Answer
$\frac{{{{\sec }^3}x}}{3} + C$
Work Step by Step
$$\eqalign{
& \text{Let }I= \int {\tan x} {\sec ^3}xdx \cr
& {\text{Split }}{\sec ^3}x{\text{ as }}{\sec ^2}x\sec x \cr
& I = \int {\tan x} {\sec ^2}x\sec xdx \cr
& = \int {{{\sec }^2}x\sec x} \tan xdx \cr
& {\text{Let }}u = \sec x,{\text{ }}du = \sec x\tan xdx \cr
& {\text{Substituting}} \cr
& I=\int {{{\sec }^2}x\left( {\sec x\tan x} \right)} dx = \int {{u^2}du} \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
&I = \int {{u^2}du} = \frac{{{u^3}}}{3} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\sec x{\text{ for }}u \cr
& I = \frac{{{{\sec }^3}x}}{3} + C \cr} $$
