Evaluating $\lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx$ Greetings I want to evaluate $\displaystyle\lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx$.
Here is my try: We have that $x\in[0,\pi]$ so $$\cos(n\pi)\le \cos(nx) \le 1.$$ Here I am not sure, but if it's correct then it gives: $$\frac{1}{2}\le\frac{1}{1+\cos^2(nx)}\le \frac{1}{1+\cos^2(n\pi)},$$ giving $$\lim_{n\rightarrow\infty} \frac{1}{2}\int_0^{\pi} \sin x dx \le \lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx \le \lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (n\pi)} dx.$$ Since $$\int_0^{\pi} \sin x dx =2$$ By squeeze theorem we may conclude that $\displaystyle \lim_{n\rightarrow\infty} \int_{0}^{\pi} \frac {\sin x}{1+ \cos^2 (nx)} dx=1$. Could you help me evaluate this, if it's wrong?
 A: One approach can be as follows:
\begin{align}
\int_0^{\pi}\frac{\sin x}{1+\cos^2nx}\,dx&\stackrel{(1)}{=}
\frac{1}{n}\int_0^{n\pi}\frac{\sin(y/n)}{1+\cos^2y}\,dy=
\frac{1}{n}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{\sin(y/n)}{1+\cos^2y}\,dy\stackrel{(2)}{=}\\
&\stackrel{(2)}{=}\frac{1}{n}\sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin\big(\frac{t+k\pi}{n}\big)}{1+\cos^2t}\,dt=
\frac{1}{\pi}\int_0^{\pi}\frac{1}{1+\cos^2t}\Big[\frac{\pi}{n}\sum_{k=0}^{n-1}\sin\Big(\frac{t+k\pi}{n}\Big)\Big]\,dt\stackrel{(3)}{\to}\\
&\stackrel{(3)}{\to}\frac{1}{\pi}\int_0^{\pi}\frac{1}{1+\cos^2t}\,dt\cdot\int_0^{\pi}\sin t\,dt=\sqrt{2}.
\end{align}
Explanations:
(1) change $y=nx$,
(2) change $t=y-k\pi$ and use that $\cos^2t$ is $\pi$-periodic,
(3) Riemann summa limit and dominated convergence theorem.

For the last integration: rewrite
$$
\cos^2x+1=\cos^2x+\cos^2x+\sin^2x=\cos^2x(2+\tan^2x),
$$
that gives
\begin{align}
\int_0^\pi\frac{1}{1+\cos^2x}\,dx&=2\int_0^{\pi/2}\frac{1}{1+\cos^2x}\,dx=2\int_0^{\pi/2}\frac{1}{2+\tan^2x}\,d\tan x=2\int_0^\infty\frac{1}{2+t^2}\,dt=\\
&=[t=s\sqrt{2}]=\sqrt{2}\int_0^\infty\frac{1}{1+s^2}\,ds=\sqrt{2}\frac{\pi}{2}=\frac{\pi}{\sqrt{2}}.
\end{align}
