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: 39



Work Step by Step

$$A=\int^{\ln\sqrt3}_{0}\frac{e^xdx}{1+e^{2x}}$$ We set $u=e^x$, which means $$du=e^xdx$$ For $x=0$, we have $$u=e^0=1$$ For $x=\ln\sqrt3$, we have $$u=e^{\ln\sqrt3}=\sqrt3$$ Therefore, $$A=\int_1^{\sqrt3}\frac{1}{1+u^2}du=\tan^{-1}u\Big]_1^{\sqrt3}$$ $$A=(\tan^{-1}\sqrt3-\tan^{-1}1)$$ $$A=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$$
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