Vtyjkyuk

$$int x^3sqrt4-x^2dx$$
I tried changing the expression like this:
$$int x^3sqrt2^2-x^2dx=int x^3sqrt(2-x)(2+x)dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.

Here's yet another way. First write: $$int x^3sqrt4-x^2 , dx = int x^2 sqrt4-x^2 , cdot x , dx$$

Now let $u = 4-x^2$, so then $du = -2x, dx$ (meaning $x,dx = -frac12du$), and $x^2 = 4 - u$. Then you get:
beginalign
int x^2 sqrt4-x^2 , cdot x , dx
&= int (4-u)sqrtu , cdot left(-frac12right) , du\[0.3cm]
&= -frac12 int (4u^1/2 - u^3/2) , du\[0.3cm]
&= -frac12left(4 cdot frac23 u^3/2 - frac25u^5/2right) + C\[0.3cm]
&= frac15(4-x^2)^5/2 - frac43(4-x^2)^3/2 + C
endalign

We can write the integral as follows
beginalign*
int x^3sqrt4-x^2dx&=int x(x^2-4)sqrt4-x^2dx+int 4xsqrt4-x^2dx\
&=int(4-x^2)sqrt4-x^2(-xdx)-2int (4-x^2)^1/2(-2xdx)\
&=int(4-x^2)^3/2(-xdx)-2int (4-x^2)^1/2(-2xdx)\
&=frac12frac(4-x^2)^5/25/2-2frac(4-x^2)^3/23/2+C\
&=frac15(4-x^2)^5/2-frac43(4-x^2)^3/2+C
endalign*

$$int x^3sqrt4-x^2spacetextdx=$$

Substitute $u=x^2$ and $textdu=2xspacetextdx$:

$$frac12int usqrt4-uspacetextdu=$$

Substitute $s=4-u$ and $textds=-spacetextdu$:

$$frac12int(s-4)sqrtsspacetextds=frac12intleft[s^frac32-4sqrtsright]spacetextds=$$
$$frac12left[int s^frac32spacetextds-4intsqrtsspacetextdsright]=frac12left[int s^frac32spacetextds-4int s^frac12spacetextdsright]=$$

Use:

$$int y^bspacetextdy=fracy^b+1b+1+textC$$

$$frac12left[frac2s^frac525-4cdotfrac2s^frac323right]+textC=frac12left[frac2s^frac525-frac8s^frac323right]+textC=fracs^frac525-frac4s^frac323+textC=$$
$$frac(4-u)^frac525-frac4(4-u)^frac323+textC=frac(4-x^2)^frac525-frac4(4-x^2)^frac323+textC$$

The single substitution $$u^2 = 4-x^2$$ will yield the simplest computation. Note this substitution implies $$x^2 = 4 - u^2, quad x , dx = -u , du,$$ hence $$x^3 sqrt4-x^2 , dx = x^2 sqrt4-x^2 cdot x , dx = (4-u^2)u(-u) , du = (u^4 - 4u^2) , du.$$ It immediately follows that $$beginalign* int x^3 sqrt4-x^2 , dx &= int u^4 - 4u^2 , du \ &= fracu^55 - frac4u^33 + C \ &= (4-x^2)^3/2 left( frac4-x^25 - frac43 right) + C \ &= -frac(4-x^2)^3/2(8+3x^2)15 + C. endalign*$$

Substituting $x=2sin t$, we get:
$$ int x^3sqrt2^2-x^2dx=32int sin^3 t cos^2 t dt=32int(1-cos^2 t)cos ^2 tsin t dt$$
$$=32int (cos^2 t-cos ^4 t)sin tdt$$
$$=-32int z^2 dz + 32int z^4 dz$$ Where $z=cos t.$