f(x)在[-π,π]连续,f(x)=x/[1cos^2(x)]∫f(x)sinxdx,求f(x).f(x)在[-π,π]连续,f(x)=x/[1+cos^2(x)]+∫f(x)sinxdx,求f(x)。漏写了个+号。刚又又少打了个加号就是不会求∫xsinx/[1+cos^2(x)]dx

Question

f(x)在[-π,π]连续,f(x)=x/[1 cos^2(x)] ∫f(x)sinxdx,求f(x).
f(x)在[-π,π]连续,f(x)=x/[1+ cos^2(x)]+ ∫f(x)sinxdx,求f(x)。漏写了个+号。刚又又少打了个加号 就是不会求∫xsinx/[1 +cos^2(x)]dx

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