Trigonometric Integral Limit My problem is to evaluate the following limit:
$$\lim_{n \to \infty}{\int^{n}_{0}{|\sin(x)|^n dx}}$$
I have no idea where to begin, but as you can tell the area under $\sin(x)^n$ decreases as $n$ approaches $\infty$. WolframAlpha can't evaluate this for $n>134$, where it equals about $9.29$. If I had to guess I would say it's probably $\infty$.
One way to tackle this could be to find a representation of $$\int^{2\pi}_{0}{|\sin(x)|^n dx}$$ and multiply by $\frac{n}{2\pi}$. This should approximate the integral for large $n$.
 A: Your idea works with some modifications. The integral turns out to be nice if $n$ is even but annoying if $n$ is odd, and we can reduce to the even case since $\int_0^{\pi} |\sin x|^n \, dx$ is monotonically decreasing as a function of $n$. (As Greg Martin observes in the comments, $|\sin x|$ is $\pi$-periodic, so we only have to integrate over an interval of length $\pi$.)
To begin, we'll rewrite the integral as
$$I(n) = \int_0^{\pi} |\sin x|^n \, dx = \int_0^{\pi} \sin^n x \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n x \, dx$$
which is a bit easier to reason about; writing $\cos x = \frac{e^{ix} + e^{-ix}}{2}$ then gives
$$I(n) = \frac{1}{2^n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum_{k=0}^n {n \choose k} e^{i(n-2k)x} \right) \, dx = \frac{1}{2^n} \sum_{k=0}^n {n \choose k} \left( \frac{e^{i(n-2k)\frac{\pi}{2}} - e^{-i(n-2k)\frac{\pi}{2}}}{i(n-2k)} \right)$$
which simplifies further to
$$I(n) = \frac{1}{2^{n-1}} \sum_{k=0}^n {n \choose k} \frac{\sin (n-2k)\frac{\pi}{2}}{n-2k}.$$
(We need a special case if $n$ is even and $n = 2k$ since then we're dividing by zero: in that case $e^{i(n-2k)x} = 1$ and the corresponding term above should be replaced with its limit as $k \to \frac{n}{2}$, which is $\frac{\pi}{2}$.)
If $n$ is even then every term disappears except the term $n = 2k$, which gives
$$I(2k) = \frac{1}{2^{2k}} {2k \choose k} \pi.$$
If $n$ is odd then the terms alternate in sign which is annoying but by the monotonicity observation from before we have $I(2k+1) \ge I(2k+2)$ which is good enough. We have
$$\int_0^n |\sin x|^n \ge \int_0^{\pi \lfloor \frac{n}{\pi} \rfloor} |\sin x|^n \, dx = \left\lfloor \frac{n}{\pi} \right\rfloor I(n)$$
so to prove that the limit is $\infty$ it suffices to prove that $I(n)$ grows faster than $\frac{1}{n}$. In fact it grows like $\frac{1}{\sqrt{n}}$: for example, Stirling's formula can be used to prove that the central binomial coefficients grow asymptotically like
$${2k \choose k} \sim \frac{2^{2k}}{\sqrt{\pi k}}$$
which gives
$$\boxed{ I(2k) \sim \sqrt{\frac{\pi}{k}} }$$
(and the same asymptotic for $I(2k+1)$ by monotonicity), and this is enough to conclude. As a sanity check, for $n = 134$ this asymptotic gives
$$\int_0^{134} |\sin x|^{134} \, dx \approx \frac{134}{\pi} \sqrt{ \frac{\pi}{67} } = 9.2 \dots$$
which roughly agrees with your calculation.

An alternative and much more general strategy for estimating integrals of this form (integrating a large power of some function) is to apply some form of Laplace's method; the idea here is that for large $n$ nearly all the contribution to the integral of $|\sin x|^n$ comes from the regions where $\sin x \approx 1$, since anything else falls off exponentially sharply as $n$ increases. So it suffices to estimate these contributions, which can generally be done using a Gaussian integral; this is where the square-root behavior comes from.
This actually gives an alternative proof of the asymptotic for the central binomial coefficients which avoids Stirling's formula, and Stirling's formula itself can be proven using a variation of this method applied to the integral $n! = \int_0^{\infty} x^n e^{-x} \, dx$, as explained e.g. in this blog post by Terence Tao.
Edit: A simple version of this idea, which I guess is an elaboration on Greg Martin's comment, already works. We'll switch to analyzing the integral $I(n) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^n x \, dx$ as before. As above, most of the contribution to this integral comes from the part where $\cos x \approx 1$ or equivalently $x \approx 0$. For $x < \sqrt{2}$ we have
$$\cos^n x \ge \left( 1 - \frac{x^2}{2} \right)^n \ge e^{- n \frac{x^2}{2}}$$
by Bernoulli's inequality, which gives
$$I(n) \ge \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{-n \frac{x^2}{2}} \, dx.$$
(It doesn't matter much but it's convenient here that $\sqrt{2} < \frac{\pi}{2}$.) This integral can be lower bounded by $\delta$ where $\delta \le \sqrt{2}$ has the property that $e^{-n \frac{x^2}{2}} \ge \frac{1}{2}$ for all $x \in [-\delta, \delta]$. This gives $\delta = \sqrt{ \frac{2 \log 2}{n} }$, which is always less than $\sqrt{2}$, so we conclude that
$$I(n) \ge \sqrt{ \frac{2 \log 2}{n} } $$
which is within a multiplicative constant of the true asymptotic and good enough. We have $2 \log 2 = 1.386 \dots$ compared to the true asymptotic  $2 \pi = 6.283 \dots$ (here we're comparing the constants under the square root).
Just for fun, we can attempt to slightly improve the asymptotic above by optimizing the constant a bit. If we ask for $e^{-n \frac{x^2}{2}} \ge r$ where $r \in [0, 1]$ we get $\delta = \sqrt{ \frac{-2 \log r}{n} }$ and
$$I(n) \ge 2r \sqrt{ \frac{-2 \log r}{n} }.$$
Now we can try to optimize the constant as a function of $r$. It suffices to minimize $r^2 \log r$, which has derivative $2r \log r + r = r(2 \log r + 1)$; setting this equal to zero gives $r = e^{- \frac{1}{2} } = 0.606 \dots$ so $\delta = \frac{1}{\sqrt{n}}$ (as suggested by Greg!) and we've improved the constant under the square root from $2 \log 2 = 1.386 \dots$ to $4r^2 = \frac{4}{e} = 1.471 \dots$, giving
$$I(n) \ge \sqrt{ \frac{4}{en} }.$$
