Answer
$$ = {\sin ^{ - 1}}\left( {\tan x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sec }^2}xdx}}{{\sqrt {1 - {{\tan }^2}x} }}} \cr
& or \cr
& \int {\frac{{{{\sec }^2}xdx}}{{\sqrt {1 - {{\left( {\tan x} \right)}^2}} }}} \cr
& {\text{substitutue }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr
& \int {\frac{{{{\sec }^2}xdx}}{{\sqrt {1 - {{\left( {\tan x} \right)}^2}} }}} = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& {\text{ use the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }}} = {\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C,{\text{ }}\left( {{\text{see page 468}}} \right),{\text{ }}a = 1 \cr
& = {\sin ^{ - 1}}\left( u \right) + C \cr
& {\text{write in terms of }}x \cr
& = {\sin ^{ - 1}}\left( {\tan x} \right) + C \cr} $$
