Prove $$\lim\_{n\to +\infty}\int\_0^{\frac{\pi}{2}}\sin^n(x)\,dx=0$$
My attempt:
$$I\_n=\int\_0^{\frac{\pi}{2}}\sin^n(x)\,dx = \left.- \int\_0^{\frac{\pi}{2}} \sin^{n-1} x \,d(\cos x) = -\sin^{n-1} x\cos x \right|\_0^{\frac{\pi}{2}} + \int\_0^{\frac{\pi}{2}} \cos x\,d(\sin^{n-1} x)$$
As $\left.-\sin^{n-1} x\cos x \vphantom{\dfrac11} \right|\_0^{\frac{\pi}{2}} = 0$, hence
$$\int\_0^{\frac{\pi}{2}}\sin^n(x)\,dx = (n-1) \int\_0^{\frac{\pi}{2}}\sin^{n-1}(x) \, dx - (n-1)\int\_0^{\frac{\pi}{2}}\sin^n(x) \, dx \Rightarrow I\_n = \frac{(n-1)I\_{n-2}}{n}$$ hence if $n=2k$: $$I\_{2k}=\frac{(2k-1)!!}{(2k)!!}\cdot\frac{\pi}{2}$$ 
if $n=2k+1$:
$$I\_{2k+1}=\frac{(2k)!!}{(2k+1)!!}$$
Got that by induction and because of $I\_1=\int\_0^{\frac{\pi}{2}}\sin x\,dx= \left.-\cos x \vphantom{\dfrac11} \right|\_0^{\frac{\pi}{2}} = 1$.
So I have to prove that if $k \to +\infty \Rightarrow I\_{2k} \to 0 \text{ and } I\_{2k+1} \to 0$. I don't know how to do that. Please help