University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 58


$\sin 2x+\cos 3x+C$

Work Step by Step

Given: $ 2 \int (\cos 2x ) dx - 3 \int (\sin 3x) dx$ Thus, $2 \int (\cos 2x ) dx - 3 \int (\sin 3x) dx=(2)(\dfrac{1}{2}) \sin 2x -3\dfrac{1}{3}(\cot 3x)+C=\sin 2x+\cos 3x+C$
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