Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 41


$$ x-\frac{1}{4}\sin 2x +\frac{1}{8}\sin4x-\frac{1}{3}\cos 3x +\cos x+C$$

Work Step by Step

We integrate as follows: \begin{align*} \int(\sin x+\cos 2 x)^{2} d x&=\int(\sin^2 x+\cos^2 2 x+2\sin x\cos 2x) d x\\ &= \int\left(\frac{1}{2}-\frac{1}{2}\cos2 x+\frac{1}{2} +\frac{1}{2}\cos4 x+\sin 3x-\sin x\right) d x\\ &= x-\frac{1}{4}\sin 2x +\frac{1}{8}\sin4x-\frac{1}{3}\cos 3x +\cos x+C \end{align*}
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