Answer
$$\frac{1}{4}\sin^{-1}(4x/3)+c$$
Work Step by Step
We have
$$\int \frac{dx}{\sqrt{9-16x^2}}=\int \frac{d(x/3)}{\sqrt{1-(4x/3)^2}}\\
=\frac{1}{4}\int \frac{d(4x/3)}{\sqrt{1-(4x/3)^2}}=\frac{1}{4}\sin^{-1}(4x/3)+c$$
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 - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 95
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