题目

求定积分（上限为1下限为0）∫[4√（1-x²）-2x√（（1-x²）]dx

提问时间：2020-07-25

答案

令x = sinθ,dx = cosθdθ
当x = 0,θ = 0,当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