University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Practice Exercises - Page 420: 3



Work Step by Step

$$A=\int^\pi_0\tan\frac{x}{3}dx=\int^\pi_0\frac{\sin\frac{x}{3}}{\cos\frac{x}{3}}dx$$ Set $u=\cos\frac{x}{3}$, which means $$du=-\frac{1}{3}\sin\frac{x}{3}dx$$ $$\sin\frac{x}{3}dx=-3du$$ - For $x=\pi$, we have $u=\cos\frac{\pi}{3}=\frac{1}{2}$ - For $x=0$, we have $u=\cos0=1$ Therefore, $$A=-3\int^{1/2}_1\frac{1}{u}du$$ $$A=-3\ln|u|\Big]^{1/2}_1$$ $$A=-3(\ln(1/2)-\ln1)$$ $$A=-3\ln(1/2)$$ $$A=-3(\ln1-\ln2)$$ $$A=-3(-\ln2)=3\ln2$$
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