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

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

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*}