Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises: 24

Answer

$\frac{1}{3}\tan^3 x+C$

Work Step by Step

$\int(\tan^2 x+\tan^4 x)\ dx$ $=\int\tan^2 x(1+\tan^2 x)\ dx$ $=\int\tan^2 x\sec^2 x\ dx$ Since the power of secant is even, save a factor of $\sec^2 x$. Let $u=\tan x$, so $du=\sec^2 x \ dx$. $=\int u^2\ du$ $=\frac{1}{3}u^3+C$ $=\boxed{\frac{1}{3}\tan^3 x+C}$
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