f(x)=1/(3+2cosx) continua pe [0,2pi] admite primitive
F:[0,2pi]->R primitiva a lui f
F(x)=int[1/(3+2cosx)]dx
substitutie tg(x/2)=t =>dx=2dt/(1+t^2) si cosx=(1-t^2)/(1+t^2)
se obtine F(x)= acolada si scrii cele de jos
2/sqrt(5)*arctg[(tg(x/2))/sqrt(5)] daca x apartine [0,pi)
pi/sqrt(5) daca x=pi
2/sqrt(5)*arctg[(tg(x/2))/sqrt(5)]+2pi/sqrt(5) daca x apartine (pi,2pi]