Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 85



Work Step by Step

We have $$\int_0^{\pi/6}\tan 2\theta d\theta=\int_0^{\pi/6}\frac{\sin2\theta}{\cos2\theta} d\theta\\ =-\frac{1}{2}\int_0^{\pi/6}\frac{(\cos2\theta)'}{\cos2\theta}=-\frac{1}{2}\ln\cos2\theta|_0^{\pi/6}\\ =-\frac{1}{2}(\ln\cos(\pi/3)-\ln \cos 0)=-\frac{1}{2}\ln(1/2)=\frac{1}{2}\ln2.$$
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