University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 69


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

Work Step by Step

$$A=\int \frac{e^{\sin^{-1}\sqrt x}dx}{2\sqrt{x-x^2}}$$ We set $a=\sin^{-1}\sqrt x$, which means $$da=\frac{(\sqrt x)'}{\sqrt{1-(\sqrt x)^2}}dx=\frac{1}{2\sqrt x\sqrt{1-x}}dx$$ $$da=\frac{1}{2\sqrt {x(1-x)}}dx=\frac{dx}{2\sqrt{x-x^2}}$$ Therefore, $$A=\int e^ada$$ $$A=e^a+C$$ $$A=e^{\sin^{-1}\sqrt x}+C$$
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