Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 87



Work Step by Step

Let $u=\sin^{-1}x$, then $du=\frac{dx}{\sqrt{1-x^2}}$, hence we have $$\int \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx=\int udu=\frac{1}{2}u^2+c\\ =\frac{1}{2}(\sin^{-1}x)^2+c.$$
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