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: 64


$$ -\frac{1}{2}(\cos^{-1}t)^2+c.$$

Work Step by Step

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