University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 28



Work Step by Step

$$A=\int^{\pi/3}_{0}\frac{4\sin\theta}{1-4\cos\theta}d\theta$$ We set $u=1-4\cos\theta$, which means $$du=-(-4\sin\theta)d\theta=4\sin\theta d\theta$$ For $\theta=\pi/3$, we have $$u=1-4\cos(\pi/3)=1-4\times\frac{1}{2}=-1$$ For $\theta=0$, we have $$u=1-4\cos0=1-4\times1=-3$$ Therefore, $$A=\int^{-1}_{-3}\frac{1}{u}du=\ln|u|\Big]^{-1}_{-3}$$ $$A=\ln1-\ln3=0-\ln3$$ $$A=-\ln3$$
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