$$\lim_{x \to \frac{\pi}{2}} \left(\frac{\int_{x^3}^{(\pi/2)^3} \left(\sin(2t^{1/3}) + \cos(t^{1/3})\right) dt}{\left(x - \frac{\pi}{2}\right)^2}\right)$$ is equal to

Form: This is a $$0/0$$ form. We apply L'Hôpital's Rule and the Leibniz Rule for differentiating integrals.

First Derivative:
Numerator: $$-(\sin(2(x^3)^{1/3}) + \cos((x^3)^{1/3})) \cdot 3x^2 = -(\sin 2x + \cos x) \cdot 3x^2$$.
Denominator: $$2(x - \pi/2)$$.
Evaluate Limit: We still have $$0/0$$ because $$\sin(2 \cdot \pi/2) + \cos(\pi/2) = 0+0=0$$. Apply L'Hôpital again.
Numerator: $$-[ (2\cos 2x - \sin x) \cdot 3x^2 + (\sin 2x + \cos x) \cdot 6x ]$$.
Denominator: $$2$$.
Plug in $$x = \pi/2$$:
$$x^2 = \pi^2/4$$. $$\cos(\pi) = -1$$, $$\sin(\pi/2) = 1$$.
$$\text{Limit} = \frac{-[ (2(-1) - 1) \cdot 3(\pi^2/4) + (0) ]}{2} = \frac{-[ -3 \cdot 3\pi^2/4 ]}{2} = \frac{9\pi^2/4}{2} = \mathbf{\frac{9\pi^2}{8}}$$.