Show that $\lim_{n \rightarrow \infty} \int_0^{2\pi} \sin(x)^n \, dx=0$ I want to show that:
$$
\lim_{n \rightarrow \infty} \int_0^{2\pi} \sin(x)^n \, dx=0
$$
and my idea was to use DCT (dominated convergenece theorem).
However, my textbook has the requirement that $u(x)=\lim_{n \rightarrow \infty} u_n(x)$ for all x. And this is not true for e.g. $x=-\frac{\pi}{2}$ which alternates between $1$ and $-1$.
Am I overseeing something using DCT or should another theorem be applied? I also thought about Fatou's lemma but the problem is that $\sin(x)^n \notin \mathcal{M}^+$.
Any hint would be appreciated.
 A: DCT only requires almost-everywhere convergence, which is the case in your limit. If the version in your textbook requires everywhere convergence, then here is a simple trick: $$\int_{0}^{2\pi}\sin^n(x)\,\mathrm{d}x=\int_{0}^{2\pi}u_n(x)\,\mathrm{d}x$$ where $$ u_n(x) = \begin{cases} \sin^n(x), & \text{if $|\sin(x)|<1$}\\0, & \text{otherwise} \end{cases}. $$ Now $u_n(x) \to 0$ everywhere and you are good to go.

For a more elementary trick, for any $\epsilon\in(0,\frac{\pi}{2})$ notice that 
$$\left|\int_{0}^{2\pi}\sin(x)^n\,\mathrm{d}x\right|\leq 4\left(\epsilon + (\tfrac{\pi}{2}-\epsilon)\cos^n(\epsilon) \right). $$
Taking limsup as $n\to\infty$, this reduces to
$$\limsup_{n\to\infty} \left|\int_{0}^{2\pi}\sin(x)^n\,\mathrm{d}x\right|\leq 4\epsilon,$$
and then the desired claim follows by letting $\epsilon \downarrow 0$.
A: $(\sin{x})^n \to 0$ a.e on $[0,2\pi]$
And $|\sin{x}|^n \leq 1 \in L^1([0,2\pi])$
So from Dominated Convergence theorem you have the conclusion.
A: You could also use the DCT to show $\int_0^\pi\sin^nxdx\to0$, then use the squeeze theorem with $\left|\int_0^{2\pi}\sin^nxdx\right|\le2\int_0^\pi\sin^nxdx$.
