f(x)=sinx + 3∫f(t)costdt(定积分0--2π)化简,求f(x)

设：∫[0,2π] f(t)cost dt = A ,则：
f(x) = sinx + 3∫[0,2π] f(t)cost dt = sinx + 3A
f(x)cosx = [sinx + 3A]cosx
A =∫[0,2π] f(t)cost dt
=∫[0,2π] f(x)cosx dx
=∫[0,2π] [sinx + 3A]cosx dx
=∫[0,2π] [sinx + 3A] d(sinx+3A)
= 1/2 [sinx + 3A]^2|[0,2π]
= 1/2 * 0
= 0
∴ f(x) = sinx

因为f的后半部分是个常数,所以f`=cosx,之后再用分部积分法将：积分f(t)costdt求出=0.5[(cost)^2+costsint].之后代入上下限即可,我算出来和一楼一样,所以要么楼主题抄错了,要么答案错了