Computing $\lim\limits_{n \to \infty} \int_0^{\frac{\pi}{2}}{\frac{(\sin(x))^{n}}{1-\sin{(x)}}\,\mathrm{d}x} $ 
$\def\d{\mathrm{d}}$I would like to compute the following limit,  $$\displaystyle{\lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \frac{(\sin(x))^{n}}{1-\sin{(x)}}\,\d x} .$$

I am looking for a high school answer.
I tried writing $$\lim_{n \to \infty} \int_0^{\frac{\pi}{2}}{\frac{(\sin(x))^{n}}{1-\sin{(x)}}\,\d x = \lim_{n \to \infty} \lim_{ε \to \frac{\pi}{2}}\int_0^ε{\frac{(\sin(x))^n}{1-\sin(x)}}\,\d x},$$ 
but it doesn't help me, since $1 - \sin(x) \leq 1, \forall x \in \left[0,  \dfrac{\pi}{2}\right]$.

The integral fails to converge for all $n$ as $1-\sin x \sim (\pi/2-x)^2$ near $\pi/2.$
