不是不能用
分部积分法,而是你运算有错:d√(a²-x²)=-2x/[2√(a²-x²)]dx=-[x/√(a²-x²)]dx;
∴∫<0,a>[x²/√(a²-x²)]dx=-∫xd√(a²-x²)=-[x√(a²-x²)]<0,a>+∫<0,a>√(a²-x²)dx
=∫<0,a>√(a²-x²)dx【令x=asinθ,则dθ=acosθdθ,x=0时θ=0;x=a时θ=π/2】
=∫<0,π/2>[√(a²-a²sin²θ)]•acosθdθ=a²∫<0,π/2>cos²θdθ
=(a²/2)∫<0,π/2>(1+cos2θ)dθ=(a²/2)[θ+(1/2)sin2θ]<0,π/2>=πa²/4;