Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 43

Answer

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

Work Step by Step

$$A=\int\frac{dx}{\sqrt{1-x^2}\sin^{-1}x}dx$$ Let $u=\sin^{-1}x$ Then we have $du=\frac{1}{\sqrt{1-x^2}}dx$. Substitute into $A$: $$A=\int\frac{1}{u}du$$ $$A=\ln|u|+C$$ $$A=\ln|\sin^{-1}x|+C$$
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