<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] 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*}
</s>