Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 303: 4


a) 2 b)2

Work Step by Step

a) let u=cos x =>du=-sin x dx=>-du=sin x dx; x=0=>u=1, x=$\pi$=>u=-1 $\int ^{\pi}_0 3\cos ^2x \sin xdx$ =$\int^{-1}_1 -3u^2 du$ =$[-u^3]^{-1}_1$ =$-(-1)^3-(-1(1)^3)=2$ --- b)Use the same substitution as in part (a); x=$2\pi$=>u=1, x=3$\pi$=>u=-1 $\int_{2\pi}^{3\pi}3cos^2x \sin x dx$ =$\int^{-1}_1 -3u^2du$ =2
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