Answer
$y=tan(x)-x+C$
Work Step by Step
$\frac{dy}{dx}=tan^{2}(x)$
$y=\int tan^{2}(x)$
$y=\int \frac{sin^{2}(x)}{cos^{2}(x)}dx$
$y=\int \frac{1-cos^{2}(x)}{cos^{2}(x)}dx$
$y=\int \frac{1}{cos^{2}(x)}-1dx$
$y=tan(x)-x+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 6 - Differential Equations - 6.1 Exercises - Page 404: 48
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