f'(x)=sinx+xcosx-sinx=xcosx>=0.
①当x>=0,需cosx>=0,故f(x)在[0,π/2],[2kπ-π/2,2kπ+π/2](k∈Z,k>0)上单调递增
②当x
函数f(x)=xsinx+cosx的增区间是
=0.①当x>=0,需cosx>=0,故f(x)在[0,π/2],[2kπ-π/2,2kπ+π"}}}'>
f'(x)=sinx+xcosx-sinx=xcosx>=0.
①当x>=0,需cosx>=0,故f(x)在[0,π/2],[2kπ-π/2,2kπ+π/2](k∈Z,k>0)上单调递增
②当x