Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 484: 4


$$\int^{\pi/2}_0\sin^5 xdx=\frac{8}{15}$$

Work Step by Step

$$A=\int^{\pi/2}_0\sin^5 xdx$$ Here we can do Integration by Parts, or more simply, try to change some factors of $\sin$ in terms of $\cos$ then apply the Substitution Rule. $$A=\int^{\pi/2}_0\sin^4 x\sin xdx$$ $$A=\int^{\pi/2}_0(\sin^2 x)^2\sin xdx$$ We already know that $\sin^2 x=1-\cos^2 x$ Therefore, we have $$A=\int^{\pi/2}_0(1-\cos^2 x)^2\sin xdx$$ Now we substitute $u=\cos x$ That makes $du=-\sin x dx$, or $\sin xdx=-du$ Also, for $x=0$, $u=1$ and for $x=\frac{\pi}{2}$, $u=0$ So, $$A=-\int^{0}_1 (1-u^2)^2du$$ $$A=-\int^{0}_1(1-2u^2+u^4)du$$ $$A=\int^0_1 (-1+2u^2-u^4)du$$ $$A=[-u+\frac{2u^3}{3}-\frac{u^5}{5}]^{0}_1$$ $$A=0-[-1+\frac{2\times1^3}{3}-\frac{1^5}{5}]$$ $$A=0-[-1+\frac{2}{3}-\frac{1}{5}]$$ $$A=\frac{8}{15}$$
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