Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 404: 63


$$\frac{1}{32}(12 x-8 \sin 2 x+\sin 4 x) +C=-\frac{1}{4}\sin^3 x\cos x-\frac{3}{8} \sin x\cos x+\frac{3}{8}x+C$$

Work Step by Step

Since \begin{align*} \sin 2x& =2\sin x\cos x\\ \cos 2x &=cos^2x-\sin ^2 x=1-2\sin^2x \end{align*} Then \begin{align*} \frac{1}{32}(12 x-8 \sin 2 x+\sin 4 x)+C&=\frac{1}{32}(12 x-16 \sin x\cos x+2\sin 2 x\cos 2x)+C\\ &=\frac{1}{32}\left(12 x-16 \sin x\cos x+4\sin x\cos x[1-2\sin^2 x]\right)+C\\ &=\frac{1}{32}\left(12 x-16 \sin x\cos x+4\sin x\cos x -8\sin^3 x\cos x \right)+C\\ &=-\frac{1}{4}\sin^3 x\cos x-\frac{3}{8} \sin x\cos x+\frac{3}{8}x+C \end{align*}
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