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 - 5.7 Exercises - Page 380: 28

Answer

$$\frac{\pi }{6} - {\sin ^{ - 1}}\left( {\frac{1}{4}} \right)$$

Work Step by Step

$$\eqalign{ & \int_{\ln 2}^{\ln 4} {\frac{{{e^{ - x}}}}{{\sqrt {1 - {e^{ - 2x}}} }}} dx \cr & {\text{Let }}u = {e^{ - x}},{\text{ }}du = - {e^{ - x}}dx,{\text{ }}dx = \frac{1}{{ - {e^{ - x}}}}du \cr & {\text{The new limits of integration are}} \cr & x = \ln 4,{\text{ }}u = {e^{ - \ln 4}} = \frac{1}{4} \cr & x = \ln 2,{\text{ }}u = {e^{ - \ln 2}} = \frac{1}{2} \cr & {\text{Substituting}} \cr & \int_{\ln 2}^{\ln 4} {\frac{{{e^{ - x}}}}{{\sqrt {1 - {e^{ - 2x}}} }}} dx = \int_{1/2}^{1/4} {\frac{{{e^{ - x}}}}{{\sqrt {1 - {u^2}} }}} \left( {\frac{1}{{ - {e^{ - x}}}}} \right)du \cr & = - \int_{1/2}^{1/4} {\frac{1}{{\sqrt {1 - {u^2}} }}} du \cr & {\text{Use the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}u + C} \cr & = - \left[ {{{\sin }^{ - 1}}u} \right]_{1/2}^{1/4} \cr & = - \left[ {{{\sin }^{ - 1}}\left( {\frac{1}{4}} \right) - {{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right] \cr & = - \left[ {{{\sin }^{ - 1}}\left( {\frac{1}{4}} \right) - \frac{\pi }{6}} \right] \cr & = \frac{\pi }{6} - {\sin ^{ - 1}}\left( {\frac{1}{4}} \right) \cr} $$
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