Answer
$$\frac{1}{2}(\sin^{-1}x)^2+c$$
Work Step by Step
Let $u=\sin^{-1}x$, then $du=\frac{dx}{\sqrt{1-x^2}}$, hence we have
$$\int \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx=\int udu=\frac{1}{2}u^2+c\\
=\frac{1}{2}(\sin^{-1}x)^2+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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