Answer
$$\frac{1}{4} \sin^{-1}(4t)+c$$
Work Step by Step
Since $u=4t$, then $du= 4dt$, and hence we have
$$\int\frac{ dt}{\sqrt{1-16t^2}}=\frac{1}{4}\int \frac{ du}{\sqrt{1-u^2}}=\frac{1}{4} \sin^{-1}u+c\\
=\frac{1}{4} \sin^{-1}(4t)+c.$$
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: 61
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
