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: 95

Answer

$$\frac{1}{4}\sin^{-1}(4x/3)+c$$

Work Step by Step

We have $$\int \frac{dx}{\sqrt{9-16x^2}}=\int \frac{d(x/3)}{\sqrt{1-(4x/3)^2}}\\ =\frac{1}{4}\int \frac{d(4x/3)}{\sqrt{1-(4x/3)^2}}=\frac{1}{4}\sin^{-1}(4x/3)+c$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.