The standard approach is to employ the substitution [tex]t=\tan\dfrac x2[/tex]. Then [tex]\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx[/tex]. You can derive the following relations:
[tex]\tan\dfrac x2=t\iff\dfrac{\sin\frac x2}{\cos\frac x2}=\dfrac{\frac t{\sqrt{1+t^2}}}{\frac1{\sqrt{1+t^2}}}[/tex]
(where the choice of [tex]\sqrt{1+t^2}[/tex] comes from the fact that it satisfies the Pythagorean theorem, [tex]1^2+t^2=(\sqrt{1+t^2})^2=1+t^2[/tex])
[tex]\sin x=\sin\dfrac{2x}2=2\sin\dfrac x2\cos\dfrac x2=2\dfrac t{\sqrt{1+t^2}}\dfrac1{\sqrt{1+t^2}}=\dfrac{2t}{1+t^2}[/tex]
and we also have
[tex]\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies2\cos^2\dfrac x2\,\mathrm dt=\mathrm dx\implies\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt[/tex]
So the integral can be written as
[tex]\displaystyle\int\frac{\sin x}{4+\sin x}\,\mathrm dx=\int\frac{\frac{2t}{1+t^2}}{4+\frac{2t}{1+t^2}}\frac2{1+t^2}\,\mathrm dt[/tex]
Simplifying the integrand gives
[tex]\displaystyle2\int\frac t{(2t^2+t+2)(t^2+1)}\,\mathrm dt[/tex]
which can be expanded into partial fractions, yielding
[tex]\displaystyle2\int\left(\frac1{t^2+1}-\frac2{2t^2+t+2}\right)\,\mathrm dt[/tex]
The first integral can be handled immediately ([tex]\arctan[/tex]), while the other needs rewriting in order to be computed similarly. Complete the square in the denominator to get
[tex]2t^2+t+2=2\left(t+\dfrac14\right)^2+\dfrac{15}8[/tex]
and cancelling a factor of 2, you'll end up with
[tex]\displaystyle2\int\left(\frac1{t^2+1}-\frac1{\left(t+\frac14\right)^2+\frac{15}{16}}\right)\,\mathrm dt[/tex]
Proper trigonometric substitutions will complete the work. (And of course you'll need to back-substitute to get the result in terms of [tex]x[/tex].)
