University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 94


$$\int^{\pi/4}_0\frac{\sec^2x}{(1+7\tan x)^{2/3}}dx=\frac{3}{7}$$

Work Step by Step

$$A=\int^{\pi/4}_0\frac{\sec^2x}{(1+7\tan x)^{2/3}}dx$$ $$A=\int^{\pi/4}_0\sec^2x(1+7\tan x)^{-2/3}dx$$ We set $u=1+7\tan x$, which means $$du=7\sec^2xdx$$ $$\sec^2xdx=\frac{1}{7}du$$ For $x=\pi/4$: $$u=1+7\tan(\pi/4)=1+7\times1=8$$ For $x=0$: $$u=1+7\tan0=1+7\times0=1$$ Therefore, $$A=\frac{1}{7}\int^8_1u^{-2/3}du$$ $$A=\Big(\frac{1}{7}\times\frac{u^{1/3}}{\frac{1}{3}}\Big)\Big]^8_1$$ $$A=\Big(\frac{3\sqrt[3]u}{7}\Big)\Big]^8_1$$ $$A=\frac{3}{7}(2-1)=\frac{3}{7}$$
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