Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 547: 7

Answer

$$e^{\pi/4 }-e^{-\pi/4 }$$

Work Step by Step

Given $$\int_{-1}^{1} \frac{e^{\tan^{-1}y}}{1+y^2}dy$$ Let $u=\tan^{-1}y\ \ \Rightarrow \ \ du=\dfrac{dy}{1+y^2}$, at $y=-1\to u=-\pi/4$, at $y=1\to u=\pi/4$ then \begin{align*} \int_{-1}^{1} \frac{e^{\tan^{-1}y}}{1+y^2}dy&=\int_{-\pi/4}^{\pi/4} e^{u}du\\ &=e^u\bigg|_{-\pi/4}^{\pi/4}\\ &=e^{\pi/4 }-e^{-\pi/4 } \end{align*}
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