How do I solve this calculus problem? (Calculus, integration w/substitution)?

Question

http://geocities.com/jsprplc2006/eq2.jpg

Answer 1

Let A = ∫ (0 to π) x f(sin x) dx
= ∫ (π to 0) (π-u) f(sin (π-u)) (-du), subs. u = π-x
= ∫ (0 to π) (π-u) f(sin u) du
= π ∫ (0 to π) f(sin u) du - ∫ (0 to π) u f(sin u) du
= π ∫ (0 to π) f(sin u) du - A
=> 2A = π ∫ (0 to π) f(sin u) du
=> A = (π/2) ∫ (0 to π) f(sin x) dx as required.

Answer 2

Let x = pi - u and notice that u = pi-x, du = -dx
and sin(pi-x ) = sin x.
I'll let you carry on from here.