本題你漏了積分限，積分限應是[0--->π]

∫[0--->π]xf(sinx)dx

做變數代換，令x=π-u，則dx=-du，u：π--->0

=-∫[π--->0](π-u)f(sin(π-u))du

=∫[0--->π](π-u)f(sinu)du

=∫[0--->π]πf(sinu)du-∫[0--->π]uf(sinu)du

定積分可隨便換積分變數

=∫[0--->π]πf(sinx)dx-∫[0--->π]xf(sinx)dx

將-∫[0--->π]xf(sinx)dx移動等式左邊與左邊合併得

2∫[0--->π]xf(sinx)dx=π∫[0--->π]f(sinx)dx

即：∫[0--->π]xf(sinx)dx=π/2∫[0--->π]f(sinx)dx