Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises: 5

Answer

$-\frac{1}{14}\cos^7(2t)+\frac{1}{5}\cos^5(2t)-\frac{1}{6}\cos^3(2t)+C$

Work Step by Step

$\int \sin^5(2t)\cos^2(2t)\ dt$ Let $u=2t$. Then $du=2\ dt$, and $dt=\frac{1}{2}du$. $=\int\sin^5 u\cos^2 u*\frac{1}{2}du$ $=\frac{1}{2}\int\sin^5 u\cos^2 u\ du$ Since the power of sine is odd, save one factor of $\sin u$ and express the remaining factors in terms of $\cos u$: $=\frac{1}{2}\int\sin^4 u\cos^2 u\sin u\ du$ $=\frac{1}{2}\int(\sin^2 u)^2\cos^2 u\sin u\ du$ $=\frac{1}{2}\int(1-\cos^2 u)^2\cos^2 u\sin u\ du$ Let $v=\cos u$. Then $dv=-\sin u\ du$, and $\sin u\ du=-dv$. $=\frac{1}{2}\int(1-v^2)^2 v^2*(-1)\ dv$ $=-\frac{1}{2}\int(1-2v^2+v^4)v^2\ dv$ $=-\frac{1}{2}\int(v^6-2v^4+v^2)\ dv$ $=-\frac{1}{2}(\frac{1}{7}v^7-\frac{2}{5}v^5+\frac{1}{3}v^3+C)$ $=-\frac{1}{14}v^7+\frac{1}{5}v^5-\frac{1}{6}v^3+C$ $=-\frac{1}{14}\cos^7 u+\frac{1}{5}\cos^5 u-\frac{1}{6}\cos^3 u+C$ $=\boxed{-\frac{1}{14}\cos^7(2t)+\frac{1}{5}\cos^5(2t)-\frac{1}{6}\cos^3(2t)+C}$
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