$\underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{1}{{\left(x-\frac{\pi }{2}\right)}^2}{\int }_{x^3}^{{\left(\frac{\pi }{2}\right)}^3}\cos \left(\frac{1}{t^3}\right)dt\right)$ is equal to

Let, $y=\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\int }_{x^3}^{{\displaystyle {\left(\frac{\pi }{2}\right)}^3}}\cos t^{{\displaystyle \frac{1}{3}}}dt}{{\left(x-{\displaystyle \frac{\pi }{2}}\right)}^2}$

$\Rightarrow y=\underset{h\to 0}{\lim }\frac{{\int }_{{\left({\displaystyle \frac{\pi }{2}-h}\right)}^3}^{{\displaystyle {\left(\frac{\pi }{2}\right)}^3}}\cos t^{{\displaystyle \frac{1}{3}}}dt}{{\left({\displaystyle \frac{\pi }{2}-h-\frac{\pi }{2}}\right)}^2}$

$\Rightarrow y=\underset{h\to 0}{\lim }\left(\frac{{\int }_{{\left({\displaystyle \frac{\pi }{2}-h}\right)}^3}^{{\displaystyle {\left(\frac{\pi }{2}\right)}^3}}\cos t^{{\displaystyle \frac{1}{3}}}dt}{h^2}\right)$

Applying L-Hospital's rule and Newton Leibnitz Theorem we get,

$\Rightarrow y=\underset{h\to 0}{\lim }\left(\frac{0+\cos \left({\displaystyle \frac{\pi }{2}-h}\right)\times 3\times {\left(\frac{\pi }{2}-h\right)}^2}{2h}\right)$

$\Rightarrow y=\underset{h\to 0}{\lim }\left(\frac{\sin {\displaystyle h}\times 3\times {\left(\frac{\pi }{2}-h\right)}^2}{2h}\right)$

We know that, $\underset{h\to 0}{\lim }\left(\frac{\sin h}{h}\right)=1$

$\Rightarrow y=\underset{h\to 0}{\lim }\left(\frac{3\times {\left(\frac{\pi }{2}-h\right)}^2}{2}\right)$

$\Rightarrow y=\frac{3{\pi }^2}{8}$