f(x)=sinx + 3∫f(t)costdt(定积分0--2π)化简,求f(x)
问题描述：
f(x)=sinx + 3∫f(t)costdt(定积分0--2π)化简,求f(x)
= =不好意思...定积分是（0--1/2π）
1楼答案好像不对...答案是f(x)=sinx - 3/4
我想知道正解的过程...
答
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