<s>[INST] 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!
 [/INST] 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 $1$.
</s>