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 - Page 480: 28

Answer

$$\frac{3}{{\ln 4}}$$

Work Step by Step

$$\eqalign{ & \int_0^{\pi /2} {{4^{\sin x}}\cos x} dx \cr & {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx \cr & {\text{The new limits of integration are:}} \cr & x = 0 \Rightarrow u = \sin 0 = 0 \cr & x = \pi /2 \Rightarrow u = \sin 0 = 1 \cr & {\text{Applying the substitution we obtain}} \cr & \int_0^{\pi /2} {{4^{\sin x}}\cos x} dx = \int_0^1 {{4^u}du} \cr & {\text{Integrating}} \cr & \int_0^1 {{4^u}du} = \left[ {\frac{{{4^u}}}{{\ln 4}}} \right]_0^1 \cr & {\text{ }} = \frac{1}{{\ln 4}}\left[ {{4^1} - {4^0}} \right] \cr & {\text{ }} = \frac{3}{{\ln 4}} \cr} $$
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