Answer
$$\int\tan^2 xdx=\tan x-x+C$$
Work Step by Step
$$A=\int\tan^2 xdx$$
Remember that $\sec^2 x=1+\tan^2 x$. So, $\tan^2 x=\sec^2 x-1$ $$A=\int(\sec^2x-1)dx$$
Also, $\int\sec^2 x=\tan x+C$. Therefore, $$A=\tan x-x+C$$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 484: 23
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