Find $\int \sec^2 x \tan x dx$

\[\int \sec^2x \tan x dx =\int \sec x (\sec x\tan x)dx.\]
Let $u=\sec x$. Then $du=\sec x \tan x dx$. Hence

\[\int \sec^2x \tan x dx =\int \sec x (\sec x\tan x)dx=\int u du=\frac{1}{2}u^2+c\]
\[=\frac{1}{2}(\sec x)^2+c.\]