University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 66


$$\int \frac{dy}{(\sin^{-1}y)\sqrt{1-y^2}}=\ln|\sin^{-1}y|+C$$

Work Step by Step

$$A=\int \frac{dy}{(\sin^{-1}y)\sqrt{1-y^2}}$$ We set $u=\sin^{-1}y$ Then $$du=\frac{dy}{\sqrt{1-y^2}}$$ Therefore, $$A=\int \frac{1}{u}du=\ln|u|+C$$ $$A=\ln|\sin^{-1}y|+C$$
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