Answer
$$\frac{1}{2}x - \frac{1}{2}\ln \left| {\sin x + \cos x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin x}}{{\sin x + \cos x}}} dx \cr
& {\text{Add and subtract }}\cos x{\text{ into the numerator}} \cr
& = \int {\frac{{\sin x + \cos x - \cos x}}{{\sin x + \cos x}}} dx \cr
& {\text{distribute}} \cr
& = \int {\left( {\frac{{\sin x + \cos x}}{{\sin x + \cos x}} - \frac{{\cos x}}{{\sin x + \cos x}}} \right)} dx \cr
& = \int {\left( {1 - \frac{{\cos x}}{{\sin x + \cos x}}} \right)} dx \cr
& = \int {dx} - \int {\frac{{\cos x}}{{\sin x + \cos x}}} dx \cr
& = \int {dx} - \frac{1}{2}\int {\frac{{\cos x + \cos x}}{{\sin x + \cos x}}} dx \cr
& {\text{Add and subtract }}\sin x{\text{ into the numerator}} \cr
& = \int {dx} - \frac{1}{2}\int {\frac{{\cos x + \cos x}}{{\sin x + \cos x}}} dx \cr
& = \int {dx} - \frac{1}{2}\int {\frac{{\cos x - \sin x + \sin x + \cos x}}{{\sin x + \cos x}}} dx \cr
& = \int {dx} - \frac{1}{2}\int {\frac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx - \frac{1}{2}\int {\frac{{\sin x + \cos x}}{{\sin x + \cos x}}} dx \cr
& = \frac{1}{2}\int {dx} - \frac{1}{2}\int {\frac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx \cr
& = \frac{1}{2}x - \frac{1}{2}\int {\frac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx \cr
& \cr
& {\text{Integrate }}\int {\frac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx{\text{ by the substitution method}} \cr
& \,\,\,u = \sin x + \cos x,\,\,\,du = \left( {\cos x - \sin x} \right)dx \cr
& \,\, - \frac{1}{2}\int {\frac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx = - \frac{1}{2}\int {\frac{{du}}{u}} = \ln \left| u \right| + C \cr
& \,\,\,{\text{substitute }}\sin x + \cos x{\text{ for }}u \cr
& \,\,\,\, = - \frac{1}{2}\int {\frac{{du}}{u}} = - \frac{1}{2}\ln \left| {\sin x + \cos x} \right| + C \cr
& \cr
& {\text{Then}} \cr
& = \frac{1}{2}x - \frac{1}{2}\int {\frac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx = \frac{1}{2}x - \frac{1}{2}\ln \left| {\sin x + \cos x} \right| + C \cr} $$
