University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 92


$$\int^{\pi/2}_{-\pi/2}15(\sin^4 3x)\cos 3xdx=-2$$

Work Step by Step

$$A=\int^{\pi/2}_{-\pi/2}15(\sin^4 3x)\cos 3xdx$$ We set $u=\sin 3x$, which means $$du=3\cos 3xdx$$ $$\cos3xdx=\frac{du}{3}$$ For $x=\pi/2$, $u=\sin\frac{3\pi}{2}=-1$ and for $x=-\pi/2$, $u=\sin(-\frac{3\pi}{2})=\sin\frac{\pi}{2}=1$ Therefore, $$A=\int^{-1}_115u^4\frac{du}{3}=\int^{-1}_15u^4du$$ $$A=\frac{5u^5}{5}\Big]^{-1}_1=u^5\Big]^{-1}_1$$ $$A=(-1)^5-1^5$$ $$A=-1-1$$ $$A=-2$$
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