∫(0~π/2) √(1 - sin2x) dx= ∫(0~π/2) √(sin²x - 2sinxcosx + cos²x) dx= ∫(0~π/2) √(sinx - cosx)² dx= ∫(0~π/2) |sinx - cosx| dx解sinx - cosx = 0,x∈[0,π/2]得tanx = 1 => x = π/4...