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: 38



Work Step by Step

$$A=\int^{\pi/4}_{\pi/6}\frac{\csc^2xdx}{1+\cot^2x}$$ We set $u=\cot x$, which means $$du=-\csc^2x dx$$ $$\csc^2xdx=-du$$ For $x=\pi/4$, we have $$u=\cot\frac{\pi}{4}=1$$ For $x=\pi/6$, we have $$u=\cot\frac{\pi}{6}=\sqrt3$$ Therefore, $$A=-\int^1_{\sqrt3}\frac{1}{1+u^2}du=-\tan^{-1}u\Big]^1_{\sqrt3}$$ $$A=-(\tan^{-1}1-\tan^{-1}\sqrt3)=\tan^{-1}\sqrt3-\tan^{-1}1$$ $$A=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$$
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