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: 63


$$\frac{1}{\sqrt 3}\sin^{-1}\left(\sqrt{\frac{3}{5}}t\right)+c.$$

Work Step by Step

Since $u=\sqrt{\frac{3}{5}}t$, then $du= \sqrt{\frac{3}{5}}dt$, and hence we have $$\int \frac{ dt}{\sqrt{5-3t^2}}=\sqrt{\frac{5}{3}}\int \frac{ du}{\sqrt{5-5u^2}}=\frac{1}{\sqrt 3}\int \frac{ du}{\sqrt{1-u^2}}\\ =\frac{1}{\sqrt 3}\sin^{-1}u+c=\frac{1}{\sqrt 3}\sin^{-1}\left(\sqrt{\frac{3}{5}}t\right)+c.$$
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.