analys - integral, missat något

Använd variabelbytet t=tan(x/2) för att bestämma primitiv till

$\frac{1}{2+\sin \left(x\right)}$

Mitt försök

$\sin \left(x\right)=\frac{2t}{1+t^2}$

$$\begin{gathered} \int \frac{1}{t^2+t+1} dt = \int \frac{{\displaystyle 1}}{{\displaystyle (t+\frac{1}{2}{)}^2+\frac{{\displaystyle 3}}{{\displaystyle 4}}}} dt =\hspace{0.17em}\frac{4}{3}\int \frac{{\displaystyle 1}}{{\displaystyle \frac{4}{3}(t+\frac{1}{2}{)}^2+1}} dt =\hspace{0.17em}\frac{{\displaystyle 4}}{{\displaystyle 3}}\int \frac{{\displaystyle 1}}{{\displaystyle (\frac{2t+1}{\sqrt{3}}{)}^2+1}}dt \\ z= \frac{{\displaystyle 2t+1}}{{\displaystyle \sqrt{3}}} ;\hspace{0.17em}dz =\hspace{0.17em}\frac{2}{\sqrt{3}} dt \\ \frac{{\displaystyle 4}}{{\displaystyle 3}}\int \frac{{\displaystyle 1}}{{\displaystyle (z{)}^2+1}}\hspace{0.17em}\frac{{\displaystyle 2}}{{\displaystyle \sqrt{3}}}dz =\hspace{0.17em}\frac{{\displaystyle 8}}{{\displaystyle 3\sqrt{3}}}arc\tan \left(z\right) + C =\hspace{0.17em}\frac{{\displaystyle 8}}{{\displaystyle 3\sqrt{3}}}arc\tan \left(\frac{2t+1}{\sqrt{3}}\right) + C=\hspace{0.17em}\frac{{\displaystyle 8}}{{\displaystyle 3\sqrt{3}}}arc\tan \left(\frac{2\tan \left({\displaystyle \frac{x}{2}}\right)+1}{\sqrt{3}}\right) + C \end{gathered}$$

Rätt svar ska vara:

$\frac{{\displaystyle 2}}{{\displaystyle \sqrt{3}}}arc\tan \left(\frac{2\tan \left({\displaystyle \frac{x}{2}}\right)+1}{\sqrt{3}}\right)+C$