<s>[INST] Another limit involving an integral How to find the following limit (if exists ) : $\lim _{x \to \infty} \dfrac 1 x \int_0^x \dfrac {t}{1+x^2 \cos^2 t}dt$ ? 
This looks quite similar to the previous To find a limit involving integral   , but the huge difference is , there is apparently no closed form formula for the indefinite integral $\int \dfrac {t}{1+x^2 \cos^2 t}dt$ , where there was a closed form formula for the previous one , which made its evaluation slightly easy . Please help . 
 [/INST] I am using the same method as in this answer you linked. Let
$$
F(x)=\frac{1}{x}\int_0^x \frac{t}{1+x^2\cos^2{t}}\mathrm{d}t.
$$
If $n$ is a positive integer, we have
$$
F(n\pi)=\frac{1}{n\pi}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t,
$$
so we can control the $t$ in the integrand :
$$\frac{1}{n}\sum_{k=0}^{n-1}k\int_{k\pi}^{(k+1)\pi} \frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}} \leq F(n\pi)\leq \frac{1}{n}\sum_{k=0}^{n-1}(k+1)\int_{k\pi}^{(k+1)\pi} \frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}}.
$$
Using the identity $\frac{1}{\pi}\int_0^\pi\frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}}=\frac{1}{\sqrt{1+n^2\pi^2}}$, that yields
$$\frac{1}{n}\frac{\pi}{\sqrt{1+n^2\pi^2}}\sum_{k=0}^{n-1}k\leq F(n\pi)\leq \frac{1}{n}\frac{\pi}{\sqrt{1+n^2\pi^2}}\sum_{k=0}^{n-1}(k+1),
$$
hence
$$F(n\pi) \to \frac{1}{2}.
$$
Now when $n\pi \leq x < (n+1)\pi$,
$$
\begin{align}
F(x) &\leq \frac{1}{n\pi}\int_0^{(n+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t \\
&=F(n\pi)+\frac{1}{n\pi}\int_{n\pi}^{(n+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t \\
&\leq F(n\pi)+\frac{n+1}{n}\int_0^\pi \frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}} \\
&= F(n\pi)+\frac{(n+1)\pi}{n\sqrt{1+n^2\pi^2}}.
\end{align}
$$
Using the same kind of inequality on the other side it shouldn't be difficult to show that
$$
F(x)\to \frac{1}{2}.
$$
</s>