So let t= 1/x, then x = 1/t, dx =-dt/t 2.
Right = ∫-(π-t) f (Sint) dt = ∫ [0,π] (π-t) f (Sint) dt
=π∫[0,π] f(sint)dt -∫[0,π] tf(sint)dt=π∫[0,π] f(sint)dt -∫[0,π] xf(sinx)dx
So 2 ∫ [0, π] xf (sinx) dx = π∫ [0, π] f (Sint) dt.
So ∫ [0, π] xf (sinx) dx = (π/2) ∫ [0, π] f (Sint) dt = (π/2) ∫ [0, π] f (sinx) dx.