Calculus 10th Edition

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: 55

Answer

$y=\frac 1 2 arctan(\frac{tanx}2)+C$

Work Step by Step

$\frac{dy}{dx}=\frac{sec^2x}{4+tan^2x}$ $y=\int \frac{sec^2x}{4+tan^2x}dx$ $u=tan, du=sec^2xdx$ $y=\frac 1 2 arctan(\frac{tanx}2)+C$
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