Evaluate $\lim_{n\to\infty} \int_0^{2 \pi} \frac{1}{x+\sin^n x+ \cos^n x} \ dx$

Evaluate $$\lim_{n\to\infty} \int_0^{2 \pi} \frac{1}{x+\sin^n x+ \cos^n x} \ dx$$ Although intuitively I may guess that the limit is $\infty$, it's a bit harder to come up with a rigorous proof. I need your support here. Thanks.

-

Intuitively, what happens is that when $n$ is large, the terms $\sin^n(x)$ and $\cos^n(x)$ contribute nothing to the denominator except when $x$ is extremely close to the points where $\sin(x) = 1$ or $\cos(x) = 1$. Note though that for every $n$, the integrand takes the value $1$ at $x=0$. So essentially the integrand will behave like $\frac 1x$ modulo what happens at the points $0$, $\pi/2$, $\pi$, $3\pi/2$ and $2\pi$. Notice that at these points, $\frac 1{x + \sin(x)^n + \cos(x)^n} > 0$ because when $\sin(x) = \pm 1$, $\cos(x) = 0$ and vice-versa, hence since $2\pi > \dots > \pi/2 > 1$, the denominator never gets close to zero.
Pick $\delta > 0$ small enough, and consider $$E = [\delta, \pi/2 - \delta] \cup [\pi/2 + \delta, \pi-\delta] \cup [\pi + \delta, 3\pi/2 - \delta] \cup [3\pi/2 + \delta, 2 \pi - \delta].$$ Over this interval, one can show that the sequence of functions $f_n(x) = \frac 1{x + \sin^n(x) + \cos^n(x)}$ converges uniformly to $\frac 1x$ ; that is because you can control what happens to $\sin(x)$ and $\cos(x)$ when you take $n$ large enough. I leave that work to you. Therefore, $$\lim_{n \to \infty} \int_E f_n(x) \, dx = \int_E \lim_{n \to \infty} f_n(x) \, dx = \int_E \frac 1x \, dx$$ which you can compute by using the sums of the integrals over the respective intervals (that will give you a bunch of logarithms involving $\delta$). Since you can choose $\delta$ such that the integral over $E$ is roughly the integral over $[0,2\pi]$ (up to a $\delta$), the integral of the $f_n$'s becomes the integral of $\frac 1x$ when $n$ is large, hence is infinite.
Hint: consider $\int_{\varepsilon}^{1}\frac{dx}{x+\sin^n{(x)}+\cos^n{(x)}}$ for some small $\varepsilon>0$.