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: 62

Answer

$$0$$

Work Step by Step

$$\eqalign{ & \int_0^\pi {{2^{\sin x}}\cos xdx} \cr & {\text{substitute }}u = \sin x,{\text{ }}du = \cos xdx \cr & {\text{express the limits in terms of }}u \cr & x = \pi {\text{ implies }}u = \sin \left( \pi \right) = 0 \cr & x = 0{\text{ implies }}u = \sin \left( 0 \right) = 0 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^\pi {{2^{\sin x}}\cos xdx} = \int_0^0 {{2^u}du} \cr & {\text{by the integral properties}} \cr & \int_0^0 {{2^u}du} = 0 \cr} $$
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