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: 44


$$\int(\tan x)^{-3/2}\sec^2 xdx=-\frac{2}{\sqrt{\tan x}}+C$$

Work Step by Step

$$A=\int(\tan x)^{-3/2}\sec^2 xdx$$ We set $a=\tan x$, then $$da=\sec^2 xdx$$ Therefore, $$A=\int a^{-3/2}da$$ $$A=\frac{a^{-1/2}}{-\frac{1}{2}}+C$$ $$A=-2a^{-1/2}+C$$ $$A=-2(\tan x)^{-1/2}+C$$ $$A=-\frac{2}{\sqrt{\tan x}}+C$$
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