Answer
$$x - \tan \left( {\frac{x}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin x}}{{\sin x + \tan x}}} dx \cr
& {\text{Use substitution }}\left( 5 \right){\text{ from page 528}}: \cr
& \,\,u = \tan \left( {\frac{x}{2}} \right){\text{,}}\,\,\,\,\,\,dx = \frac{2}{{1 + {u^2}}}du,\,\,\,\,\,\,\sin x = \frac{{2u}}{{1 + {u^2}}}\,\,\,\,\,{\text{ and}}\,{\text{ }}\,\,\cos x = \frac{{1 - {u^2}}}{{1 + {u^2}}} \cr
& \cr
& \int {\frac{{\sin x}}{{\sin x + \tan x}}} dx = \int {\frac{{\frac{{2u}}{{1 + {u^2}}}}}{{\frac{{2u}}{{1 + {u^2}}} + \frac{{2u}}{{1 - {u^2}}}}}\left( {\frac{2}{{1 + {u^2}}}du} \right)} \cr
& {\text{simplify}} \cr
& = \int {\frac{{\frac{{2u}}{{1 + {u^2}}}}}{{\frac{{2u\left( {1 - {u^2}} \right) + 2u\left( {1 + {u^2}} \right)}}{{\left( {1 + {u^2}} \right)\left( {1 - {u^2}} \right)}}}}\left( {\frac{2}{{1 + {u^2}}}du} \right)} \cr
& = \int {\frac{{2\left( {1 - {u^2}} \right)du}}{{2u - 2{u^3} + 2u + 2{u^3}}}} \cr
& = \int {\frac{{4u\left( {1 - {u^2}} \right)du}}{{4u\left( {1 + {u^2}} \right)}}} \cr
& = \int {\frac{{1 - {u^2}}}{{1 + {u^2}}}} du \cr
& = \int {\left( {\frac{2}{{1 + {u^2}}} - 1} \right)} du \cr
& {\text{Integrate}} \cr
& = 2{\tan ^{ - 1}}u - u + C \cr
& \cr
& {\text{write in terms of }}x{\text{; substitute }}u = \tan \left( {\frac{x}{2}} \right) \cr
& = 2{\tan ^{ - 1}}\left( {\tan \left( {\frac{x}{2}} \right)} \right) - \tan \left( {\frac{x}{2}} \right) + C \cr
& = 2\left( {\frac{x}{2}} \right) - \tan \left( {\frac{x}{2}} \right) + C \cr
& = x - \tan \left( {\frac{x}{2}} \right) + C \cr} $$
