Limit of a trigonometric integral: $\lim\limits_{x \to \infty} \int_{0}^{\pi} \frac{\sin (t)}{1+\cos^{2}(xt)} \, \mathrm dt$ For all $x \in \mathbb{R}$, let 
$$
{\rm f}\left(x\right)
=\int_{0}^{\pi}\frac{\sin\left(t\right)}{1 + \cos^{2}\left(xt\right)}\,{\rm d}t
$$
Compute the limit when $x\rightarrow +\infty$.
My attempt :
I tried the substitution $u=\sin(t)$ (and $u=\cos^2(xt)$) but it seems worse:
$$
\int _{0}^{1}\!{\frac {u}{ \left( 1+ \left( \cos \left( x\arcsin
 \left( u \right)  \right)  \right) ^{2} \right) \sqrt {1-{u}^{2}}}}{du}
$$
I tried to use a subsequence $(x_n)$ which is tends to $+\infty$ and use the dominated convergence theorem but it didn't work either.
Sorry if my attempt doesn't make much sense.
Thank you in advance.
 A: We can use a generalization of the Riemann-Lebesgue lemma that states that if $f(x)$ is a continuous function on $[a,b]$, where $0 \le a < b$, and $g(x)$ is a continuous $T$-periodic function on $[0, \infty)$, then $$\lim_{n \to \infty}\int_{a}^{b} f(x) g(nx) \, dx = \left(\frac{1}{T} \int_{0}^{T} g(x) \, dx \right) \left(\int_{a}^{b} f(x) \, dx \right).$$
A proof can be found in THIS PAPER.
The proof uses the fact that a primitive of $g(x)$ can be expressed as $$ \int_{0}^{x} g(t) \, dt = \frac{x}{T} \int_{0}^{T} g(t) \ dt + h(x),$$ where $h(x)$ is $T$-periodic function. (Lemma 2.2)
The regular Riemann-Lesbegue lemma is the case where the average value of $g(x)$ is $0$.
Applying the lemma to this particular integral, we get
$$ \lim_{x \to \infty} \int_{0}^{\pi} \frac{\sin t}{1+\cos^{2}(xt)} \ dt = \left( \frac{1}{\pi} \int_{0}^{\pi} \frac{1}{1+\cos^{2} t} \ dt \right) \left( \int_{0}^{\pi} \sin t \ dt  \right). $$
The first integral can be evaluated by making the substitution $u = \tan t$.
This particular example actually appears in the paper, so I'll omit the details.
We end up with 
$$ \lim_{x \to \infty} \int_{0}^{\pi} \frac{\sin t}{1+\cos^{2}(xt)} \ dt =  \frac{1}{\pi} \left(\frac{\pi}{\sqrt{2}} \right) \left(2\right) = \sqrt{2} .$$
