∫x^2⼀√1-x^6dx

2025-01-07 23:39:05

推荐回答（1个）

回答1：

令u=x³，du=3x²dx，dx=1/(3x²) du
∫x³/√(1-x^6) dx = ∫x²/√[1-(x³)²] dx
= (1/3)∫du/√(1-u²) ...*
= (1/3)arcsin(u) + C
= (1/3)arcsin(x³) + C

若不明白由*处到答案的过程，可设u=sinz，du=cosz dz
(1/3)∫du/√(1-u²)
= (1/3)∫cosz/√(1-sin²z) dz
= (1/3)∫cosz/√(cos²z) dz
= (1/3)∫cosz/cosz dz
= (1/3)z + C
= (1/3)arcsin(u) + C
= (1/3)arcsin(x³) + C