What has flipping a coin to do with $\int_0^{2 \pi}\sin^{2n}(x)\,dx$? If you toss a coin $2n$ times the chance of getting heads $n$ times is
$$\frac{1}{4^n} \binom{2n}{n} $$ (simple combinatorics problem).
Now I've made a seemingly obscure observation:
$$\int_0^{2 \pi}\sin^{2n}(x)\,dx = \frac{2 \pi}{4^n} \binom{2n}{n} $$
The identities only differ by a factor of $2 \pi$. Is there an explanation for this phenomenon or just a cool coincidence?
UPDATE:
As Aloizio Macedo pointed out, a fairly related question has been answered with a combinatorial proof by Qiaochu Yuan here.
 A: I thought I had a real approach to this question, but after working a bit, I realized that I needed some Fourier analysis at least.
In a generating function approach, the number of ways to get exactly $k$ heads and $2n-k$ tails would be $\left[x^k\right](1+x)^{2n}$ where each choice of an $x$ represents a head and each choice of a $1$ a tail. Since
$$
\frac1{2\pi}\int_0^{2\pi}e^{ikx}\,\mathrm{d}x=\left\{\begin{array}{}1&\text{if }k=0\\0&\text{if }k\ne0\end{array}\right.
$$
$\left[x^n\right](1+x)^{2n}$ is simply
$$
\begin{align}
\frac1{2\pi}\int_0^{2\pi}\left(1+e^{ix}\right)^{2n}e^{-inx}\,\mathrm{d}x
&=\frac1{2\pi}\int_0^{2\pi}\left(e^{-ix/2}+e^{ix/2}\right)^{2n}\,\mathrm{d}x\tag1\\
&=\frac{4^n}{2\pi}\int_0^{2\pi}\cos^{2n}(x/2)\,\mathrm{d}x\tag2\\
&=\frac{4^n}{\pi}\int_0^{\pi}\cos^{2n}(x)\,\mathrm{d}x\tag3\\
&=\frac{4^n}{2\pi}\int_0^{2\pi}\cos^{2n}(x)\,\mathrm{d}x\tag4\\
&=\frac{4^n}{2\pi}\int_0^{2\pi}\sin^{2n}(x)\,\mathrm{d}x\tag5
\end{align}
$$
Explanation:
$(1)$: distributive property
$(2)$: $2\cos(x/2)=e^{ix/2}+e^{-ix/2}$
$(3)$: substitute $x\mapsto2x$
$(4)$: $\cos^{2n}(x)=\cos^{2n}(x+\pi)$
$(5)$: $\cos^{2n}(x)=\sin^{2n}(x+\pi/2)$  
Since there are $2^{2n}=4^n$ possible outcomes, we get the probability of exactly $n$ heads to be
$$
\frac1{2\pi}\int_0^{2\pi}\sin^{2n}(x)\,\mathrm{d}x
$$
A: This is still not a combinatorial proof, purely, but it avoids complex numbers, at least, and is phrased in terms of probability.
First note that your integral is the same as $\int_{0}^{2\pi}\cos^{2n}x\,dx$.
Now we use the property that $\cos^2 x =\frac{1+\cos 2x}{2}.$ If $X$ is a uniform random variable in $[0,2\pi]$, then the random variables $U=\cos X$ and $V=\cos 2X$ have the same distribution, so it means that $$E\left[U^{2n}\right]=\frac{1}{2^n}E\left[(1+V)^n\right]=\frac{1}{2^n}E\left[(1+U)^n\right]$$ and thus you get an inductive formula. Since $E\left[U^k\right]=0$ when $k$ is odd, you get:
$$E\left[U^{2n}\right]=\frac{1}{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}E\left[U^{2k}\right]$$
We want to prove $$E\left[U^{2n}\right]=\frac{1}{4^n}\binom{2n}{n}.$$
When $n=0$ we get $E[U^0]=1=\frac1{2^0}\binom{0}{0}.$ 
Inductively, then, we need to show:
$$\frac{1}{4^n}\binom{2n}{n}=\frac{1}{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\frac{1}{2^{2k}}\binom{2k}{k}$$
which is equivalent to:
$$\binom{2n}{n}=\sum_{k=0}^{\lfloor n/2\rfloor}2^{n-2k}\binom{n}{n-2k}\binom{2k}{k}\tag{1}$$
which can be proved combinatorially.
