This statement is false and cannot be proven.

Let $I = int_0^pi xf(sin x) dx$. Substitute $u = pi - x$, then $I = int_0^pi (pi - u)f(sin(pi - u)) du = int_0^pi (pi - x)f(sin x) dx = piint_0^pi f(sin x) dx - int_0^pi xf(sin x) dx = piint_0^pi f(sin x) dx - I$. Hence $2I = piint_0^pi f(sin x) dx$, so $I = rac{pi}{2}int_0^pi f(sin x) dx$.

Let $I = int_0^pi xf(sin x) dx$. Substitute $u = pi + x$, then $I = int_0^pi (pi + u)f(sin(pi + u)) du = int_0^pi (pi + x)f(sin x) dx = piint_0^pi f(sin x) dx + int_0^pi xf(sin x) dx = piint_0^pi f(sin x) dx + I$. Hence $0 = piint_0^pi f(sin x) dx$, which leads to the conclusion.

Let $I = int_0^pi xf(sin x) dx$. Substitute $u = -x$, then $I = int_0^pi (-u)f(sin(-u)) du = -int_0^pi xf(-sin x) dx$.

Let $I = int_0^pi xf(sin x) dx$. Substitute $u = 2x$, then $I = int_0^pi (u/2)f(sin(u/2)) du = (1/2)int_0^pi uf(sin(u/2)) du$