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 integral would diverge if $x^2\sin^2(x)$ was "quite small" when $x$ is near a multiple of $\pi$. Using the periodicity of $\sin(x)$, we can examine the behavior near $k\pi$ by shifting and looking at
$$ (x+k\pi)^2 \sin^2(x) \approx (x+k\pi)^2 x^2$$
for $x$ near $0$. This is valid for very small $x$ through the Taylor polynomial, in a way that could be made rigorous.
For $\lvert x \rvert < \frac{1}{2k\pi}$ (for $k \geq 1$), we have that
$$(x+k\pi)^2 \sin^2(x) \leq 2.$$
So for $\lvert x \rvert < \frac{1}{2k\pi}$, we have
$$ \frac{1}{1+(x+k\pi)^2 \sin^2(x)} \geq \frac{1}{3}.$$
With this in mind, we can write
$$\begin{align}
\int_0^\infty \frac{1}{1 + x^2\sin^2(x)}dx &= \sum_{k \geq 0} \int_{k\pi}^{(k+1)\pi} \frac{1}{1 + x^2\sin^2(x)}dx  \\
&= \sum_{k \geq 0} \int_{0}^{\pi} \frac{1}{1 + (x+k\pi)^2\sin^2(x)}dx.
\end{align}$$
Now as everything in sight is positive, 
$$ \begin{align}
\sum_{k \geq 0} \int_0^\pi \frac{1}{1 + (x+\pi)^2\sin^2(x)}dx &\geq \sum_{k \geq 1} \int_0^{\frac{1}{2\pi k}} \frac{1}{1 + (x+k\pi)^2\sin^2(x)}dx \\
&\geq \sum_{k \geq 1} \int_0^{\frac{1}{2\pi k}} \frac{1}{3} dx\\
&\geq \sum_{k \geq 1} \frac{1}{6\pi k}, 
\end{align}$$
which diverges.
In other words, I believe that $x^2\sin^2(x)$ is "quite small" when $x$ is near $k\pi$, and it is "quite small" in a region that decays linearly with $k$. And this is enough to show that the integral diverges.
