这个积分怎么求的
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∫(0,sinθ)√(1-ρ^2)ρdρ
=(-1/2)*∫(0,sinθ)√(1-ρ^2)d(1-ρ^2)
=(-1/3)*(1-ρ^2)^(3/2)|(0,sinθ)
=(1/3)*(1-cos^3θ)
原式=(1/2)*∫(0,π/2)(1/3)*(1-cos^3θ)dθ
=(1/6)*[∫(0,π/2)dθ-∫(0,π/2)cos^3θdθ]
=(1/6)*[π/2+∫(0,π/2)(1-sin^2θ)d(sinθ)]
=(1/6)*{π/2+[sinθ-(1/3)*sin^3θ]|(0,π/2)}
=(1/6)*(π/2+1-1/3)
=π/12+1/9
=(-1/2)*∫(0,sinθ)√(1-ρ^2)d(1-ρ^2)
=(-1/3)*(1-ρ^2)^(3/2)|(0,sinθ)
=(1/3)*(1-cos^3θ)
原式=(1/2)*∫(0,π/2)(1/3)*(1-cos^3θ)dθ
=(1/6)*[∫(0,π/2)dθ-∫(0,π/2)cos^3θdθ]
=(1/6)*[π/2+∫(0,π/2)(1-sin^2θ)d(sinθ)]
=(1/6)*{π/2+[sinθ-(1/3)*sin^3θ]|(0,π/2)}
=(1/6)*(π/2+1-1/3)
=π/12+1/9
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