$$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$$ is equal to

<p>$$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$$</p> <p>Using Newton Leibniz theorem</p> <p>$$\begin{aligned} & =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{\sin \left(2 \times \frac{\pi}{2}\right) \cdot 0-\sin (2 x) \cdot 3 x^2+\left(\cos \frac{\pi}{2}\right) \cdot 0-\cos x \cdot 3 x^2}{2\left(x-\frac{\pi}{2}\right)}\right] \\ & =\lim _\limits w{x \rightarrow \frac{\pi}{2}} \frac{-3 x^2 \sin 2 x-3 x^2 \cos x}{2\left(x-\frac{\pi}{2}\right)}\left(\frac{0}{0}\right) \text { form } \\ & =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{-6 x \sin 2 x-6 x^2 \cos 2 x-6 x \cos x+3 x^2 \sin x}{2}\right] \\ & =\frac{6 \times \frac{\pi^2}{4}+3 \times \frac{\pi^2}{4}}{2} \\ & =\frac{9 \pi^2}{8} \end{aligned}$$</p>