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: 87


$$\int^{3\pi}_{\pi}\cot^2\frac{x}{6} dx=6\sqrt3-2\pi$$

Work Step by Step

$$A=\int^{3\pi}_{\pi}\cot^2\frac{x}{6} dx$$ Recall the identity: $$\cot^2a=\csc^2a-1$$ Apply the identity here, we have $$A=\int^{3\pi}_{\pi}(\csc^2\frac{x}{6}-1)dx$$ $$A=\Big(-6\cot\frac{x}{6}-x\Big)\Big]^{3\pi}_{\pi}$$ $$A=\Big(-6\cot\frac{\pi}{2}-3\pi\Big)-\Big(-6\cot\frac{\pi}{6}-\pi\Big)$$ $$A=(-6\times0-3\pi)-(-6\sqrt3-\pi)$$ $$A=-3\pi+6\sqrt3+\pi$$ $$A=6\sqrt3-2\pi$$
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