f(x)=π-x/2=a0/2+∑ancosnωx+bnsinnωx；∑的n从1~∞
T=π,ω=2π/T=2
an=2/π ∫f(x)cosnωxdx n=0,1,2...
bn=2/π ∫f(x)sinnωxdx n=1,2,3...
an=2/π ∫(π-x/2)cosn2xdx
bn=2/π ∫(π-x/2)sinn2xdx
注：定积分下限为-π/2,上限为π/2
an=2∫cosn2xdx-2 /π∫x/2cosn2xdx=2/n sinnπ-1 /4n^2π∫2nxdsin2nx
采用分部积分法
∫2nxdsin2nx=2nxsin2nx-∫sin2nxd2nx=2nxsin2nx+∫dcos2nx=2nπsinnπ
an=3/2n sinnπ
an=0
n=0,2,4...
an=3/2n
n=1,3,5...
bn=2∫sinn2xdx-2/π∫x/2sinn2xdx=-1/4n^2π∫2nxsinn2xd2nx
采用分部积分法
∫2nxsinn2xd2nx=-∫2nxdcos2nx=-2nxcos2nx+∫dsin2nx=-2nπcosnπ
bn=cosnπ/2n
bn=1/2n
n=1,3,5...
bn=-1/2n
n=2,4,6...
希望能解决您的问题.