Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 61


$$\frac{1}{4} \sin^{-1}(4t)+c$$

Work Step by Step

Since $u=4t$, then $du= 4dt$, and hence we have $$\int\frac{ dt}{\sqrt{1-16t^2}}=\frac{1}{4}\int \frac{ du}{\sqrt{1-u^2}}=\frac{1}{4} \sin^{-1}u+c\\ =\frac{1}{4} \sin^{-1}(4t)+c.$$
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