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: 10


$\ln (1+\sqrt 2)i+(1-\dfrac{\pi}{4})j+(\dfrac{\pi}{4\sqrt 2}-\dfrac{1}{\sqrt 2})k$

Work Step by Step

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