Answer
$y=\frac 1 2 arctan(\frac{tanx}2)+C$
Work Step by Step
$\frac{dy}{dx}=\frac{sec^2x}{4+tan^2x}$
$y=\int \frac{sec^2x}{4+tan^2x}dx$
$u=tan, du=sec^2xdx$
$y=\frac 1 2 arctan(\frac{tanx}2)+C$
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 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 513: 55
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