Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 75

Answer

$$ - \ln 3$$

Work Step by Step

$$\eqalign{ & \int_{ - \pi }^0 {\frac{{\sin x}}{{2 + \cos x}}dx} \cr & {\text{substitute }}u = 2 + \cos x,{\text{ }}du = - \sin xdx \cr & {\text{express the limits in terms of }}u \cr & x = - \pi {\text{ implies }}u = 2 + \cos \left( { - \pi } \right) = 1 \cr & x = 0{\text{ implies }}u = 2 + \cos \left( 0 \right) = 3 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_{ - \pi }^0 {\frac{{\sin x}}{{2 + \cos x}}dx} = - \int_1^3 {\frac{{du}}{u}} \cr & {\text{find the antiderivative}} \cr & = - \left. {\left( {\ln \left| u \right|} \right)} \right|_1^3 \cr & {\text{use the fundamental theorem}} \cr & = - \left( {\ln \left| 3 \right| - \ln \left| 1 \right|} \right) \cr & {\text{simplify}} \cr & = - \ln 3 \cr} $$
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