Find the average of $\sin^{100} (x)$ in 5 minutes?
Since $$\sin x=\frac{e^{ix}-e^{-ix}}{2i} $$ and, for $k\in\mathbb Z $, $$\int_0^{2\pi}e^{ikx}\,dx=\left\{\begin{array}{cl}0&k\ne0\\ 2\pi&k=0\end{array}\right.,$$ we have $$\int_0^{2\pi}\sin^{100}x\,dx=\frac1{2^{100}}\sum_{k=0}^{100}\binom{100}{k}\int_0^{2\pi}e^{ikx}(-1)^{100-k}e^{-i(100-k)x}\,dx=\frac{\binom{100}{50}}{2^{100}}2\pi,$$ and the average value is $$\frac{\binom{100}{50}}{2^{100}}.$$
I suspect Michael Lugo's answer was the intended one, but for what it's worth, Andres Caicedo's combinatorial answer has a combinatorial proof. The Riemann sum
$$\frac{1}{n} \sum_{k=0}^{n-1} \cos^{100} \frac{2 \pi k}{n}$$
counts the probability that you return to where you started in a random walk of length $100$ on $\mathbb{Z}/n\mathbb{Z}$ (where adjacent residues are connected by an edge); this is because the adjacency matrix is $P + P^{-1}$ where $P$ is a permutation matrix describing an $n$-cycle, so the eigenvalues of the adjacency matrix are $e^{ \frac{2 \pi i k}{n} } + e^{- \frac{2 \pi i k}{n} } = 2 \cos \frac{2 \pi k}{n}$.
For $n > 100$ this probability is clearly just $\frac{1}{2^{100}} {100 \choose 50}$, and taking the limit as $n \to \infty$ we obtain our result. And combinatorialists know that ${2n \choose n} \approx \frac{4^n}{\sqrt{\pi n}}$ by one of various methods (Stirling's formula, singularity analysis, the central limit theorem...).
Actually (I had forgotten that I knew this) one can get the integral directly without looking at a Riemann sum. The key is that $\mathbb{Z}$ (the graph where adjacent integers are connected by an edge) is the representation graph of $\text{SO}(2)$ acting on its standard representation $V$, so
$$\int_0^{2\pi} (2 \cos x)^{100} \, dx$$
is precisely the multiplicity of the trivial representation in $V^{\otimes 100}$, which is precisely the number of walks of length $100$ from the origin to itself on $\mathbb{Z}$.
Assuming Arnold means to find an approximate value -- I'd do this as follows: first, we may as well find the average of $\cos^{100} x$. I'll do this over a half-period, $-\pi/2 \le x \le \pi/2$. But $\cos x \approx 1-x^2/2$, so $\cos^{100} x \approx (1-x^2/2)^{100}$. If $x$ is small -- which it will have to be for $\cos x$ to be large (i. e. near 1) -- then $1-x^2/2 \approx e^{-x^2/2}$. So $\cos^{100} x \approx e^{-50x^2}$.
So the number we're looking for is about $$ {1 \over \pi} \int_{-\pi/2}^{\pi/2} e^{-50x^2} \: dx. $$ But the integrand is so small far from zero that the limits of integration can be replaced with $-\infty$ and $\infty$ without changing much. That gives $$ {1 \over \pi} \int_{-\infty}^\infty e^{-50x^2} \: dx. $$ Change variables, $u = x/\sqrt{50}$, to get $$ {1 \over \pi \sqrt{50}} \int_{-\infty}^\infty e^{-u^2} \: du. $$ Finally, that integral is well-known to be $\sqrt{\pi}$; the approximate answer is $1/\sqrt{50\pi}$.
This method has the advantage that it works for high powers of any function and isn't specialized to the trig functions. One source that explains this trick (and uses it to approximate the same integral) is Sanjoy Mahajan's book Street-Fighting Mathematics (link goes to Creative-Commons downloadable version of the book).