Find $\lim\limits_{n \to \infty} \int\limits_0^n \frac1{1 + n^2 \cos^2 x} dx$. I have to find the limit:
$$\lim\limits_{n \to \infty} \displaystyle\int_0^n \dfrac{1}{1 + n^2 \cos^2 x} dx$$
How should I approach this?
I kept looking for some appropriate bounds (for the Squeeze Theorem) that I could use to determine the the limit, but I didn't come up with anything useful.

Let 
$$ S_n=\int_0^\pi\frac{1}{1+n^2\cos^2x}\,dx $$
Then
$$\lim_{n\to\infty}S_n=0  $$
Assume that
$$ L=\lim_{n\to \infty} \int_0^n \dfrac{1}{1 + n^2 \cos^2 x} dx $$
Then
$$ \left\lfloor\frac{n}{\pi} \right\rfloor S_n\le L\le\left(\left\lfloor\frac{n}{\pi} \right\rfloor+1\right)S_n $$
So
$$ L=\lim_{n\to\infty}\left\lfloor\frac{n}{\pi} \right\rfloor S_n $$
So you need to find
$$ \lim_{n\to\infty}\left\lfloor\frac{n}{\pi} \right\rfloor S_n  $$
Using an $\arctan$ substitution gives $S_n=\frac{\pi}{\sqrt{n^2+1}}$ which gives $L=1$.
Addendum: Because of the discontinuity at $\frac{\pi}{2}$ and given the symmetry of the function, it will be better to express
$$ S_n=2\int_0^\frac{\pi}{2} \frac{dx}{1+n^2\cos^2x}$$
Then
\begin{eqnarray}
S_n&=&2\int_0^\frac{\pi}{2} \frac{dx}{\sin^2x+\cos^2x+n^2\cos^2x}\\
&=&2\int_0^\frac{\pi}{2} \frac{dx}{\sin^2x+(1+n^2)\cos^2x}\\
&=&2\int_0^\frac{\pi}{2} \frac{\sec^2x\,dx}{\tan^2x+(\sqrt{1+n^2})^2}\\
&=&2\int_0^\infty \frac{du}{u^2+(\sqrt{1+n^2})^2}\quad\text{where }u=\tan x\\
&=&\frac{2}{\sqrt{n^2+1}}\left[\arctan\left(\frac{u}{\sqrt{n^2+1}}\right)\right]_0^\infty\\
&=&\frac{2}{\sqrt{n^2+1}}\cdot\frac{\pi}{2}\\
&=&\frac{\pi}{\sqrt{n^2+1}}
\end{eqnarray}

