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: 28

Answer

$\frac{1}{7}\sec^7 x+\frac{2}{5}\sec^5 x+\frac{1}{3}\sec^3 x+C$

Work Step by Step

$\int\tan^5 x\sec^3 x\ dx$ Since the power of tangent is odd, save a factor of $\sec x\tan x$ and express the remaining factors in terms of $\sec x$: $=\int\tan^4 x\sec^2 x\sec x\tan x\ dx$ $=\int(\tan^2 x)^2\sec^2 x\sec x\tan x\ dx$ $=\int(1+\sec^2 x)^2\sec^2 x\sec x\tan x\ dx$ Let $u=\sec x$. Then $du=\sec x\tan x\ dx$. $=\int(1+u^2)^2 u^2\ du$ $=\int(1+2u^2+u^4)u^2\ du$ $=\int(u^6+2u^4+u^2)\ du$ $=\frac{1}{7}u^7+\frac{2}{5}u^5+\frac{1}{3}u^3+C$ $=\boxed{\frac{1}{7}\sec^7 x+\frac{2}{5}\sec^5 x+\frac{1}{3}\sec^3 x+C}$
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