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: 85



Work Step by Step

Let $u=4x$, then $du =4dx$. Now, we have $$\int \frac{dx}{\sqrt{1-16x^2}}=\frac{1}{4}\int \frac{du}{\sqrt{1-u^2}}\\ =\frac{1}{4}\sin^{-1}u+c=\frac{1}{4}\sin^{-1}(4x)+c.$$
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