Prove that $\lim_{n\to \infty}\int_{0}^{\pi/6}{\sin}^{n}x~dx$ = 0 Prove that the $\lim_{n\to \infty}\int_{0}^{\pi/6}{\sin}^{n}x~dx = 0$. Make use of the following theorem:
Let $f_n$ be a sequence of functions in $R[a, b]$. Suppose the sequence converges uniformly on $[a, b]$ to the function f . Then $f\in R[a, b]$ and $\int_{a}^{b}f_n(x)dx \Rightarrow \int_{a}^{b}f(x)~dx$
You may assume $\sin(x)$ is continuous. 
I am not sure exactly how to start this proof. I think I would need to first find the pointwise limit, which I believe is $0$.
 A: Alternatively, one may observe that
$$
0\le \sin x \leq \frac12,\qquad x \in \left[0,\frac{\pi}6\right],
$$ giving
$$
0\le \int_0^{\pi/6}\sin^n x \:dx\leq \int_0^{\pi/6}\frac1{2^n}\:dx.
$$
A: Note that for $x\in [0,\pi/6]$, we have 
$$|\sin^n(x)|\le \frac1{2^n}<\epsilon$$
whenever $n>\frac{-\log(\epsilon)}{\log(2)}$.  Therefore, $\sin^n(x)$ converges uniformly to $0$ for $x\in[0,\pi/6]$.  
Finally, using the given theorem we have
$$\begin{align}
\lim_{n\to \infty}\int_0^{\pi/6}\sin^n(x)\,dx&=\int_0^{\pi/6}\lim_{n\to \infty}\sin^n(x)\,dx\\\\
&=\int_0^{\pi/6}(0)\,dx\\\\
&=0
\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

This is a nice appplication of $\underline{Laplace\,'s\ Method}$:

\begin{align}
&\lim_{n\to \infty}\int_{0}^{\pi/6}\sin^{n}\pars{x}\,\dd x =
\lim_{n\to \infty}\int_{0}^{\pi/6}\sin^{n}\pars{{\pi \over 6} - x}\,\dd x
\\[5mm] = &\
\lim_{n\to \infty}\int_{0}^{\pi/6}
\exp\pars{n\ln\pars{\sin\pars{{\pi \over 6} - x}}}\,\dd x =
\lim_{n\to \infty}\int_{0}^{\pi/6}
\exp\pars{n\ln\pars{{1 \over 2} - {\root{3} \over 2}\,x}}\,\dd x
\\[5mm] = &\
\lim_{n\to \infty}\int_{0}^{\pi/6}
\exp\pars{-n\ln\pars{2} - n\root{3}x}\,\dd x =
\lim_{n\to \infty}\bracks{{1 \over 2^{n}}\int_{0}^{\infty}
\exp\pars{-\root{3}nx}\,\dd x} = \bbx{\ds{0}}
\end{align}
