Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 17


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

Work Step by Step

Given $$\int \tan ^{5} x \sec x d x$$ Then \begin{align*} \int \tan ^{5} x \sec x d x&=\int \tan ^{4} x \sec x\tan x d x\\ &= \int (\sec^2 x-1) ^{2} x \sec x\tan x d x,\ \ \ \ \text{let } u = \sec x,\ \ d\sec x\tan xdx\\ &= \int (u^2-1)^2du\\ &= \int (u^4-2u^2+1)du\\ &= \frac{1}{5}u^5-\frac{2}{3}u^3+u+C\\ &= \frac{1}{5}\sec^5x-\frac{2}{3}\sec^3x+\sec x+C \end{align*}
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