∫x3⼀√(1-x2)dx

原式=½∫x²dx²/√(1-x²）
令t=√(1-x²），则t²=1-x²，x²=1-t²
原式=½∫(1-t²)d(1-t²)/t=∫(t²-1)dt=1/3t³-t+C
x=0时，t=1；x=1时，t=0
∴∫（0~1）[x^3/√(1-x^2)]dx = ∫(1~0)(t²-1)dt=0-1/3+1=2/3

令x=sint，t属于(0,π/2)。
∫（0~1）[x^3/√(1-x^2)]dx = ∫(0~π/2) [(sint)^3/cost] cost dt
= ∫(0~π/2) (sint)^3 dt
= - ∫(0~π/2) [1-(cost)^2] d(cost)
= - [cost - 1/3*(cost)^3] | (0,π/2)
= - { [cos0 - 1/3*(cos0)^3] - [cos(π/2) - 1/3*(cos(π/2))^3]}
= - { [ 1 - 1/3] - [ 0 - 0] }
= 2/3

解:
设x=sint,t=arcsinx,1-x^2=(cost)^2,dx=costdt
原式=∫[0,π/2](sint)^3dt
=∫[0,π/2](sint)^2dcost
=∫[0,π/2](1-cos^2t)dcost
=∫[0,π/2]dcost-∫[0,1]cos^2tdcost
=cost|[0,π/2]-1/3cos^3t|[0,π/2]
=-1-1/3(0-1)
=-1+1/3=-2/3