求（0→π/2）∫√（1-2sinx）dx定积分

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/16 11:45:16

被积函数应该是√(1 - sinx)吧?否则当x = π/2时√(1 - 2sinx)根本不存在呢
∫(0→π/2) √(1 - sinx) dx
= ∫(0→π/2) √[sin²(x/2) - 2sin(x/2)cos(x/2) + cos²(x/2)] dx
= ∫(0→π/2) √[sin(x/2) - cos(x/2)]² dx
= ∫(0→π/2) |sin(x/2) - cos(x/2)| dx
= ∫(0→π/2) |√2sin(x/2 - π/4)| dx
= ∫(0→π/2) - 2√2sin(x/2 - π/4) d(x/2 - π/4),被积函数在(0,π/2)内 < 0
= 2√2cos(x/2 - π/4) |(0→π/2)
= 2√2(1 - 1/√2)
= 2√2 - 2

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