谁知道这一个怎么求不定积分 根号下x-x^2的不定积分,用换元法啊啊
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√(x - x^2) = √[(1/2)^2 - (x - 1/2)^2]
令x - 1/2 = (1/2)siny、dx = (1/2)cosy dy
siny = 2x - 1、cos^2(y) = 1 - (2x - 1)^2 = 1 - (4x^2 - 4x + 1) = 4x - 4x^2 → cosy = 2√(x - x^2)
√(x - x^2) = √[(1/2)^2 - (1/2)^2sin^2(y)] = (1/2)cosy
则∫ √(x - x^2) dx
= ∫ [(1/2)cosy]^2 dy
= (1/4)∫ cos^2(y) dy
= (1/8)∫ (1 + cos2y) dy
= (1/8)[y + (1/2)sin2y] + C
= (1/8)y + (1/8)sinycosy + C
= (1/8)arcsin(2x - 1) + (1/8)(2x - 1) * 2√(x - x^2) + C
= (1/8)arcsin(2x - 1) + (1/4)(2x - 1)√(x - x^2) + C
令x - 1/2 = (1/2)siny、dx = (1/2)cosy dy
siny = 2x - 1、cos^2(y) = 1 - (2x - 1)^2 = 1 - (4x^2 - 4x + 1) = 4x - 4x^2 → cosy = 2√(x - x^2)
√(x - x^2) = √[(1/2)^2 - (1/2)^2sin^2(y)] = (1/2)cosy
则∫ √(x - x^2) dx
= ∫ [(1/2)cosy]^2 dy
= (1/4)∫ cos^2(y) dy
= (1/8)∫ (1 + cos2y) dy
= (1/8)[y + (1/2)sin2y] + C
= (1/8)y + (1/8)sinycosy + C
= (1/8)arcsin(2x - 1) + (1/8)(2x - 1) * 2√(x - x^2) + C
= (1/8)arcsin(2x - 1) + (1/4)(2x - 1)√(x - x^2) + C
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