∫(0,π)(1-sinx)^0.5.dx

Question

∫(0,π)(1- sin x)^0.5.dx

Answer 1

∫ √(1 - sinx) dx、令y^2 = 1 - sinx、x = arcsin(1 - y^2)、dx = - 2/√(2 - y^2) dy
= ∫ y * (- 2)/√(2 - y^2) dy
= ∫ 1/√(2 - y^2) d(2 - y^2)
= 2√(2 - y^2) + C
= 2√[2 - (1 - sinx)] + C
= 2√(1 + sinx) + C
∴
∫(0→π) √(1 - sinx) dx
= 2∫(0→π/2) √(1 - sinx) dx.x∈[0,π/2]与x∈[π/2,π]的区域相等
= 4√(1 + sinx) |(0→π/2)
= 4(√2 - 1)