Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 18


$$ \frac{1}{9}\tan^9 x+c$$

Work Step by Step

Given $$\int \tan ^{8} x \sec ^{2} x d x$$ Let $$u=\tan x\ \ \to \ \ du=\sec ^{2} x d x$$ \begin{align*} \int \tan^{8} x \sec ^{2} x d x&=\int u^{8} du\\ &=\frac{1}{9}u^9 +c\\ &= \frac{1}{9}\tan^9 x+c \end{align*}
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