Proving the Limit of a Sequence of without the Dominated Convergence Theorem Show:  

$$ \lim_{n\to \infty} \int_{0}^{\frac{\pi }{2}} \sin^n(x) = 0. $$

I showed this using the dominated convergence theorem. I want other methods to prove it. Thank you.
 A: The sequence converges uniformly to zero over any interval bounded away from $\pi/2$. Thus given $\varepsilon>0$ choose an interval $[0,\eta]\subsetneq [0,\pi/2]$ so that the integral over the remaining interval is tiny, and use uniform convergence over what remains.
A: One approach is to use the $\beta$ function which gives the answer

$$ I=\,{\frac {\sqrt {\pi}\,\Gamma \left( \frac{n}{2}+\frac{1}{2} \right) }{2\,\Gamma \left( \frac{n}{2}+1 \right) }}.$$

Now, we appeal to the Stirling approximation 

$$ n! =\Gamma(n+1) \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n  $$ 

to finish the problem.
Note: The $\beta$ function is given by

$$ \beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,\mathrm{d}\theta, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0. $$

A: For any $a \in (0,\frac{\pi}{2})$,
  Let $d = cos(a) < 1$
  $0 \le \sin(x)^n \le d^n$ for any $x \in [0,\frac{\pi}{2}-a]$, and $0 \le \sin(x)^n \le 1$ for any $x \in [\frac{\pi}{2}-a,\frac{\pi}{2}]$
  $0 \le \int_0^{\frac{\pi}{2}} \sin(x)^n \text{ dx} \le (\frac{\pi}{2}-a) d^n + a \to a$ as $n \to \infty$
  $0 \le \limsup_{n\to\infty} \int_0^{\frac{\pi}{2}} \sin(x)^n \text{ dx} \le a$
Therefore $\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \sin(x)^n \text{ dx} = 0$
A: You could also use monotone convergence where $\sin^n(x)$ converges to $0$ almost everywhere on the interval $[0,\pi /2]$.  This is of course extending the notion of monotone convergence to a decreasing sequence of measurable functions.  I know this isn't a great answer but it's my thoughts on the issue.
A: You can evaluate this explicitly; cases vary with even and odd.  Do the following.
$$\int_0^{\pi/2} \sin^n(x)\,dx = \int_0^{\pi/2} \sin^{n-1}(x)\sin(x)\,dx$$
Integrate by parts with $dv = \sin(x)\,dx$.  Use the fact that $\sin^2(x) 
+ \cos^2(x)$ to get rid of the $\cos^2(x).$  Now solve for the original integral and get a nice reduction formula.
