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: 13


$$ \frac{\sin ^{4} x \cos x}{5}+\frac{1}{5}\left( -\cos x+\frac{1}{3}\cos^3x\right)+C$$

Work Step by Step

Since $$\int \sin ^{m} x \cos ^{n} x d x=\frac{\sin ^{m+1} x \cos ^{n-1} x}{m+n}+\frac{n-1}{m+n} \int \sin ^{m} x \cos ^{n-2} x d x$$ Then for $m=3,\ n= 2$, we get \begin{align*} \int \sin ^{3} x \cos ^{2} x d x&=\frac{\sin ^{4} x \cos x}{5}+\frac{1}{5} \int \sin ^{3} x d x\\ &= \frac{\sin ^{4} x \cos x}{5}+\frac{1}{5}\left( \int (1-\cos^2 x)\sin xdx\right)\\ &= \frac{\sin ^{4} x \cos x}{5}+\frac{1}{5}\left( -\cos x+\frac{1}{3}\cos^3x\right)+C \end{align*}
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