If we think of $\binom{2n}{n}$ as the number of ways of picking an $n$-subset of $\{1,\dots,n\}\times\{1,2\}$ then $k$ is the number of elements $m\in\{1,\dots,n\}$ such that $(m,1)$ and $(m,2)$ are both in the $n$-subset. That means we need to pick an $n-2k$-subset $P$ of $\{1,\dots,n\}$, then for each $p\in P$, choose either $(p,1)$ or $(p,2)$, and then choose $k$ elements of the $2k$ elements not in $P$, and add $(q,1)$ and $(q,2)$ to our subset.
The interesting thing about this proof of $(1)$ is that the meaning of the $\binom{2n}{n}$ on the left side is kind-of different from the meaning of $\binom{2k}{k}$ on the right side.

A power series approach. Even less combinatorial than the first approach.
Consider for $|x|<1$ the function $f(x,t)=\frac{1}{1-x\cos t}=\sum_{k=0}^{\infty} x^k\cos^kt$
The integral $$F(x)=\frac1{2\pi}\int_{-\pi}^{\pi} f(x,t)\,dt$$
can be found by using the Weierstrass substitution. This gives that:
$$\begin{align}F(x)&=\frac{1}{\sqrt{1-x^2}}
\end{align}$$
So if $T$ is a uniform random variable in $[0,2\pi]$ you get:
$$\sum_{k=0}^{\infty} x^kE\left[\cos^k T\right]=E[f(x,T)]=F(x)=\frac{1}{\sqrt{1-x^2}}=\sum_{k=0}^{\infty}\frac{1}{4^k}\binom{2k}{k}x^k$$
That last step requires you to know the generating function for the central binomial coefficients, $\binom{2k}{k}$. (Alternatively, if you prove the equality elsewhere, you can use this to prove the central binomial coefficients generating function.)

Finally, maybe the underlying reason is due to the characteristic function for a random variable. Unfortunately, this is essentially an encoding of the proof that starts with $\cos(x)=\frac{1}{2}\left(e^{ix}+e^{-ix}\right).$ 
Given a random variable, $X$, we have the characteristic function:
$$G_X(t)=E\left[e^{itX}\right]$$
If $X$ is a random integer variable, and $T$ is a uniform random variable in $[0,2\pi]$ then we have a fundamental result:
$$E\left[G_X(T)\right]=P(X=0).$$
Now if $X_1,\dots,X_{n}$ are independent coin tosses returning values $-1$ and $1$, then $G_{X_i}(t)=\cos t$ and $$G_{X_1+\cdots+X_n}=G_{X_1}G_{X_2}\cdots G_{X_n}$$
So if $X=X_1+\cdots+X_n$, then $G_X(t)=\cos^n(t)$ and $$P(X=0)=E[G_X[T]]=\frac{1}{2\pi}\int_{0}^{2\pi} \cos^n t\,dt$$
That's just really fundamentally hiding the usual approach of writing $\cos x=
\frac{1}{2}(e^{ix}+e^{-ix})$ and using the algebraic property of the characteristic function.
A: Well, if you write
$$
\sin^{2n} x = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2n}
,
$$
expand using the binomial theorem, and integrate term by term, everything will be zero except the contribution from the middle (constant) term,
$$
\int_0^{2\pi} \binom{2n}{n} (e^{ix})^n (-e^{-ix})^n \,\left(\frac{1}{2i}\right)^{2n} \, dx
= 2 \pi \, \binom{2n}{n} \left(\frac{1}{2}\right)^{2n}
,
$$
which is a bit similar to how the coefficients in the binomial expansion of $(p+q)^{2n}$ tell you the probabilities of getting $k$ heads and $2n-k$ tails when flipping a coin with probability $p$ and $q$ for head and tail.
But whether you would count this as an “explanation”, I don't know...
