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: 21


$\dfrac{1}{\pi }e^{\sec \pi t} +c$

Work Step by Step

Solve $\int e^{\sec \pi t} \sec \pi t \tan \pi tdt$ Let us $p=\sec \pi t$ and $dp=(\sec \pi t \tan \pi t) \pi dt$ This implies $\int e^{\sec \pi t} \sec \pi t \tan \pi tdt=\dfrac{1}{\pi}\int e^p dp$ $=\dfrac{1}{\pi} e^p +c$ $=\dfrac{1}{\pi }e^{\sec \pi t} +c$ Hence,$\int e^{\sec \pi t} \sec \pi t \tan \pi tdt=\dfrac{1}{\pi }e^{\sec \pi t} +c$
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