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: 60


$$\frac{1}{6} \tan^{-1}\frac{8}{3}$$

Work Step by Step

Since $u=\frac{2}{3}t$, then $du= \frac{2}{3}dt$, and hence when $t:0\to 4 $, then $u:0\to 8/3$. Now, we have $$\int_ 0^4\frac{ dt}{4t^2+9}=\frac{1}{6}\int_0^{8/3} \frac{ du}{u^2+1}=\frac{1}{6} \tan^{-1}u|_0^{8/3}\\ =\frac{1}{6} \tan^{-1}\frac{8}{3}-\frac{1}{6} \tan^{-1}0=\frac{1}{6} \tan^{-1}\frac{8}{3}.$$
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