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: 15



Work Step by Step

$$A=\int^{\pi/2}_0\sin^7ydy$$ This is Case 1, so we need to find a way to rewrite $$(\sin^2y)^k\sin y=(1-\cos^2y)^k\sin y$$ We have, $$A=\int^{\pi/2}_0(\sin^2y)^3\sin ydy$$ $$A=-\int^{\pi/2}_0(1-\cos^2y)^3d(\cos y)$$ We would take $u=\cos y$ - For $y=0$, $u\cos0=1$ - For $y=\pi/2$, $u=\cos\pi/2=0$ $$A=-\int^0_1(1-u^2)^3du=\int^1_0(1-u^2)^3du$$ $$A=\int^1_0(1-3u^2+3u^4-u^6)du$$ $$A=\Big(u-u^3+\frac{3u^5}{5}-\frac{u^7}{7}\Big)\Big]^1_0$$ $$A=1-1^3+\frac{3\times1^5}{5}-\frac{1^7}{7}$$ $$A=\frac{3}{5}-\frac{1}{7}=\frac{16}{35}$$
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