University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 88


$$\int^{\pi}_{0}\tan^2\frac{\theta}{3} d\theta=3\sqrt3-\pi$$

Work Step by Step

$$A=\int^{\pi}_{0}\tan^2\frac{\theta}{3} d\theta$$ Recall the identity: $$\tan^2a=\sec^2a-1$$ Apply the identity here, we have $$A=\int^{\pi}_{0}(\sec^2\frac{\theta}{3}-1)d\theta$$ $$A=\Big(3\tan\frac{\theta}{3}-\theta\Big)\Big]^{\pi}_{0}$$ $$A=\Big(3\tan\frac{\pi}{3}-\pi\Big)-\Big(3\tan0-0\Big)$$ $$A=3\sqrt3-\pi-(0-0)$$ $$A=3\sqrt3-\pi$$
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.