∫(1,-1) x^4√(1-x^2) dx 求该定积分
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∫(-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
= 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
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