求定积分(0,a) ∫x²√(a²-x²) dx原式=(0,a)∫(ax²√[1-(x/a)²]dx令x/a=sint,则dx=acostdt,x=0时,t=0；x=a时,t=π/2.故原式=(0,π/2)a⁴∫sin²tcos²tdt=(0,π/2)(a⁴/4...