Question

积分区间[0,派]被积函数为根号下1-sinx

Answer 1

简单分析一下，详情如图所示

Answer 2

(0，π)∫√(1-sinx)dx
= (0，π)∫√[sin²(x/2)+cos²(x/2)-2sin(x/2)cos(x/2)]dx
= (0，π)∫ | sin(x/2)-cos(x/2) | dx
= (0，π/2)∫ [cos(x/2)-sin(x/2)] dx + (π/2，π)∫ [sin(x/2)-cos(x/2)] dx
= 2[sin(x/2)+cos(x/2)] | (0，π/2) + 2[-cos(x/2)-sin(x/2)] | (π/2，π)
= 2[sin(x/2)+cos(x/2)] | (0，π/2) - 2[cos(x/2)+sin(x/2)] | (π/2，π)
= 2{(√2/2+√2/2)-(0+1)} - 2{(0+1)-(√2/2+√2/2)}
= 2(√2-1) - 2(1-√2)
= 4(√2-1)

Answer 3