University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Practice Exercises - Page 474: 40


$\dfrac{\sec^5 x}{5}-\dfrac{\sec^3 x}{5}+c$

Work Step by Step

Consider the integral $\int \tan^3 x \sec^3 x dx$ The given integral can be re-written as: $\int \tan^2 x \sec^2 x (\sec x \tan x) dx=\int (\sec^2 x -1) \sec^2 x (\sec x\tan x dx) $ Plug in $a= \sec x \implies da= \sec x \tan x dx$ We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$ This implies that $\int (a^4 -a^2) da =\dfrac{a^5}{5}-\dfrac{a^3}{3}+C=\dfrac{\sec^5 x}{5}-\dfrac{\sec^3 x}{5}+c$
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