calculus - 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!

Answer

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

Let $u = x^2$ and $\mathrm{d}u = 2x\,dx$ and let $dv = \cos\frac x2\,dx \implies v = 2 \sin \frac x2$.

$$2x^2 \sin \frac x2 - \int 2x\cdot 2\sin \frac x2\,dx$$

You'll only need to do integration by parts one additional time. Let me know if you get stuck after that.