Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$

Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$

I am trying to compute the following integral:
$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
I know this requires integration by parts multiple times but I am having
trouble figuring out what to do once you have integrated twice. This is
what I have done:
Let $u = \cos \frac{x}{2}$ and $\mathrm{d}u =
\frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ and $\mathrm{d}v =
x^2$ and $v= \frac{x^3}{3}$.
\begin{align} &\int x^2 \cos \frac{x}{2} \\ &\cos \frac{x}{2} \cdot
\frac{x^3}{3} - \int \frac{x^3}{3} \cdot
\frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x \end{align}
So now I integrate $\int \frac{x^3}{3} \cdot
\frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ to get:
\begin{align} \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \cdot
\frac{x^4}{12} - \int \frac{x^4}{12} \cdot \frac{\cos x}{2} \end{align}
Now, this is where I get stuck. I know if I continue, I will end up with
$\frac{-\sin\left(\dfrac{x}{2}\right)}{2}$ again when I integrate $\cos
\frac{x}{2}$. So, where do I go from here?
Thanks!