tan(x/3)+cot(x/3)=
sin(x/3)/cos(x/3)+cos(x/3)/sin(x/3)=
(sin²(x/3)+cos²(x/3))/(sin(x/3)cos(x/3))=
2/sin(2x/3).
So ∫4csc²(2x/3)dx is the same integral=
-4cot(2x/3)×(3/2)=-6cot(2x/3)=-6/tan(2x/3)=
-6(1-tan²(x/3))/(2tan(x/3)))=
3(tan²(x/3)-1)/tan(x/3)=3tan(x/3)-3cot(x/3).
Now add in the constant of integration:
3tan(x/3)-3cot(x/3)+c, answer option III.