Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 31

Answer

$\displaystyle\int\limits_{\pi/6}^{\pi/2}\csc t\cot tdt=1$

Work Step by Step

$\displaystyle\int\limits_{\pi/6}^{\pi/2}\csc t\cot tdt$ We know that $\dfrac{d}{dx}\csc t=-\cot t\csc t$. So we can integrate this expression directly and then apply the second part of the fundamental theorem of algebra: $\displaystyle\int\limits_{\pi/6}^{\pi/2}\csc t\cot tdt=-\csc t\Big|_{\pi/6}^{\pi/2}=-(\csc\dfrac{\pi}{2}-\csc\dfrac{\pi}{6})=-(1-2)=1$
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