How is the formula derived?

Prove: let x=π-t, then x is from 0 to π, t is from π to 0, and DX =-DT.

The original formula is labeled I.

Then I=- (integral interval π to 0)∫(π-t)f(sin(π-t)dt.

=-(integer range π to 0)∫(π-t)f(sin(t)dt

= (integration interval 0 to π)∫(π-t)f(sin(t)dt

= (integration interval 0 to π)∫πf(sin(t)dt-I

So 2I= (integration interval 0 to π)∫πf(sin(t)dt.

That is, I = (π/2) ∫ f (Sint) dt = (π/2) ∫ f (sinx) dx.