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