anzz
幼苗
共回答了17个问题采纳率:100% 举报
∫ √(2x - x²) dx
= ∫ √[1 - (x - 1)²] dx
令x - 1 = sinz,dx = cosz dz
= ∫ cos²z dz
= ∫ (1 + cos2z)/2 dz
= (1/2)z + (1/2)sinzcosz + C
= (1/2)arcsin(x - 1) + (1/2)(x - 1)√[1 - (x - 1)²] + C
= (1/2)arcsin(x - 1) + (1/2)(x - 1)√(2x - x²) + C
1年前
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