∫(0,9)x^2√(a^2-x^2)dx

你的题目估计写错了,积分区间应该是(0,a),或将a^2改成9；这是很简单的问题,直接利用三角换元法：
令x=asint (-π/2＜t＜π/2),则√(a^2-x^2) = acost,dx = acost dt
当 x = 0 ,t = 0 ；当 x = a ,t = π/2
∫(0,a) x^2√(a^2-x^2)dx
=∫(0,π/2) (asint)^2·acost·acost dt
=a^4·∫(0,π/2) (sint)^2·(cost)^2 dt
=a^4/4·∫(0,π/2) (sin2t)^2 dt
=a^4/8·∫(0,π/2) (1 - cos4t) dt
=a^4/8·(t - 1/4 sin4t)|(0,π/2)
=π·a^4/16