如何用分部积分法（二重积分法）解决？

Question

如题所述

Answer 1

let
x=2sinu
dx=2cosu du
∫ x^2.√(4-x^2) dx
=-(1/3)∫ x d(4-x^2)^(3/2)

=-(1/3)x(4-x^2)^(3/2) + (1/3)∫ (4-x^2)^(3/2) dx
=-(1/3)x(4-x^2)^(3/2) + (16/3)∫ (cosu)^4 du
=-(1/3)x(4-x^2)^(3/2) + (16/3)[(1/4)sinu.(cosu)^3 -(3/8)sinu.cosu +(3/8)u] + C
=-(1/3)x(4-x^2)^(3/2) + (4/3)sinu.(cosu)^3 -2sinu.cosu +2u + C
=-(1/3)x(4-x^2)^(3/2) + (4/3)(x/2).(√(4-x^2)/2)^3 -2(x/2)(√(4-x^2)/2)+2arcsin(x/2)+ C
=-(1/3)x(4-x^2)^(3/2) + (1/12)x.(4-x^2)^(3/2) -(1/2)x.√(4-x^2)+2arcsin(x/2)+ C
=-(1/4)x(4-x^2)^(3/2) -(1/2)x.√(4-x^2)+2arcsin(x/2)+ C
//
∫ (cosu)^4 du
=∫ (cosu)^3 dsinx
=sinx.(cosu)^3 +3∫ (cosu)^2 .(sinx)^2 dx
=sinx.(cosu)^3 +3∫ (cosu)^2 .[1-(cosx)^2] dx
4∫ (cosu)^4 du =sinx.(cosu)^3 -3∫ (cosu)^2 dx

∫ (cosu)^4 du
=(1/4)sinu.(cosu)^3 -(3/4)∫ (cosu)^2 du
=(1/4)sinu.(cosu)^3 -(3/4)[ (1/2)sinu.cosu -(1/2)∫ du ]
=(1/4)sinu.(cosu)^3 -(3/8)sinu.cosu +(3/8)∫du
=(1/4)sinu.(cosu)^3 -(3/8)sinu.cosu +(3/8)u + C