University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 12 - Section 12.2 - Integrals of Vector Functions; Projectile Motion - Exercises - Page 654: 9



Work Step by Step

Plug $p= 2t \implies dp=dt$ and $\cos 2t =1-2 \sin^2 t$ We need to integrate the given integral as follows: $[\int_{0}^{\pi/2} \cos t dt]i-[\int_{0}^{\pi/2} \sin 2t dt]j+[\int_{0}^{\pi/2} \sin^2 t dt]k=[\int_{0}^{\pi/2} (1-2 \sin^2 t) dt]i-[\int_{0}^{\pi/2} \sin p dp]j+[\int_{0}^{\pi/2} (\dfrac{1}{2}-\dfrac{1}{2} \cos 2t dt]k$ or, $[\int_{0}^{\pi/2} (1-2 \sin^2 t) dt]i-[\int_{0}^{\pi/2} \sin p dp]j+[\int_{0}^{\pi/2} (\dfrac{1}{2}-\dfrac{1}{2} \cos 2t dt]k=[\sin t]_{0}^{\pi/2}i+[\dfrac{1}{2} \cos 2t ]_{0}^{\pi/2}j+[(\dfrac{1}{2}t-\dfrac{1}{4} \sin 2t]_{0}^{\pi/2}$ Thus, $[\sin t]_{0}^{\pi/2}i+[\dfrac{1}{2} \cos 2t ]_{0}^{\pi/2}j+[(\dfrac{1}{2}t-\dfrac{1}{4} \sin 2t]_{0}^{\pi/2}=i-j+\dfrac{\pi}{4}k$
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