Answer
$$\frac{{{2^{\tan x}}}}{{\ln 2}} + C $$
Work Step by Step
$$\eqalign{
& \int {{2^{\tan x}}{{\sec }^2}x} dx \cr
& {\text{use substitution}}{\text{: }} \cr
& {\text{ }}u = \tan x,{\text{ then }}\frac{{du}}{{dx}} = {\sec ^2}x,\,\,\,\,\frac{{du}}{{{{\sec }^2}x}} = dx \cr
& {\text{then}} \cr
& \int {{2^{\tan x}}{{\sec }^2}x} dx = \int {{2^u}{{\sec }^2}x} \left( {\frac{{du}}{{{{\sec }^2}x}}} \right) \cr
& = \int {{3^u}} du \cr
& {\text{integrate using }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& = \frac{{{2^u}}}{{\ln 2}} + C \cr
& {\text{replace }}\tan x{\text{ for }}u \cr
& = \frac{{{2^{\tan x}}}}{{\ln 2}} + C \cr} $$
