∫(1-x^2)^m/2dx

Question

∫(1-x^2)^m/2dx

Answer 1

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