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 338: 37


$$\int^{\pi/2}_{-\pi/2}\frac{2\cos\theta d\theta}{1+(\sin\theta)^2}=\pi$$

Work Step by Step

$$A=\int^{\pi/2}_{-\pi/2}\frac{2\cos\theta d\theta}{1+(\sin\theta)^2}$$ We set $u=\sin\theta$, which means $$du=\cos\theta d\theta$$ For $x=\pi/2$, we have $$u=\sin\frac{\pi}{2}=1$$ For $x=-\pi/2$, we have $$u=\sin\Big(-\frac{\pi}{2}\Big)=-1$$ Therefore, $$A=2\int^1_{-1}\frac{1}{1+u^2}du=2\tan^{-1}u\Big]^1_{-1}$$ $$A=2(\tan^{-1}1-\tan^{-1}(-1))$$ $$A=2\Big(\frac{\pi}{4}-\Big(-\frac{\pi}{4}\Big)\Big)$$ $$A=2\Big(\frac{\pi}{2}\Big)=\pi$$
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