Question

求不定积分∫x2根号4-x2dx

Answer 1

令x = 2sinθ,dx = 2cosθdθ
∫ x²/√(4 - x²) dx
= ∫ 4sin²θ/√(4 - 4sin²θ) * (2cosθdθ)
= ∫ 4sin²θ/|2cosθ| * (2cosθdθ)
= 4∫ sin²θ dθ
= 2∫ (1 - cos2θ) dθ
= 2θ - 2(1/2)sin2θ + C
= 2θ - 2sinθcosθ + C
= 2arcsin(x/2) - 2(x/2)√(4 - x²)/2 + C
= 2arcsin(x/2) - (x/2)√(4 - x²) + C

Answer 2

设x = 2sint,
4(sint)^2*2costdt = 8(sint)^2dsint

Answer 3