Question

∫x3√(2-x2)dx求积分

∫x3√(2-x2)dx求积分

Answer 1

令x = √2sinz，dx = √2coszdz
√(2 - x²) = √(2 - 2sin²z) = √2cosz => cosz = √(2 - x²)/√2
∫ x³√(2 - x²) dx
= ∫ 2^(3/2)sin³z · √2cosz · √2cosz dz
= 4√2∫ sin²zcos²z d(- cosz)
= 4√2∫ (cos²z - 1)cos²z d(cosz)
= 4√2∫ (cos⁴z - cos²z) d(cosz)
= (4√2)(1/5 · cos⁵z - 1/3 · cos³z) + C
= (4√2)[1/5 · (2 - x²)^(5/2)/2^(5/2) - 1/3 · (2 - x²)^(3/2)/2^(3/2)] + C
= (- 1/15)(3x² + 4)(2 - x²)^(3/2) + C