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: 61


$$\int \frac{e^{\sin^{-1}x}dx}{\sqrt{1-x^2}}=e^{\sin^{-1}x}+C$$

Work Step by Step

$$A=\int \frac{e^{\sin^{-1}x}dx}{\sqrt{1-x^2}}$$ We set $u=\sin^{-1}x$ Then $$du=\frac{1}{\sqrt{1-x^2}}dx$$ Therefore, $$A=\int e^udu=e^u+C$$ $$A=e^{\sin^{-1}x}+C$$
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