University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 7


$$\int\sin^5xdx=\frac{2(\cos x)^3}{3}-\cos x-\frac{(\cos x)^5}{5}+C$$

Work Step by Step

$$A=\int\sin^5xdx$$ Following the integral form $\int\sin^mx\cos^nxdx$, this is Case 1, where $m=5$ and $n=0$. $$A=\int\sin^4x(\sin xdx)$$ $$A=-\int(1-\cos^2x)^2d(\cos x)$$ We set $u=\cos x$. $$A=-\int (1-u^2)^2du$$ $$A=-\int(1-2u^2+u^4)du$$ $$A=-\Big(u-\frac{2u^3}{3}+\frac{u^5}{5}\Big)+C$$ $$A=\frac{2u^3}{3}-u-\frac{u^5}{5}+C$$ $$A=\frac{2(\cos x)^3}{3}-\cos x-\frac{(\cos x)^5}{5}+C$$
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