Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 16

Answer

$$\ln 2$$

Work Step by Step

$$\eqalign{ & \int_0^{\pi /2} {\frac{{\sin x}}{{1 + \cos x}}} dx \cr & {\text{substitute }}u = 1 + \cos x,{\text{ }}du = - \sin xdx \cr & {\text{express the limits in terms of }}u \cr & x = 0{\text{ implies }}u = 1 + \cos 0 = 2 \cr & x = \pi /2{\text{ implies }}u = 1 + \cos \left( {\pi /2} \right) = 1 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^{\pi /2} {\frac{{\sin x}}{{1 + \cos x}}} dx = - \int_2^1 {\frac{{du}}{u}} \cr & {\text{find the antiderivative}} \cr & = - \left. {\left( {\ln \left| u \right|} \right)} \right|_2^1 \cr & {\text{use the fundamental theorem}} \cr & = - \left( {\ln \left| 1 \right| - \ln \left| 2 \right|} \right) \cr & {\text{simplify}} \cr & = - \left( {0 - \ln 2} \right) \cr & = \ln 2 \cr} $$
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