1. Let $I(x)=\int_{x^3}^{(\pi/2)^3}\left(\sin(2t^{1/3})+\cos(t^{1/3})\right)dt.$ We need $\lim_{x\to \pi/2}\frac{I(x)}{\left(x-\frac\pi2\right)^2}.$

2. First simplify the integrand by substituting $u=t^{1/3}\implies t=u^3,\quad dt=3u^2\,du.$ When $t=x^3$, $u=x$; when $t=(\pi/2)^3$, $u=\pi/2$. So $I(x)=\int_x^{\pi/2} 3u^2\left(\sin 2u+\cos u\right)du.$ Thus the limit becomes $\lim_{x\to \pi/2}\frac{\int_x^{\pi/2}3u^2(\sin 2u+\cos u)\,du}{\left(x-\frac\pi2\right)^2}.$

3. Let $f(u)=3u^2(\sin 2u+\cos u).$ Then $\int_x^{\pi/2} f(u)\,du$ is of the standard form. Since the denominator is quadratic, first check $f\left(\frac\pi2\right)$: $f\left(\frac\pi2\right)=3\left(\frac\pi2\right)^2\left(\sin \pi+\cos \frac\pi2\right)=3\left(\frac\pi2\right)^2(0+0)=0.$ So the numerator vanishes to at least second order, and the limit is finite.

4. Use the expansion near $x=a$ where $a=\frac\pi2$: $\int_x^a f(u)\,du \sim -\frac{f'(a)}{2}(x-a)^2$ when $f(a)=0$. Hence $\lim_{x\to a}\frac{\int_x^a f(u)\,du}{(x-a)^2}=-\frac{f'(a)}{2}.$ So we only need $f'(a)$.

5. Differentiate: $f(u)=3u^2(\sin 2u+\cos u).$ Using product rule, $f'(u)=6u(\sin 2u+\cos u)+3u^2(2\cos 2u-\sin u).$ Now put $u=\frac\pi2$:

• $\sin \pi=0$
• $\cos \frac\pi2=0$
• $\cos \pi=-1$
• $\sin \frac\pi2=1$

Therefore, $f'\left(\frac\pi2\right)=6\cdot \frac\pi2\cdot 0+3\left(\frac\pi2\right)^2\left(2(-1)-1\right).$ So $f'\left(\frac\pi2\right)=3\cdot \frac{\pi^2}{4}\cdot (-3)=-\frac{9\pi^2}{4}.$

6. Hence