Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises - Page 524: 9


$$\frac{3\pi}{8} $$

Work Step by Step

Given $$ \int_{0}^{\pi}\cos^4(2t)dt$$ since $$\cos^2x=\frac{1}{2}(1+\cos 2x) $$ then \begin{align*} \int_{0}^{\pi}\cos^4(2t)dt&=\frac{1}{4}\int_{0}^{\pi}[1+\cos (4t)]^2dt\\ &=\frac{1}{4}\int_{0}^{\pi}[1+2\cos (4t)+\cos^2(4t)] dt\\ &=\frac{1}{4}\int_{0}^{\pi}[1+2\cos (4t)+\frac{1}{2}(1+\cos(8t))] dt\\ &=\frac{1}{4}\int_{0}^{\pi}[\frac{3}{2}+2\cos (4t)+ \cos(8t) ] dt\\ &=\frac{1}{4}\left(\frac{3}{2}t+\frac{1}{2}\sin (4t)+\frac{1}{8} \sin(8t)\right)\bigg|_{0}^{\pi}\\ &=\frac{3\pi}{8} \end{align*}
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