∫(1,-1) x^4√(1-x^2) dx 求该定积分 要过程

发布网友 发布时间：1小时前

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热心网友 时间：1分钟前

∫(-1->1) x⁴√(1 - x²) dx
= 2∫(0->1) x⁴√(1 - x²) dx
x = sinz，dx = cosz dz
x = 0，z = 0，x = 1，z = π/2
=> 2∫(0->π/2) (sin⁴z)(cosz)(cosz) dz
= 2∫(0->π/2) sin⁴z(1-sin²z) dz
= 2(0->π/2) sin⁴z dz - 2(0->π/2) sin^6z dz
= 2*(4-1)!!/(4!!)*π/2 - 2*(6-1)!!/(6!!)*π/2
= 3π/8 - 5π/16
= π/16

热心网友 时间：2分钟前

原函数是偶函数，
令x=cosa，x=0,a=π/2，x=1,a=0
∫[-1,1] x^4√(1-x^2) dx
=2∫[0,1] x^4√(1-x^2) dx
=2∫[π/2,0] cos^4 asina dx
=-2/5cos^5 a[π/2,0]
=-2/5

热心网友 时间：1分钟前

∫(1,-1) x^4√(1-x^2) dx =2∫(0~1) x^4√(1-x^2) dx
令x=sint,dx=costdt
x=0时t=0；x=1时t=π/2
2∫(0~1) x^4√(1-x^2) dx =2∫(0~π/2) (sint)^4(cost)^2dt
=2∫(0~π/2) (sint)^4(1-(sint)^2)dt
=2[∫(0~π/2) (sint)^4dt-∫(0~π/2) (sint)^6dt]
=2[(3/4)(1/2)(π/2)-(5/6)(3/4)(1/2)(π/2)]=π/16