Does $\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ \operatorname dx$ converge? I have been trying to prove the following integral: 

$$\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ dx$$ 

diverges (please correct me if I am mistaken).
I have tried to use different comparison tests (as this is an integral of a positive function) with no success.
Any ideas?

The idea is to bound the integral below on intervals where $\displaystyle \frac{1}{1+x^2\left(\sin x\right)^2}$ has spikes, that is to say, it suffices to find some $\varepsilon_k$ such that $$\sum_{k\geq1}\int_{k\pi -\varepsilon_k}^{k\pi +\varepsilon_k}\frac{1}{1+x^2\left(\sin x\right)^2}dx$$ diverges.
On each of these intervals, since $\sin^2(x)$ is $\pi$-periodic, $1+x^2\sin^2(x)\leq 1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)$, hence $$\sum_{k=1}^N\int_{k\pi -\varepsilon_k}^{k\pi +\varepsilon_k}\frac{1}{1+x^2\left(\sin x\right)^2}dx \geq \sum_{k=1}^N \frac{2 \varepsilon_k}{1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)}$$
Some rough asymptotics suggest $$\frac{2 \varepsilon_k}{1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)}\sim \frac{2 \varepsilon_k}{\pi^2k^2\epsilon^2_k}=\frac 2\pi \frac{1}{k^2\varepsilon_k} $$
Setting $\varepsilon_k=\frac 1k$ seems therefore like a sound idea, since we would get something like the harmonic series, which diverges.
Indeed, $$\begin{align}\frac{ \frac 2k}{1+(k\pi + \frac 1k)^2\sin^2(\frac 1k)}&=\frac 2k \frac{1}{1+(\pi^2k^2+2\pi +\frac{1}{k^2})(\frac{1}{k^2}+o(\frac{1}{k^2}))} \\
&=\frac 2k \frac{1}{1+\pi^2 + o(1)}\\
&\sim \frac{2}{1+\pi^2}\frac{1}k
\end{align}$$
With this choice of $\varepsilon_k$, $\displaystyle \sum_{k\geq 1}^\infty \frac{2 \varepsilon_k}{1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)}$ diverges, which concludes the proof.
