题目

计算∫（下限-1,上限1）x^2[sinx/(1+x^4)+√(1-x^2)]dx

提问时间：2021-03-18

答案

∫(- 1→1) x²[sinx/(1 + x⁴) + √(1 - x²)] dx
= ∫(- 1→1) x²sinx/(1 + x⁴) dx + ∫(- 1→1) x²√(1 - x²) dx
= 0 + 2∫(0→1) x²√(1 - x²) dx,第一个是奇函数,第二个是偶函数
令x = sinθ,dx = cosθ dθ
当x = 0,θ = 0
当x = 1,θ = π/2
= 2∫(0→π/2) sin²θcos²θ dθ
= 2∫(0→π/2) (1/2 * sin2θ)² dθ
= (1/2)∫(0→π/2) sin²(2θ) dθ
= (1/4)∫(0→π/2) (1 - cos4θ) dθ
= (1/4)[θ - (1/4)sin4θ] |(0→π/2)
= (1/4)(π/2)
= π/8