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: 25

Answer

$\frac{1}{9}\tan^9 x+\frac{2}{7}\tan^7 x+\frac{1}{5}\tan^5 x+C$

Work Step by Step

$\int\tan^4 x\sec^6 x\ dx$ Since the power of secant is even, we save a factor of $\sec^2 x$ and express the rest in terms of $\tan x$: $=\int \tan^4 x (\sec^2 x)^2\sec^2 x\ dx$ $=\int \tan^4 x(\tan^2 x+1)^2\sec^2 x\ dx$ Let $u=\tan x$. Then $du=\sec^2 x\ dx$. $=\int u^4(u^2+1)^2\ du$ $=\int u^4(u^4+2u^2+1)\ du$ $=\int(u^8+2u^6+u^4)\ du$ $=\frac{1}{9}u^9+\frac{2}{7}u^7+\frac{1}{5}u^5+C$ $=\boxed{\frac{1}{9}\tan^9 x+\frac{2}{7}\tan^7 x+\frac{1}{5}\tan^5 x+C}$
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