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: 22


$$\int^{\pi/2}_0\sin^22\theta\cos^32\theta d\theta=0$$

Work Step by Step

$$A=\int^{\pi/2}_0\sin^22\theta\cos^32\theta d\theta$$ $$A=\frac{1}{2}\int^{\pi/2}_0\sin^22\theta\cos^32\theta d(2\theta)$$ We set $a=2\theta$. For $\theta=\pi/2$, $a=\pi$ and for $\theta=0$, $a=0$ $$A=\frac{1}{2}\int^\pi_0\sin^2a\cos^3ada$$ This is Case 2, so we would rewrite $$\cos^3a=\cos^2a\cos a=(1-\sin^2a)\cos a$$ Therefore, $$A=\frac{1}{2}\int^\pi_0\sin^2a(1-\sin^2a)\cos ada$$ $$A=\frac{1}{2}\int^\pi_0(\sin^2a-\sin^4a)d(\sin a)$$ Take $u=\sin a$ For $a=\pi$, $u=\sin\pi=0$ and for $a=0$, $u=0$ Therefore, $$A=\frac{1}{2}\int^0_0(u^2-u^4)du=0$$
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