You are wrong with g(x). If g'(x)= cos(x/2) then g(x)=2sin(x/2)
I know this can be done long hand & several ways.
I started from integrating by parts for definite integral (as I will be adding the limits later)
∫ f(x)g'(x) dx = f(x)g(x) - ∫f'(x)g(x) dx
Let f(x) = (3x-∏) so f'(x) = 3
then g'(x) = cos (x/2) & from standard integral table g(x) = -1/2 sin(x/2)
then I used substitution for ∫-1/2 sin(x/2) dx
let u = x/2
so -1/2∫ sin u du = -3/2 cos u +c = -3/2 cos(x/2) +c
Now ∫ f(x)g'(x) dx = (3x-∏) (-1/2 sin(x/2)) - 3(-3/2 cos(x/2)) + c
= (3x-∏) (-1/2 sin(x/2))- (1/2(2 cos(x/2)) +c
where did i start going wrong?