Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 484: 24


$$\int(\tan^2 x+\tan^4 x)dx=\frac{\tan^3 x}{3}+C$$

Work Step by Step

$$A=\int(\tan^2 x+\tan^4 x)dx$$ $$A=\int\tan^2 x(1+\tan^2 x)dx$$ $$A=\int\tan^2 x\sec^2 xdx$$ ($1+\tan^2 x=\sec^2 x$) Now let $u=\tan x$ Then $du=\sec^2 xdx$. Therefore, $$A=\int u^2du$$ $$A=\frac{u^3}{3}+C$$ $$A=\frac{\tan^3 x}{3}+C$$
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