since $ u=\tan x $, then $ du=\sec^2x dx $ and hence, $$
\int \sec^2x\tan xdx =\int u d u= \frac{1}{2}u^2 +c\\
=\frac{1}{2}\tan^2x +c
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 20
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