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: 44


$$\int\sec^6xdx=\frac{\tan^5x}{5}+\frac{2\tan^3x}{3}+\tan x+C$$

Work Step by Step

$$A=\int\sec^6xdx$$ $$A=\int(\sec^2x)^2(\sec^2xdx)$$ $$A=\int(\tan^2x+1)^2d(\tan x)$$ We set $u=\tan x$. $$A=\int(u^2+1)^2du$$ $$A=\int(u^4+2u^2+1)du$$ $$A=\frac{u^5}{5}+\frac{2u^3}{3}+u+C$$ $$A=\frac{\tan^5x}{5}+\frac{2\tan^3x}{3}+\tan x+C$$
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