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: 3


a)$\frac{1}{2}$ b)$\frac{-1}{2}$

Work Step by Step

a)Let u=$\tan x$=> du=$sec^2xdx$ x=0=>u=0,x=$\frac{\pi}{4}=>u=1$ =$\int ^{\pi/4}_0tan xsec^2xdx$ =$\int^1_0udu$ =$[\frac{u^2}{2}]^1_0$ =$\frac{1^2}{2}-0$ =$\frac{1}{2}$ --- b)Use the same substitution as in part (a) ;x=$\frac{\pi}{4}=>u=-1, x=0=>u=0$ =$\int^0_{-\pi/4}\tan xsec^2xdx$ =$\int^0_{-1}udu$ =$[\frac{u^2}{2}]^0_{-1}$ =0-1/2 =$\frac{-1}{2}$
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