Find the limit $\lim\limits_{n\to \infty} \int_{0}^{n} {1\over{1+n^2\cos^2x}}\,dx$ $$\lim_{n\to \infty} \int_{0}^{n} {1\over{1+n^2\cos^2x}}\, dx$$
Some help please. I don't have any idea. Thank you!
 A: When $n\in(\pi k,\pi(k+1))$, note that
$$\lim_{k\to\infty}\int_{\pi k}^{\pi(k+1)}\frac1{1+\pi^2k^2\cos^2(x)}~\mathrm dx=0$$
Thus, the limit is equal to the limit when $n=\pi k$.
When $n=\pi k$, we have due to the period of $\cos^2(x)$,
$$\begin{align}L&=\lim_{k\to\infty}\int_0^{\pi/2}\frac{2k}{1+\pi^2k^2\cos^2(x)}~\mathrm dx\\&=\lim_{k\to\infty}\int_0^{\pi/2}\frac{2k\sec^2(x)}{1+\pi^2k^2+\tan^2(x)}~\mathrm dx\\&=\lim_{k\to\infty}\int_0^{+\infty}\frac{2k}{1+\pi^2k^2+u^2}~\mathrm du\\&=\lim_{k\to\infty}\frac{\pi k}{\sqrt{1+\pi^2 k^2}}\\&=\boxed{1}\end{align}$$
A: Here's what I think is really going on in this problem: On $[0,\pi],$ $\cos^2 x$ has a unique minimum of $0$ at $\pi/2,$ and at that point the second derivative is positive. Those properties alone guarantee a limit that we can evaluate.
More precisely, suppose $f$ is $C^2$ and periodic with period $p.$ Suppose further that $0<a<p,$ $f(a)=0,$ and $f>0$ on $[0,p]\setminus \{a\}.$ Claim: If $f''(a)>0,$ then
$$\tag 1 \lim_{n\to \infty} \int_0^{np} \frac{dx}{1+(np)^2f(x)} = \frac{\pi}{p}\left (\frac{2}{f''(a)}\right )^{1/2}.$$
Because the second derivative of $\cos^2 x$ at $\pi/2$ is $2,$ $(1)$ tell us the limit in our specific problem is $1,$ just as the other answers say.
Proof of claim (sketch): By periodicty,
$$\tag 2 \int_0^{np} \frac{dx}{1+(np)^2f(x)} = n\int_0^{p} \frac{dx}{1+(np)^2f(x)}.$$
Now $f(a) = f'(a)=0.$ Thus Taylor says $f(x) \approx f''(a)(x-a)^2/2$ is a very good approximation to $f(x)$ in some $I=[a-\delta, a +\delta].$ Below I'll just assume $f(x) = f''(a)(x-a)^2/2$ in $I$ to keep things simple (this is a sketch after all).
Now the integral on the right of $(2)$ is the integral over $I$ plus the integral over $[0,p]\setminus I.$ Because $f$ has a positive minimum over the latter set, the integgral over it is on the order of $1/n^2,$ which means we can forget about it. So we're left with
$$\int_{I} \frac{dx}{1+(np)^2f''(a)(x-a)^2/2}= \int_{[-\delta,\delta]} \frac{dx}{1+(np)^2f''(a)x^2/2} = 2\int_{[0,\delta]} \frac{dx}{1+(np)^2f''(a)x^2/2}.$$
Make the change of variables $x = y/(np(f''(a)/2)^{1/2}).$ Then the last integral equals
$$2\frac{1}{np(f''(a)/2)^{1/2}}\int_0^{\delta np(f''(a)/2)^{1/2}} \frac{dy}{1+y^2}.$$
Now multiply that by $n.$ The $n$'s cancel, and as $n\to \infty,$ the integral $\to \pi/2.$ We're done (modulo simplifying assumptions).
A: Using the formula for the tangent of a sum, we get
$$
\arctan(\alpha\tan(x))=x+\arctan\left(\frac{(\alpha-1)\tan(x)}{1+\alpha\tan^2(x)}\right)
$$
Since $\left|\frac{(\alpha-1)\tan(x)}{1+\alpha\tan^2(x)}\right|\le\frac{|\alpha-1|}{2\sqrt{\alpha}}$, arctan never has to go through a singularity, so this is a nice, smooth function.
$$
\require{cancel}
\begin{align}
\int_0^n\frac1{1+n^2\cos^2(x)}\,\mathrm{d}x
&=\int_0^n\frac{\sec^2(x)}{1+n^2+\tan^2(x)}\,\mathrm{d}x\\
&=\cancel{\left.\frac1{\sqrt{n^2+1}}\arctan\left(\frac{\tan(x)}{\sqrt{n^2+1}}\right)\right]_0^n}\\
&=\left.\frac1{\sqrt{n^2+1}}\left(x-\arctan\left(\frac{\left(\sqrt{n^2+1}-1\right)\tan(x)}{\sqrt{n^2+1}+\tan^2(x)}\right)\right)\right]_0^n\\
&=\frac1{\sqrt{n^2+1}}\left(n-\arctan\left(\frac{\left(\sqrt{n^2+1}-1\right)\tan(n)}{\sqrt{n^2+1}+\tan^2(n)}\right)\right)
\end{align}
$$
Since $\arctan$ is between $-\frac\pi2$ and $\frac\pi2$, the limit as $n\to\infty$ is $1$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\lim_{n \to \infty}\int_{0}^{n}{\,\dd x \over 1 + n^{2}\cos^{2}\pars{x}}  =
\lim_{n \to \infty}\int_{0}^{\left\lfloor n/\pi\right\rfloor\pi\ +\ \braces{n/\pi}\pi}{\,\dd x \over 1 + n^{2}\cos^{2}\pars{x}}
\\[5mm] = &\
\lim_{n \to \infty}\braces{%
\left\lfloor\,{n \over \pi}\,\right\rfloor
\int_{0}^{\pi}{\,\dd x \over 1 + n^{2}\cos^{2}\pars{x}} +
\int_{0}^{\braces{n/\pi}\pi}{\,\dd x \over 1 + n^{2}\cos^{2}\pars{x}}}
\end{align}

$$
\mbox{Note that}\quad\left\{\begin{array}{rcl}
\ds{\int_{0}^{\pi}{\,\dd x \over 1 + n^{2}\cos^{2}\pars{x}}} & \ds{=} & \ds{\pi \over \root{n^{2} + 1}}
\\[3mm]
\ds{0 < \int_{0}^{\braces{n/\pi}\pi}{\,\dd x \over 1 + n^{2}\cos^{2}\pars{x}}} & \ds{<} & \ds{\int_{0}^{\pi}{\,\dd x \over 1 + n^{2}\cos^{2}\pars{x}} =
{\pi \over \root{n^{2} + 1}}}
\end{array}\right.
$$
such that
$\ds{\bbx{\lim_{n \to \infty}\int_{0}^{\braces{n/\pi}\pi}
{\,\dd x \over 1 + n^{2}\cos^{2}\pars{x}} = 0}}$ and
\begin{align}
&\lim_{n \to \infty}\int_{0}^{n}{\,\dd x \over 1 + n^{2}\cos^{2}\pars{x}}  =
\lim_{n \to \infty}\braces{%
\left\lfloor\,{n \over \pi}\,\right\rfloor\,{\pi \over \root{n^{2} + 1}}}
\\[5mm] = &\
\lim_{n \to \infty}\pars{%
{n \over \root{n^{2} + 1}} - {\braces{n/\pi}\pi \over \root{n^{2} + 1}}} =
\bbx{\large 1}
\end{align}
