Question

X²乘以根号下4-x²的不定积分？

Answer 1

第二类换元法，令x=2sint，t∈[-π/2，π/2]，则dx=2costdt，√(4-4sin²t)=2√cos²t=2|cost|=2cost，∴原式=……

Answer 2

先正佘弦函数换元，再降成一次，再换回原来的元。

看过程体会

满意，请及时采纳。谢谢！

Answer 3

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

Answer 4

记x=2sint,则积分变成S16(sint)^2(cost)^2dt=2S(sin2t)^2d(2t)=S(1-cos4t)d(2t)=2t+sin4t/2+C.将t=arcsin(x/2)代入上式做适当变形就可以了。

Answer 5

令x＝2cost