University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 342: 42


$y=\arctan (x) -2 \arcsin x+2$

Work Step by Step

We need to integrate: $y=\int \dfrac{dx}{1+x^2} \int \dfrac{2dx}{\sqrt{1-x^2}}$ We use the differential trigonometric formula: $\int \dfrac{dx}{\sqrt{x^2-a^2}}=\arcsin (x/a)+C$ and $\int \dfrac{1}{x^2+a^2}=\arctan (x/a)+C$ $y=\arctan (x) -2 \arcsin x+c$ Take initial conditions: $y(0)=2$ Then $y(0)=\arctan (0) -2 \arcsin 0+c =2 \implies c=2$ Thus, we have $y=\arctan (x) -2 \arcsin x+2$
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