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: 25


$$\int^{\pi/4}_{0}(1+e^{\tan\theta})\sec^2\theta d\theta=e$$

Work Step by Step

$$A=\int^{\pi/4}_{0}(1+e^{\tan\theta})\sec^2\theta d\theta$$ We set $u=\tan\theta$, which means $$du=\sec^2\theta d\theta$$ For $\theta=\pi/4$, we have $$u=\tan(\pi/4)=1$$ For $\theta=0$, we have $$u=\tan0=0$$ Therefore, $$A=\int^1_0(1+e^u)du=\int^1_01du+\int^1_0e^udu$$ $$A=u\Big]^1_0+e^u\Big]^1_0$$ $$A=1+(e-e^0)$$ $$A=1+e-1=e$$
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