Question

f(x)在[-π,π]连续,f(x)=x/[1 cos^2(x)] ∫f(x)sinxdx,求f(x).

Answer 1

如果你的题目是f(x)=x/[1 +cos^2(x)]+ ∫f(x)sinxdx那么

两边同时乘以sinx得f(x)sinx={x/[1 +cos^2(x)]+ ∫f(x)sinxdx}sinx

令∫f(x)sinxdx=t

->f(x)sinx={x/[1 +cos^2(x)]+ t}sinx两边积分

∫f(x)sinxdx=∫xsinx/[1 +cos^2(x)]dx+∫ tsinxdx

即t=∫xsinx/[1 +cos^2(x)]dx+t∫ sinxdx

->t=∫xsinx/[1 +cos^2(x)]dx

把t带入原式得

f(x)=x/[1 +cos^2(x)]+ t

如果是f(x)=x/[1 -cos^2(x)]+ ∫f(x)sinxdx方法一样的,而且那个积分还简单点