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: 27


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

Work Step by Step

Given: $dy=\sqrt y \cos^2 \sqrt y dy$ Now separate the differentials and integrate. $\int dx=\int \dfrac{\sec^2 y dy}{\sqrt y}$ Plug in $a=\sqrt y \implies 2da=\dfrac{1}{\sqrt y}dy$ Then $\int \dfrac{\sec^2 y dy}{\sqrt y}=x+c \implies 2 \int sec^2 (a) da=x+c$ or, $2 \tan a=x+c \implies 2 \tan \sqrt y=x+c$ This implies that $\sqrt y=\tan^{-1} (x+c/2)$ Thus, $y=[\tan^{-1} (\dfrac{x+c}{2})]^2$
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