Calculate the following limit $$\lim_{n \to \infty} \int_0^\pi \frac{2(2\sin{x}-\sin{((2n+1)x)}+\sin{((2n-1)x)}}{5-4\cos{(2nx)}-\cos{(4nx)}}dx$$
After some trigonometry manipulation (writing the numerator in terms of $\sin{(nx)}$ and the denominator in terms of $\sin^2{(nx)}$ by double angle cosine formulas) we get to the much nicer expression
$$\lim_{n \to \infty} \int_0^\pi \frac{\sin(x)}{1+\cos^2{(nx)}}dx$$
This looks to me like the way to go, I have not found any easy way to compute this integral. The first ideea that comes to mind is to do the substitution $u = \cos{x}$ and the expression becomes $$\lim_{n \to \infty}\int_{-1}^1 \frac{1}{1+T_{2n}(u)}du$$ where $T_m(x)$ is a Chebyshev polynomial of the first kind(i.e. it satisfies $T_m(\cos(x))=\cos(mx)$). But this lead me nowhere since computing the integral using the formula $$T_n(x)=\frac{1}{2}\left( \left( x-\sqrt{x^2-1} \right)^n + \left( x+\sqrt{x^2-1} \right)^n \right)$$ but it seems horrible to plug in the formula and solve it? If you were to go down that rabbit hole how would you start? $u=\sec{\theta}$ would bring you back to trig functions.
Another approach I have tried is the following $$\int_0^\pi\ \frac{\sin(x)}{1+\cos^2{(nx)}}dx=\frac{1}{2} \int_0^\pi \frac{\sin(x)}{1-\frac{\sin^2{(nx)}}{2}}=\sum_{k=0}^\infty \frac{1}{2^n}\int_0^\pi \sin{(x)}\sin^{2n}{(nx)}dx$$ However the integral in the summation is not very nice and it can't be evaluated with $\sin(t)=\text{Im}{e^{it}}$, since $z^m \neq (\text{Im}z)^m$ in general.
So let's take $$J_n=\int_0^\pi \sin{(x)}\sin^{2n}{(nx)}dx$$ Maybe it is possible to solve using integration by parts and finding a recursive formula, but it did not look promising to me.
I have also thought about using complex analysis, but again it went no where. If I take a contour a semicircle that contains the segment $[-pi, pi]$, it would be atrocious to calculate the integral on the "circle part", even though the contour integral might be doable.
In despair, I have tried using substitution $u=nx$, so the integral becomes $$\lim_{n \to \infty} \int_0^\pi \frac{\sin(x)}{1+\cos^2{(nx)}}dx=\lim_{n \to \infty} \int_0^{n\pi} \frac{1}{n}\frac{\sin(\frac{u}{n})}{1+\cos^2{(u)}}du=^\text{L'Hopital and Leibniz rule}=\lim_{n \to \infty}\int_0^{n\pi} \frac{1}{n}\frac{\cos(\frac{u}{n})}{1+\cos^2{(u)}}du=\lim_{n \to \infty}\int_0^\pi \frac{\cos(x)}{1+\cos^2{(nx)}}dx$$
How should I approach this problem?
