令a=3即可，详情如图所示
∫[1/√(9-x²)]dx【令x=3sinu，则dx=3cosudu】=∫[1/√(9-9sin²u)]3cosudu=∫du=u+C=3arcsin(x/3)+C;
∫dx/√(9-x²)=∫d(x/3)/√(1-x²/9)=arcsin(x/3)+C。
letx=3sinudx=3cosu du∫dx/√(9-x^2)=∫ du=u + C=arcsin(x/3) +C
