University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 17


$$\int^{\pi/6}_0\cos^{-3}2\theta\sin2\theta d\theta=\frac{3}{4}$$

Work Step by Step

$$A=\int^{\pi/6}_0\cos^{-3}2\theta\sin2\theta d\theta$$ We set $u=\cos2\theta$, which means $$du=-2\sin2\theta d\theta$$ $$\sin2\theta d\theta=-\frac{1}{2}du$$ For $\theta=0$, we have $u=\cos0=1$ For $\theta=\pi/6$, we have $u=\cos(\pi/3)=1/2$ Therefore, $$A=-\frac{1}{2}\int^{1/2}_1u^{-3}du=-\frac{1}{2}\times\frac{u^{-2}}{-2}\Big]^{1/2}_1$$ $$A=\frac{1}{4u^2}\Big]^{1/2}_1$$ $$A=\frac{1}{4}\Big(\frac{1}{\frac{1}{2^2}}-1\Big)=\frac{1}{4}(4-1)$$ $$A=\frac{3}{4}$$
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