求积分∫√(1-x^2)^3dx上限1下限0 分母全在根号内
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求积分∫√(1-x^2)^3dx上限1下限0 分母全在根号内
令x = sinθ,dx = cosθ dθ
x∈[0,1] → θ∈[0,π/2]
∫(0~1) (1 - x²)^(3/2) dx
= ∫(0~π/2) cos³θ • cosθ dθ
= ∫(0~π/2) (cos²θ)² dθ
= ∫(0~π/2) [(1/2)(1 + cos2θ)]² dθ
= (1/4)∫(0~π/2) (1 + 2cos2θ + cos²2θ) dθ
= (1/4)∫(0~π/2) (1 + 2cos2θ) dθ + (1/8)∫(0~π/2) (1 + cos4θ) dθ
= (1/4)[θ + sin2θ] |(0~π/2) + (1/8)[θ + 1/4 • sin4θ] |(0~π/2)
= (1/4)(π/2) + (1/8)(π/2)
= 3π/16
x∈[0,1] → θ∈[0,π/2]
∫(0~1) (1 - x²)^(3/2) dx
= ∫(0~π/2) cos³θ • cosθ dθ
= ∫(0~π/2) (cos²θ)² dθ
= ∫(0~π/2) [(1/2)(1 + cos2θ)]² dθ
= (1/4)∫(0~π/2) (1 + 2cos2θ + cos²2θ) dθ
= (1/4)∫(0~π/2) (1 + 2cos2θ) dθ + (1/8)∫(0~π/2) (1 + cos4θ) dθ
= (1/4)[θ + sin2θ] |(0~π/2) + (1/8)[θ + 1/4 • sin4θ] |(0~π/2)
= (1/4)(π/2) + (1/8)(π/2)
= 3π/16
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