integration - How prove this $lim_{ntoinfty}int_{0}^{frac{pi}{2}}sin{x^n}dx=0$

show that

$$\lim_{n\to\infty}\int_{0}^{\dfrac{\pi}{2}}\sin{x^n}dx=0$$

I have see this similar problem

$$\lim_{n\to\infty}\int_{0}^{\dfrac{\pi}{2}}\sin^n{x}dx=0$$
poof:
$\forall \xi>0,0<\delta<\xi/2$,and there is $N$,such $0<\sin^n{\pi/2-\delta}<\xi/\pi(n\ge N)$
then we have
$$\int_{0}^{\pi/2}\sin^n{x}dx=\left(\int_{0}^{\pi/2-\delta}+\int_{\pi/2-\delta}^{\pi/2}\right)\sin^n{x}dx=I_{1}+I_{2}$$
then
$$|I_{1}|\le\left(\sin{\pi/2-\delta}\right)^n(\pi/2-\delta)<\xi/\pi\cdot\pi/2=\xi/2$$

and
$$|I_{2}|\le\left(\pi/2-(\pi/2-\delta)\right)=\delta<\xi/2$$
and This problem have many other methods,

But for this

$$\lim_{n\to\infty}\int_{0}^{\dfrac{\pi}{2}}\sin{x^n}dx=0$$
I can't prove it,Thank you