1. Observe the upper and lower bounds. On the left is x to 1, and on the right is 1 to 1/x (from 1/x to 1 in turn). Pay attention to x->; 1/x， 1->； 1 can be solved by inverse function.

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.