Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 21

Answer

\begin{aligned} \int 8 \cos ^{3} 2 \theta \sin 2 \theta \ d \theta =-\cos ^{4} 2 \theta+C \end{aligned}

Work Step by Step

Given $$ \int 8 \cos ^{3} 2 \theta \sin 2 \theta \ d \theta $$ So, we have \begin{aligned} I&=\int 8 \cos ^{3} 2 \theta \sin 2 \theta \ d \theta\\ &=8\int (-\frac{1}{2}) (\cos 2 \theta )^3(-2\sin 2 \theta) \ d \theta\\ &=-4\frac{\cos ^{4} 2 \theta}{4}+C\\ &=-\cos ^{4} 2 \theta+C \end{aligned}
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