What's the best approach?

Rewrite the considered integral as $\frac{1}{2}\int_0^\pi x^3\sin(2x)\,dx$ Now consider $\int_0^\pi \cos(ax)\,dx = \frac{\sin(a\pi)}{a}$ Differentiating w.r.t. $a$ three times and putting $a=2$ so that $\int_0^\pi x^3\sin(2x)\,dx=\frac{\partial^3}{\partial a^3}\left[\frac{\sin(a\pi)}{a}\right]_{a=2}=\frac{\pi}{4}\left(3-2\pi^2\right)$ Hence $\int_0^\pi x^3\sin(x)\cos(x)\,dx=\frac{\pi}{8}\left(3-2\pi^2\right)\approx-6.57347192497878$ Remark : I think differentiating multiple times (although it's tedious) is easier than applying integration by parts multiple times.

I solved this problem using complex numbers. $\begin{aligned} I &=\int_{0}^{\pi} \cos {x} \ \sin {x} \ x^3 \mathrm{d}x\\ &=\dfrac{1}{2} \int_{0}^{\pi} \sin {2x} \ \ x^3 \mathrm{d}x \\ &= \dfrac{1}{2}\Im \int_{0}^{\pi} e^{2ix} \ x^3 \mathrm{d}x \end{aligned}$

Now, substitute $2ix=t$ to get $\begin{aligned} I &= \dfrac{1}{32} \Im \int_{0}^{2i\pi} e^t \ t^3 \mathrm{d}t\\ &=\dfrac{1}{32} \Im \int_{0}^{2i\pi} e^t (t^3+3t^2-3t^2-6t+6t+6-6) \mathrm{d}t \ \ \ \ \ \ \ \ \ \ (\star)\\ &= \dfrac{1}{32} \Im\left[e^t(t^3-3t^2+6t-6)\right]_{0}^{2i\pi} \end{aligned}$

Substitute and get the imaginary part. You will find that $I=\dfrac{3\pi-2\pi^3}{8} \approx -6.573$

$\text{Note :}$ I arrived at $(\star)$ by using the fact that $\int_{a}^{b} e^x(f(x)+f'(x))=\left[e^x f(x)\right]_{a}^{b}$