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 - Page 525: 31


$$\frac{1}{4}\tan^3x-\frac{1}{2}\tan^2x+\ln \sec x+c$$

Work Step by Step

Given $$\int \tan^5xdx $$ Since $$1+\tan^2x=\sec^2x $$ Then \begin{align*} \int \tan^5xdx &=\int \tan^3x \tan^2xdx \\ &=\int \tan^3x (\sec^2x-1)dx \\ &=\int( \tan^3x \sec^2x-\tan x(\sec^2x-1))dx \\ &=\int( \tan^3x \sec^2x-\tan x \sec^2x +\tan x )dx \\ &=\frac{1}{4}\tan^3x-\frac{1}{2}\tan^2x+\ln \sec x+c \end{align*}
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