Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises: 16

Answer

$$ - \frac{{{{\cos }^6}x}}{6} + \frac{{{{\cos }^8}x}}{8} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\sin }^3}x{{\cos }^5}x} dx \cr & {\text{split off }}\sin x \cr & = \int {{{\sin }^2}x{{\cos }^5}x} \sin xdx \cr & {\text{pythagorean identity}} \cr & = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^5}x} \sin xdx \cr & u = \cos x,{\text{ }}du = - \sin xdx \cr & = - \int {\left( {1 - {u^2}} \right){u^5}} du \cr & = - \int {\left( {{u^5} - {u^7}} \right)} du \cr & {\text{evaluate the integral}} \cr & = - \frac{{{u^6}}}{6} + \frac{{{u^8}}}{8} + C \cr & {\text{replace }}u = \cos x \cr & = - \frac{{{{\cos }^6}x}}{6} + \frac{{{{\cos }^8}x}}{8} + C \cr} $$
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