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: 78

Answer

$${e^{1/2}} - 1$$

Work Step by Step

$$\eqalign{ & \int_0^{\pi /4} {{e^{{{\sin }^2}x}}\sin 2x} dx \cr & \int_0^{\pi /4} {{e^{{{\sin }^2}x}}\left( {2\sin x\cos x} \right)} dx \cr & {\text{substitute }}u = {\sin ^2}x,{\text{ }}du = 2\sin x\cos xdx \cr & {\text{express the limits in terms of }}u \cr & x = \pi /4{\text{ implies }}u = {\sin ^2}\left( {\pi /4} \right) = 1/2 \cr & x = 0{\text{ implies }}u = {\sin ^2}0 = 0 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^{\pi /4} {{e^{{{\sin }^2}x}}\left( {2\sin x\cos x} \right)} dx = \int_0^{1/2} {{e^u}du} \cr & {\text{find the antiderivative}} \cr & = \left. {\left( {{e^u}} \right)} \right|_0^{1/2} \cr & {\text{use the fundamental theorem}} \cr & = {e^{1/2}} - {e^0} \cr & {\text{simplify}} \cr & = {e^{1/2}} - 1 \cr} $$
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