Answer
$$\int\tan^2x\sec^2xdx=\frac{\tan^3x}{3}+C$$
Work Step by Step
$$A=\int\tan^2x\sec^2xdx$$
We set $u=\tan x$
Then $$du=\sec^2xdx$$
Therefore, $$A=\int u^2du=\frac{u^3}{3}+C$$ $$A=\frac{\tan^3x}{3}+C$$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
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: 24
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