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: 34


$$\int^{\pi/2}_{\pi/4}\cot tdt=\ln\sqrt2$$

Work Step by Step

$$A=\int^{\pi/2}_{\pi/4}\cot tdt=\int^{\pi/2}_{\pi/4}\frac{\cos t}{\sin t}dt$$ We set $u=\sin t$, which means $$du=\cos tdt$$ For $t=\pi/2$, we have $$u=\sin\frac{\pi}{2}=1$$ For $t=\pi/4$, we have $$u=\sin\frac{\pi}{4}=\frac{\sqrt2}{2}$$ Therefore, $$A=\int_{\sqrt2/2}^1\frac{1}{u}du$$ $$A=\ln|u|\Big]_{\sqrt2/2}^1$$ $$A=\ln1-\ln\frac{\sqrt2}{2}=-\ln\frac{\sqrt2}{2}$$ $$A=\ln\Big(\frac{\sqrt2}{2}\Big)^{-1}$$ $$A=\ln\frac{2}{\sqrt2}=\ln\sqrt2$$
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