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: 24


$\sin \pi -\sin 1$ or, $-\sin 1$

Work Step by Step

Solve $\int_{0}^{\sqrt{\ln \pi}} 2xe^{x^2} \cos (e^{x^2}) dx$ Let us $p=e^{x^2}$ and $dp=2pe^{x^2}dx$ This implies $\int_{0}^{\sqrt{\ln \pi}} 2xe^{x^2} \cos (e^{x^2}) dx=\int_{0}^{\sqrt{\ln \pi}} \cos p dp$ $=[sin p]_{0}^{\sqrt{\ln \pi}} +c$ $=[sin e^{x^2}]_{0}^{\sqrt{\ln \pi}} +c$ Also, $=[\sin e^{(\sqrt{\ln \pi})^2}- \sin e^0] +c$ or, $=\sin \pi -\sin 1$ or, $=-\sin 1$
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