Limit of $\int_{0}^{2\pi} |\sin(x)|^{n} \ dx$ How does one show that as $n\to \infty,$ $$\int_{0}^{2\pi} |\sin(x)|^{n} \ \to 0?$$ 
Also what about the integral $\int_{0}^{2n\pi} |\sin(x)|^{n} \ dx$ ? I am not able to proceed. I tried using the fact that $|\sin(x)| < |x|$ but that really doesn't help. 
 A: Since $|\sin(x)|^n \le 1$ and we have $\sin(x)^n \to 0$ for almost every $x\in(0,2\pi)$ as $n\to\infty$ this follows from Lebesgue's dominated convergence theorem.
Edit: 
Alternatively, without the use of the dominated convergence theorem, note from the graph of $\sin(x)$ that
$$\int_0^{2\pi} |\sin(x)|^n dx = 4 \int_0^{\pi/2} \sin(x)^n dx$$
Notice that $\int_0^{\pi/2} \sin(x)^n dx\ge 0$.
Then let $0<t<\pi/2$ be small and write
$$\int_0^{\pi/2} \sin(x)^n dx = \int_0^{\pi/2-t} \sin(x)^n dx + \int_{\pi/2-t}^{\pi/2} \sin(x)^n dx \le \frac\pi{2} \sin(\pi/2-t)^n + t$$
Now take $\limsup_{n\to\infty}$ on both sides to see that
$$\limsup_{n\to\infty} \int_0^{\pi/2} \sin(x)^n dx \le t$$
Since this holds for all $t>0$ we have
$$\limsup_{n\to\infty} \int_0^{\pi/2} \sin(x)^n dx=0$$
Thus the limit exists and equals $0$.
A: Hint: use the definition of limits. Given an $\epsilon>0$, find an $n$ (not necessarily the first $n$, that's irrelevant) such that $\int_0^{2\pi}|\sin x|^ndx<\epsilon$.
Now, assuming $\epsilon$ is small enough, let $x_1, x_2$ be so that $$0<x_1<\frac\pi2<x_1+\frac\epsilon4<x_2<\frac{3\pi}2<x_2+\frac\epsilon4<2\pi$$
and let $n$ be large enough that on $[0,x_1]\cap[x_1+\epsilon/4, x_2]\cap[x_2+\frac\epsilon4]$, we have $\sup |\sin x|^n < \frac{\epsilon}{4\pi}$. This makes $\int_0^{2\pi}|\sin x|^ndx < \epsilon$.
A: As for $\int_{0}^{2n\pi} |\sin(x)|^{n} \ dx,$ note this is the same as $n\int_{0}^{2\pi} |\sin(x)|^{n} \ dx= n \int_{0}^{2\pi} |\cos(x)|^{n} \ dx.$
Now for $x \ge 0, \cos x \ge 1-x^2.$ So
$$ \int_{0}^{2\pi} |\cos(x)|^{n} \ dx > \int_{0}^{1/\sqrt n} \cos^n(x)\ dx \ge \int_{0}^{1/\sqrt n} (1-x^2)^n \ dx \ge (1/\sqrt n) (1-1/n)^n \sim 1/(e\sqrt n).$$
Therefore $\int_{0}^{2n\pi} |\sin(x)|^{n} \ dx$ is bounded below by an expression asymptotic to $\sqrt n/e.$ Hence the limit in question is $\infty.$
A: We'll show that the integral over $[0, \pi]$ converges to zero, which is equivalent to the desired result. Let $I = [0, \frac{\pi}{2} - \epsilon] \cup [\frac{\pi}{2} + \epsilon, \pi]$, for any $\epsilon > 0$ sufficiently small. It is easy to verify that $|sin(x)|$ attains a maximum that is less than one on $I$. Since $f(t, n) = t^n$ ($t$, $n >0$) is an increasing function of $t$ and goes to zero as $n \to \infty$ for $t < 1$, this implies that convergence of $sin(x)^n$ to zero as $n \to \infty$ is uniform on $I$. Therefore the portion of the integral taken over $I$ vanishes. For all $n$ the integrand is bounded above by one on $[0, \pi] - I$. Therefore as $\epsilon \to 0$ and $n \to \infty$, the portion of the integral taken over $[0, \pi] - I$ also vanishes. It follows that the integral over $[0, \pi]$ converges to 0 as $n \to \infty$, as claimed.
