在x∈[0,2π]内解sin(x + 1) = 0解得x = π - 1,x = 2π - 1在x∈[0,π - 1]和[2π - 1,2π],sin(x + 1) > 0在x∈[π - 1,2π - 1],sin(x + 1) < 0∴∫(0→2π) |sin(x + 1)| dx= ∫(0→π - 1) sin(x + 1) dx - ∫(...