Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 35

$$\frac{1}{2}\tan^2 x+c$$

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