f（x）=sin²x+2sinxcosx-cos²x
=sin2x-cos2x
=√2sin(2x-π/4)
2kπ-π/2≤2x-π/4≤2kπ+π/2
2kπ-π/4≤2x≤2kπ+3π/4
kπ-π/8≤x≤kπ+3π/8
所以单调递增区间：[kπ－π/8,kπ＋3π/8],k∈Z