Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 57


$$-\frac{1}{5} \sin ^{4} x \cos x-\frac{4}{15} \sin ^{2} x \cos x-\frac{8}{15} \cos x+C$$

Work Step by Step

Given $$\int \sin ^{5} x d x $$ By using $$\int \sin ^{n} x d x=-\frac{1}{n} \sin ^{n-1} x \cos x+\frac{n-1}{n} \int \sin ^{n-2} x d x$$ With ($n=5$) \begin{align*} \int \sin ^{5} x d x&=-\frac{1}{5} \sin ^{4} x \cos x+\frac{4}{5} \int \sin ^{3} x d x\\ &= -\frac{1}{5} \sin ^{4} x \cos x+\frac{4}{5}\left(-\frac{1}{3} \sin ^{2} x \cos x+\frac{2}{3} \int \sin x d x\right)\\ &= -\frac{1}{5} \sin ^{4} x \cos x+\frac{4}{5}\left(-\frac{1}{3} \sin ^{2} x \cos x-\frac{2}{3} \cos x\right)+C\\ &=-\frac{1}{5} \sin ^{4} x \cos x-\frac{4}{15} \sin ^{2} x \cos x-\frac{8}{15} \cos x+C \end{align*}
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