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.
 A: Here is a direct approach. Let $f(x)$ be the integrand, and note that $f(x)$ has a period $\pi$. Let $k$ be the largest positive integer such that $(2k+1)\frac{\pi}{2}<n$. Then:
$$\begin{align}
\int_0^nf(x)\,dx&=\int_0^{\pi/2}f(x)\,dx+\int_{\pi/2}^{3\pi/2}f(x)\,dx+\dots+\int_{(2k+1)\pi/2}^nf(x)\,dx \\
&=\int_0^{\pi/2}f(x)\,dx+k\int_{\pi/2}^{3\pi/2}f(x)\,dx+\int_{(2k+1)\pi/2}^nf(x)\,dx
\end{align} $$
Each of those can be evaluated with the substitution $t=\tan(x) \Rightarrow \cos^2(x)=\frac{1}{1+\tan^2(x)}=\frac{1}{1+t^2}$ and $dx=\frac{dt}{1+t^2}$
$$\begin{align}
\int\frac{1}{1+n^2\cos^2(x)}\,dx&=\int\frac{1}{1+n^2\frac{1}{1+t^2}}\frac{1}{1+t^2}\,dt\\
&=\int\frac{1}{t^2+n^2+1}\,dt\\
&=\frac{1}{\sqrt{n^2+1}}\tan^{-1}\left(\frac{t}{\sqrt{n^2+1}}\right)+C
\end{align} $$
Then:
$$\begin{align}
\int_0^{\pi/2}f(x)\,dx&=\frac{1}{\sqrt{n^2+1}}\tan^{-1}\left(\frac{t}{\sqrt{n^2+1}}\right)\Big|_0^\infty \\
&=\frac{\pi}{2\sqrt{n^2+1}}\overset{n\to\infty}{\to} 0
\end{align}$$
and
$$\begin{align}
\int_{(2k+1)\pi/2}^nf(x)\,dx &= \frac{1}{\sqrt{n^2+1}}\tan^{-1}\left(\frac{t}{\sqrt{n^2+1}}\right)\Big|_{-\infty}^{\tan(n)} \\
&=\frac{1}{\sqrt{n^2+1}}\left(\tan^{-1}\left(\frac{\tan(n)}{\sqrt{n^2+1}}\right)+\frac{\pi}{2}\right)\\
&\overset{n\to\infty}{\to} 0
\end{align}$$
because $1/\sqrt{n^2+1}\to 0$ and the expression in the parentheses is bounded. Finally,
$$\begin{align}
k\int_{\pi/2}^{3\pi/2}f(x)\,dx &= k\frac{1}{\sqrt{n^2+1}}\tan^{-1}\left(\frac{t}{\sqrt{n^2+1}}\right)\Big|_{-\infty}^{\infty} \\
&= \frac{\pi k}{\sqrt{n^2+1}}
\end{align}$$
So you just have to find
$$\lim_{n\to\infty}\frac{\pi k}{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{\pi k}{n} $$
Note that the choice of $k$ implies $(2k+1)\frac{\pi}{2}<n<(2k+3)\frac{\pi}{2}$.
A: 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}

