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.

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 $$\begin{align*} \int x^3 \sqrt{4-x^2} \, dx &= \int u^4 - 4u^2 \, du \\ &= \frac{u^5}{5} - \frac{4u^3}{3} + C \\ &= (4-x^2)^{3/2} \left( \frac{4-x^2}{5} - \frac{4}{3} \right) + C \\ &= -\frac{(4-x^2)^{3/2}(8+3x^2)}{15} + C. \end{align*}$$
