University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Practice Exercises - Page 421: 29


$y^2=\sin^{-1} (2 \tan x+c) $

Work Step by Step

Given: $y \dfrac{dy}{dx}=\sec y^2\sec^2 x$ Separate the differentials and integrate. $\int \sec^2 xdx=\int y \cos y^2 dy$ or, $\int \sec^2 xdx=(\dfrac{1}{2} \int \cos y^2 (2y) dy$ This implies that $(\dfrac{1}{2}) \sin y^2=\tan x+c$ Thus, $\sin^{1} (\sin y^2)=\sin^{-1} (2 \tan x+c) \implies y^2=\sin^{-1} (2 \tan x+c) $
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