Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
 A: We can write the integral as follows
\begin{align*}
\int x^3\sqrt{4-x^2}dx&=\int x(x^2-4)\sqrt{4-x^2}dx+\int 4x\sqrt{4-x^2}dx\\
&=\int(4-x^2)\sqrt{4-x^2}(-xdx)-2\int (4-x^2)^{1/2}(-2xdx)\\
&=\int(4-x^2)^{3/2}(-xdx)-2\int (4-x^2)^{1/2}(-2xdx)\\
&=\frac{1}{2}\frac{(4-x^2)^{5/2}}{5/2}-2\frac{(4-x^2)^{3/2}}{3/2}+C\\
&=\frac15(4-x^2)^{5/2}-\frac43(4-x^2)^{3/2}+C
\end{align*}
A: $$\int x^3\sqrt{4-x^2}\space\text{d}x=$$

Substitute $u=x^2$ and $\text{d}u=2x\space\text{d}x$:

$$\frac{1}{2}\int u\sqrt{4-u}\space\text{d}u=$$

Substitute $s=4-u$ and $\text{d}s=-\space\text{d}u$:

$$\frac{1}{2}\int(s-4)\sqrt{s}\space\text{d}s=\frac{1}{2}\int\left[s^{\frac{3}{2}}-4\sqrt{s}\right]\space\text{d}s=$$
$$\frac{1}{2}\left[\int s^{\frac{3}{2}}\space\text{d}s-4\int\sqrt{s}\space\text{d}s\right]=\frac{1}{2}\left[\int s^{\frac{3}{2}}\space\text{d}s-4\int s^{\frac{1}{2}}\space\text{d}s\right]=$$

Use:
$$\int y^b\space\text{d}y=\frac{y^{b+1}}{b+1}+\text{C}$$

$$\frac{1}{2}\left[\frac{2s^{\frac{5}{2}}}{5}-4\cdot\frac{2s^{\frac{3}{2}}}{3}\right]+\text{C}=\frac{1}{2}\left[\frac{2s^{\frac{5}{2}}}{5}-\frac{8s^{\frac{3}{2}}}{3}\right]+\text{C}=\frac{s^{\frac{5}{2}}}{5}-\frac{4s^{\frac{3}{2}}}{3}+\text{C}=$$
$$\frac{(4-u)^{\frac{5}{2}}}{5}-\frac{4(4-u)^{\frac{3}{2}}}{3}+\text{C}=\frac{(4-x^2)^{\frac{5}{2}}}{5}-\frac{4(4-x^2)^{\frac{3}{2}}}{3}+\text{C}$$
A: Here's yet another way.  First write: $$\int x^3\sqrt{4-x^2} \, dx = \int x^2 \sqrt{4-x^2} \, \cdot x \, dx$$
Now let $u = 4-x^2$, so then $du = -2x\, dx$ (meaning $x\,dx = -\frac{1}{2}du$), and $x^2 = 4 - u$.  Then you get:
\begin{align}
  \int x^2 \sqrt{4-x^2} \, \cdot x \, dx
    &= \int (4-u)\sqrt{u} \, \cdot \left(-\frac{1}{2}\right) \, du\\[0.3cm]
    &= -\frac{1}{2} \int (4u^{1/2} - u^{3/2}) \, du\\[0.3cm]
    &= -\frac{1}{2}\left(4 \cdot \frac{2}{3} u^{3/2} - \frac{2}{5}u^{5/2}\right) + C\\[0.3cm]
    &= \frac{1}{5}(4-x^2)^{5/2} - \frac{4}{3}(4-x^2)^{3/2} + C
\end{align}
A: 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 $$x^3 \sqrt{4-x^2} \, dx = x^2 \sqrt{4-x^2} \cdot x \, dx = (4-u^2)u(-u) \, du = (u^4 - 4u^2) \, du.$$  It immediately follows that $$\begin{align*} \int x^3 \sqrt{4-x^2} \, dx &= \int u^4 - 4u^2 \, du \\ &= \frac{u^5}{5} - \frac{4u^3}{3} + C \\ &= (4-x^2)^{3/2} \left( \frac{4-x^2}{5} - \frac{4}{3} \right) + C \\ &= -\frac{(4-x^2)^{3/2}(8+3x^2)}{15} + C. \end{align*}$$
A: Substituting $x=2\sin t$, we get:
$$ \int x^3\sqrt{2^2-x^2}dx=32\int \sin^3 t \cos^2 t dt=32\int(1-\cos^2 t)\cos ^2 t\sin t dt$$
$$=32\int (\cos^2 t-\cos ^4 t)\sin tdt$$
$$=-32\int z^2 dz + 32\int z^4 dz$$ Where $z=\cos t.$ 
