Calculus 10th Edition

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

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 335: 56

Answer

$$\frac{1}{2}\ln \left( {\sqrt 2 + 1} \right) + \ln \left( {\frac{1}{{\sqrt 2 }}} \right)$$

Work Step by Step

$$\eqalign{ & \int_{\pi /8}^{\pi /4} {\left( {\csc 2\theta - \cot 2\theta } \right)} d\theta \cr & {\text{Distribute}} \cr & = \int_{\pi /8}^{\pi /4} {\csc 2\theta } d\theta - \int_{\pi /8}^{\pi /4} {\cot 2\theta } d\theta \cr & = \frac{1}{2}\int_{\pi /8}^{\pi /4} {\csc 2\theta } \left( 2 \right)d\theta - \frac{1}{2}\int_{\pi /8}^{\pi /4} {\cot 2\theta } \left( 2 \right)d\theta \cr & {\text{Integrate by the formulas of the page 333}} \cr & = - \frac{1}{2}\left[ {\ln \left| {\csc 2\theta + \cot 2\theta } \right|} \right]_{\pi /8}^{\pi /4} - \frac{1}{2}\left[ {\ln \left| {\sin 2\theta } \right|} \right]_{\pi /8}^{\pi /4} \cr & {\text{Evaluating}} \cr & = - \frac{1}{2}\left[ {\ln \left| {\csc \frac{\pi }{2} + \cot \frac{\pi }{2}} \right|} \right] + \frac{1}{2}\left[ {\ln \left| {\csc \frac{\pi }{4} + \cot \frac{\pi }{4}} \right|} \right] \cr & - \frac{1}{2}\left[ {\ln \left| {\sin \frac{\pi }{2}} \right| - \ln \left| {\sin \frac{\pi }{4}} \right|} \right] \cr & {\text{Simplifying}} \cr & = - \frac{1}{2}\left[ {\ln \left| 1 \right|} \right] + \frac{1}{2}\left[ {\ln \left| {\sqrt 2 + 1} \right|} \right] + \ln \left( 1 \right) - \ln \left( {\sqrt 2 } \right) \cr & = \frac{1}{2}\ln \left( {\sqrt 2 + 1} \right) + \ln \left( {\frac{1}{{\sqrt 2 }}} \right) \cr} $$
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