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: 39


$$\frac{1}{7}\tan^7x+ \frac{1}{9}\tan^9x+c$$

Work Step by Step

\begin{align*} \int \tan^6x\sec^4 xdx&= \int \tan^6x\sec^2 x\sec^2 x dx\\ &= \int \tan^6x(1+\tan^2x)\sec^2 x dx\\ &= \int (\tan^6x+\tan^8x)\sec^2 x dx\\ &=\frac{1}{7}\tan^7x+ \frac{1}{9}\tan^9x+c \end{align*}
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