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.
Best Answer
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 $(2k+1)\frac{\pi}{2}<n<(2k+3)\frac{\pi}{2}$.