记:∫[0,π/2]f(t)dt=k（常数）则f(x)=xcosx+∫ [0,π/2]f(t)dt可化为f(x)=xcosx+k两边在[0,π/2]积分有∫[0,π/2]f(t)dt=∫[0,π/2]tcostdt+k∫[0,π/2]dt【分部积分】k=tsint[0,π/2]-∫[0,π/2]sintdt+kπ/2k=π/...