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.
The Digital Mystery of Hutulo's Works
The expression of Hutuluo's calligraphy is the numerical relationship, which inevitably reflects the ancient mathematical leve