University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 23



Work Step by Step

Solve $\int_{\ln(\pi/6)}^{\ln(\pi/2)} 2e^{v} \cos e^v dv$ Let us $p=e^{v}$ and $dp=e^{v} dv$ This implies $\int_{\ln(\pi/6)}^{\ln(\pi/2)} 2e^{v} \cos e^v dv=2\int_{\ln(\pi/6)}^{\ln(\pi/2)} \cos p dp$ $=2[sin p]_{\ln(\pi/6)}^{\ln(\pi/2)} +c$ $=2[sin e^{v}]_{\ln(\pi/6)}^{\ln(\pi/2)} +c$ Also, $=2[sin e^{\ln(\pi/2)}-sin e^{\ln(\pi/6)}] +c$ or, $=2[sin (\pi/2)-sin (\pi/6)] +c$ $=2(1-0.5)$ $=1$
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