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

This is a $$\frac{0}{0}$$ form. We apply L'Hôpital's Rule and the Leibniz Rule for differentiating integrals.

Let $$L$$ be the limit:

$$L = \lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx} \int_{x^3}^{(\frac{\pi}{2})^3} \cos(t^{1/3}) \, dt}{\frac{d}{dx} (x - \frac{\pi}{2})^2}$$

The derivative of the integral is $$-\cos((x^3)^{1/3}) \cdot 3x^2 = -3x^2 \cos(x)$$.

$$L = \lim_{x \to \frac{\pi}{2}} \frac{-3x^2 \cos(x)}{2(x - \frac{\pi}{2})}$$

Applying L'Hôpital's again (since $$\cos(\pi/2) = 0$$):

$$L = \lim_{x \to \frac{\pi}{2}} \frac{-[6x \cos(x) - 3x^2 \sin(x)]}{2}$$

Substitute $$x = \frac{\pi}{2}$$:

$$L = \frac{-(6(\frac{\pi}{2})(0) - 3(\frac{\pi}{2})^2(1))}{2} = \frac{3 \cdot \frac{\pi^2}{4}}{2} = \frac{3\pi^2}{8}$$