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: 61



Work Step by Step

Given $$ \int_{0}^{\pi/6}\sin 2 x\cos 4xdx$$ Use $$ \sin (a x) \cos (b x)=\frac{1}{2} \sin ((a-b) x)+\frac{1}{2} \sin ((a+b) x) $$ Then \begin{align*} \int_{0}^{\pi/6}\sin 2 x\cos 4xdx&=\frac{1}{2}\int_{0}^{\pi}(\sin (- 2x)+ \sin 6x)dx\\ &=\frac{1}{2}\int_{0}^{\pi}( \sin 6x-\sin 2x)dx\\ &=\frac{ -1}{2}\left(\frac{1}{6}\cos6 x-\frac{1}{2}\cos2 x\right)\bigg|_{0}^{\pi} \\ &=\frac{1}{24} \end{align*}
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