Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 354: 117


$\frac{e^{sin \left(\frac{\pi^{2}}{2}\right)}-1}{\pi}$

Work Step by Step

Solve for the indefinite integral $\int e^{sin \pi x} cos(\pi x)dx$ let $ sin (\pi x) =u$ $\pi cos(\pi x) dx=du$ $ cos(\pi x) dx=\frac{du}{\pi}$ $\int e^{sin \pi x} cos(\pi x)dx$ $=\frac{1}{\pi} \int e^udu$ $=\frac{1}{\pi}e^u+C$ $=\frac{1}{\pi}e^{sin (\pi x)}+C$ Solve for the definite integral $\int e^{sin \pi x} cos(\pi x)dx$ $[0, \frac{\pi}{2}]$ $=\frac{1}{\pi}e^{sin (\pi x)}$ $[0, \frac{\pi}{2}]$ $=\frac{1}{\pi}e^{sin (\frac{\pi^{2}}{2} )}-\frac{1}{\pi}e^{sin (0)}$ $=\frac{e^{sin (\frac{\pi^{2}}{2})}-1}{\pi}$
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