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: 10



Work Step by Step

$$A=\int^{\pi/6}_03\cos^53xdx$$ $$A=\int^{\pi/2}_0\Big(3\cos^53x\Big)\frac{1}{3}d(3x)$$ $$A=\int^{\pi/2}_0\cos^53xd(3x)$$ We set $a=3x$. $$A=\int^{\pi/2}_0\cos^5ada$$ Following the integral form $\int\sin^mx\cos^nxdx$, we would apply Case 2 here, where $m=0$ and $n=5$. $$A=\int^{\pi/2}_0(\cos^2a)^2\cos ada$$ $$A=\int^{\sin(\pi/2)}_{\sin0}(1-\sin^2a)^2d(\sin a)$$ We set $u=\sin a$. Limits of integration: $\sin(\pi/2)=1$ and $\sin0=0$ $$A=\int^1_0(1-u^2)^2du$$ $$A=\int^1_0(1-2u^2+u^4)du$$ $$A=\Big(u-\frac{2u^3}{3}+\frac{u^5}{5}\Big)\Big]^1_0$$ $$A=\Big(1-\frac{2}{3}+\frac{1}{5}\Big)-0$$ $$A=\frac{8}{15}$$
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