Answer
$$\frac{{\sqrt 2 }}{2}{\tan ^{ - 1}}\left( {\sqrt 2 \tan x} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 + {{\sin }^2}x}}} \cr
& {\text{Use the Pythagorean Identity co}}{{\text{s}}^2}x + {\sin ^2}x = 1 \cr
& {\text{Then si}}{{\text{n}}^2}x = 1 - {\cos ^2}x \cr
& \int {\frac{{dx}}{{1 + {{\sin }^2}x}}} = \int {\frac{{dx}}{{2 - {{\cos }^2}x}}} = \int {\frac{{dx}}{{2 - \frac{1}{{{{\sec }^2}x}}}}} \cr
& = \int {\frac{{{{\sec }^2}xdx}}{{2{{\sec }^2}x - 1}}} \cr
& {\text{Using the Pythagorean Identity }}{\sec ^2}x - 1 = {\tan ^2}x \cr
& = \int {\frac{{{{\sec }^2}xdx}}{{2\left( {1 + {{\tan }^2}x} \right) - 1}}} \cr
& = \int {\frac{{{{\sec }^2}xdx}}{{2 + 2{{\tan }^2}x - 1}}} \cr
& = \int {\frac{{{{\sec }^2}xdx}}{{1 + 2{{\tan }^2}x}}} \cr
& {\text{Using the substitution }}u = \tan x,\,\,\,du = {\sec ^2}xdx \cr
& = \int {\frac{{du}}{{1 + 2{u^2}}}} \cr
& = \frac{1}{2}\int {\frac{{du}}{{{u^2} + 1/2}}} = \frac{1}{2}\int {\frac{{du}}{{{u^2} + {{\left( {1/\sqrt 2 } \right)}^2}}}} \cr
& {\text{Integrating by the formula }}\int {\frac{{du}}{{{u^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& = \frac{1}{2}\left( {\frac{1}{{1/\sqrt 2 }}} \right){\tan ^{ - 1}}\left( {\frac{u}{{1/\sqrt 2 }}} \right) + C \cr
& = \frac{{\sqrt 2 }}{2}{\tan ^{ - 1}}\left( {\sqrt 2 u} \right) + C \cr
& {\text{Write in terms of }}x{\text{; replace }}u = \tan x \cr
& = \frac{{\sqrt 2 }}{2}{\tan ^{ - 1}}\left( {\sqrt 2 \tan x} \right) + C \cr} $$
