University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 3


a) $$\int^{\pi/4}_0\tan x\sec^2xdx=\frac{1}{2}$$ b) $$\int_{-\pi/4}^0\tan x\sec^2xdx=-\frac{1}{2}$$

Work Step by Step

Part a) $$A=\int^{\pi/4}_0\tan x\sec^2xdx$$ Set $u=\tan x$, we then have $$du=\sec^2xdx$$ For $x=0$, we have $u=0$ and for $x=\pi/4$, we have $u=1$. Therefore, $$A=\int^1_0udu$$ $$A=\frac{u^2}{2}\Big]^1_0$$ $$A=\frac{1}{2}(1^2-0^2)=\frac{1}{2}$$ Part b) $$A=\int_{-\pi/4}^0\tan x\sec^2xdx$$ We carry out a similar substitution as in part a). For $x=0$, we have $u=0$ and for $x=-\pi/4$, we have $u=-1$. Therefore, $$A=\int^0_{-1}udu$$ $$A=\frac{u^2}{2}\Big]^0_{-1}$$ $$A=\frac{1}{2}(0^2-1^2)=\frac{1}{2}(-1)=-\frac{1}{2}$$
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