Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 4

Answer

$$-\frac{1}{2} \cos ^{2} x+\frac{1}{2} \cos ^{4} x-\frac{1}{6} \cos ^{6} x+c$$

Work Step by Step

\begin{aligned} \int \sin ^{5} x \cos x d x &=\int \sin ^{4} x \sin x \cos x d x \\ &=\int\left(1-\cos ^{2} x\right)^{2} \sin x \cos x d x \\ &=\int\left(1-2 \cos ^{2} x+\cos ^{4} x\right) \sin x \cos x d x \\ &=\int\left(\cos x-2 \cos ^{3} x+\cos ^{5} x\right) \sin x d x \\ &=-\frac{1}{2} \cos ^{2} x+\frac{1}{2} \cos ^{4} x-\frac{1}{6} \cos ^{6} x+c \end{aligned}
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