University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 45


$$\int4\tan^3xdx=2\tan^2x-4\ln|\sec x|+C$$

Work Step by Step

$$A=\int4\tan^3xdx=\int4\tan x\tan^2xdx$$ $$A=\int4\tan x(\sec^2x-1)dx$$ $$A=4\Big(\int\tan x\sec^2xdx-\int\tan xdx\Big)$$ $$A=4\Big(\int\tan xd(\tan x)-\ln|\sec x|\Big)$$ We set $u=\tan x$. $$A=4\Big(\int udu-\ln|\sec x|\Big)$$ $$A=4\Big(\frac{u^2}{2}-\ln|\sec x|\Big)+C$$ $$A=2u^2-4\ln|\sec x|+C$$ $$A=2\tan^2x-4\ln|\sec x|+C$$
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