∫1/√(2-3x^2) dx
let
x =√(2/3) .siny
dx = √(2/3) cosy dy
∫1/√(2-3x^2) dx
=∫√(2/3) cosy dy/(√2 cosy)
=(1/√3)∫dy
= (1/√3) y + C
=(1/√3) arcsin(√3x/√2) + C(√2 cosy)這個(gè)是？？？怎么得出來的？謝謝！√(2-3x^2) = √[2-2(siny)^2] = √2. cosy