Let t=π-x

dt = -dx

replace

I = ∫ (π-t) f (sin (π-t) dt = ∫ π f (Sint) dt-∫ TF (Sint) dt = π ∫ f (Sint) dt-I integral limits are all [0, π].

So I = π/2∫f(sint)dt.

(2) Since the latter integral is constant, let the integral result be c..

therefore

f(x) = x/( 1+cos^2x) + C

X/( 1+cos 2x) * sinx is an even function.

Csinx is a strange function.

So this equation can be written as

C = 2 ∫ xsinx/( 1+cos 2x) dx integration limit is [0, π].

Let f (x) = x/[2-x 2]

Then f (sinx) = sinx/(2-sinx2) = sinx/(1+cos2x).

Use conclusion

c =π∫sinx/( 1+cos^2x)dx =-π∫ 1/( 1+cos^2x)dcosx =-π*[arctan(cosx)]= π^2/2

So f (x) = x/( 1+cos 2x)+π 2/2.