<s>[INST] 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.
 [/INST] 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}
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