Answer
$$ -\frac{1}{2}(\cos^{-1}t)^2+c.$$
Work Step by Step
Since $ u=\cos^{-1}t $, then $ du=\frac{-1}{\sqrt{1-t^2}}dt $.
Now, we have
$$\int \frac{\cos^{-1}t}{\sqrt{1-t^2}}dt =-\int udu\\
=-\frac{1}{2}u^2+c=-\frac{1}{2}(\cos^{-1}t)^2+c.$$
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: 64
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