TITLE: How to compute $\lim\limits_{n\to\infty}\int_0^\pi \frac{\sin x}{1+3\cos^2(nx)}\text dx$
QUESTION [5 upvotes]: In one of our exam  class we were given the following limit to compute
  $$
\lim_{n \to \infty}\int_{0}^{\pi}
\frac{\sin\left(x\right)}{1 + 3\cos^{2}\left(nx\right)}\,\mathrm{d}x
$$

patently is not advisable to use the convergent dominate which I failles to apply here since the integrand does not even converges.
The wonderful think with is that
$\lim\limits_{n \to \infty}\cos^{2}\left(nx\right)$ does not and this problem kept me awake for a while now. 
Does anyone has some tips how to attack this problem ?.

REPLY [3 votes]: Let $$\alpha=\frac1{2\pi}\int_0^{2\pi}\frac{dx}{1+3\cos^2(x)}.$$I have no idea how to find $\alpha$; find a really good calculus student.
Then if $f\in C[a,b]$ it follows that $$\lim_{n\to\infty}\int_a^b\frac{f(x)\,dx}{1+3\cos^2(nx)}=\alpha\int_a^bf(x)\,dx.$$
Hint: Since $f$ is uniformly continuous, if $n$ is large enough then $f$ is within $\epsilon$ of being constant on each period of $\cos(nx)$. It follows that the difference of the two integrals above is less than $\epsilon(b-a)$ or $2\epsilon(b-a)$ or something...
(This is more or less the same as pisco125's answer. The proof he gave is slicker; what's above works if $f$ is just Riemann integrable.)