Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 29


$$\frac{4\sqrt 3}{9}.$$

Work Step by Step

Since $(\tan\left(3 t-\frac{\pi}{6}\right) )'=3\sec^{2}\left(3 t-\frac{\pi}{6}\right) $, we have $$\int_{0}^{ \pi/6 }\sec ^{2}\left(3 t-\frac{\pi}{6}\right) d t= \frac{1}{3}\tan\left(3 t-\frac{\pi}{6}\right)|_{0 }^{ \pi/6 }\\ =\frac{1}{3} \tan\left(\frac{\pi}{2}-\frac{\pi}{6}\right)+\frac{1}{3}\tan\left(\frac{\pi}{6}\right)\\ =\frac{1}{3}\tan\left(\frac{\pi}{3}\right)+\frac{1}{3}\tan\left(\frac{\pi}{6}\right)=\frac{1}{3}[\sqrt 3+\frac{\sqrt 3}{3}]\\ =\frac{4\sqrt 3}{9}.$$
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