∫<0,1>(1-x^2)^m/2dx

如题所述

令x = sinθ，dx = cosθ dθ
x = 0 --> θ = 0
x = 1 --> θ = π/2
∫(0→1) (1 - x²)^(m/2) dx
= ∫(0→π/2) (1 - sin²θ)^(m/2) * (cosθ) dθ
= ∫(0→π/2) (cos²θ)^(m/2) * (cosθ) dθ
= ∫(0→π/2) |cosθ|^m * cosθ dθ
= ∫(0→π/2) (cosθ)^(m + 1) dθ，|cosθ| = cosθ 当 θ∈[0，π/2]
若m + 1是奇数，则结果
= [(m + 1) - 1]!!/(m + 1)!!
= m!!/(m + 1)!!
若m + 1是偶数，则结果
= [(m + 1) - 1]!!/(m + 1)!! * π/2
= m!!/(m + 1)!! * π/2

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