Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.5 Exercises - Page 302: 68

Answer

$$\frac{1}{2}$$

Work Step by Step

$$\eqalign{ & \int_{\pi /12}^{\pi /4} {\csc 2x\cot 2x} dx \cr & {\text{Let }}u = 2x,{\text{ }}du = 2dx \cr & {\text{The new limits of integration are:}} \cr & x = \frac{\pi }{4} \to u = 2\left( {\frac{\pi }{4}} \right) = \frac{\pi }{2} \cr & x = \frac{\pi }{{12}} \to u = 2\left( {\frac{\pi }{{12}}} \right) = \frac{\pi }{6} \cr & {\text{Substituting}} \cr & \int_{\pi /12}^{\pi /4} {\csc 2x\cot 2x} dx = \int_{\pi /6}^{\pi /2} {\csc u\cot u\left( {\frac{1}{2}} \right)} du \cr & = \frac{1}{2}\int_{\pi /6}^{\pi /2} {\csc u\cot u} du \cr & {\text{Integrating}} \cr & = - \frac{1}{2}\left[ {\csc u} \right]_{\pi /6}^{\pi /2} \cr & = - \frac{1}{2}\left[ {\csc \left( {\frac{\pi }{2}} \right) - \csc \left( {\frac{\pi }{6}} \right)} \right] \cr & = - \frac{1}{2}\left( {1 - 2} \right) \cr & = \frac{1}{2} \cr} $$
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