Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 18


$$\int e^{3x}cos4x\,dx=\frac{1}{25}e^{3x}(4sin4x+3cos4x)+C$$

Work Step by Step

$$\int e^{3x}cos4x\,dx=\frac{1}{4}e^{3x}sin4x+\frac{3}{16}e^{3x}cos4x-\frac{9}{16}\int e^{3x}cos4x\,dx$$ $$\frac{25}{16}\int e^{3x}cos4x\,dx=\frac{1}{16}e^{3x}(4sin4x+3cos4x)$$ $$\int e^{3x}cos4x\,dx=\frac{1}{25}e^{3x}(4sin4x+3cos4x)+C$$
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