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 338: 40



Work Step by Step

$$A=\int^{e^{\pi/4}}_{1}\frac{4dt}{t(1+\ln^2t)}$$ We set $u=\ln t$, which means $$du=\frac{dt}{t}$$ For $t=1$, we have $$u=\ln1=0$$ For $t=e^{\pi/4}$, we have $$u=\ln(e^{\pi/4})=\frac{\pi}{4}$$ Therefore, $$A=4\int_0^{\pi/4}\frac{1}{1+u^2}du=4\tan^{-1}u\Big]_0^{\pi/4}$$ $$A=4(\tan^{-1}\frac{\pi}{4}-\tan^{-1}0)$$ $$A=4\tan^{-1}\frac{\pi}{4}$$ *Note: Do not confuse $\tan\frac{\pi}{4}=1$ and $\tan^{-1}\frac{\pi}{4}\approx38.146^{o}$
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