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  • 求定積分(上限為1下限為0)∫[4√(1-x²)-2x√((1-x²)]dx

    求定積分(上限為1下限為0)∫[4√(1-x²)-2x√((1-x²)]dx
    數(shù)學(xué)人氣:336 ℃時(shí)間:2019-11-04 08:58:23
    優(yōu)質(zhì)解答
    令x = sinθ,dx = cosθdθ
    當(dāng)x = 0,θ = 0,當(dāng)x = 1,θ = π/2
    ∫(0-->1) [4√(1 - x²) - 2x√(1 - x²)] dx
    = ∫(0-->π/2) (4cosθ - 2sinθcosθ)(cosθdθ)
    = ∫(0-->π/2) (4cos²θ - 2sinθcos²θ) dθ
    = 4∫(0-->π/2) cos²θ dθ - 2∫(0-->π/2) sinθcos²θ dθ
    = 4∫(0-->π/2) (1 + cos2θ)/2 dθ + 2∫(0-->π/2) cos²θ d(cosθ)
    = 2(θ + 1/2 · sin2θ) + (2/3)cos³θ |(0-->π/2)
    = 2(π/2) - 0 + (2/3)(0) - (2/3)
    = π - 2/3
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