Does $\int_{0}^{\infty}\frac{dx}{1+(x\sin5x)^2}$ converge? I would like your help with deciding whether the following integral converges or not:
$$\int_{0}^{\infty}\frac{dx}{1+(x\sin5x)^2}.$$
I tried to compare it to other functions and to change the variables, but it didn't work for me.
Thanks a lot! 

Consider the intervals $I_k=[(k-\frac{1}{2})\frac{\pi}{5},(k+\frac{1}{2})\frac{\pi}{5}]$. The $I_k$ are the periods of $\sin^2(5x)$.  Therefore,
$$
\begin{align}
\int_{I_k}\frac{\mathrm{d}x}{1+x^2\sin^2(5x)}
&\ge\left|\int_{I_k}\frac{\cos(5x)\;\mathrm{d}x}{1+(k+\frac{1}{2})^2\pi^2/25\;\sin^2(5x)}\right|\\
&=\frac{1}{(k+\frac{1}{2})\pi}\int_{-1}^1\frac{(k+\frac{1}{2})\pi/5\;\mathrm{d}t}{1+(k+\frac{1}{2})^2\pi^2/25\;t^2}\\
&=\frac{2}{(k+\frac{1}{2})\pi}\tan^{-1}\left(\frac{(k+\frac{1}{2})\pi}{5}\right)
\end{align}
$$
Since $\tan^{-1}\left(\frac{(k+\frac{1}{2})\pi}{5}\right)>\frac{\pi}{4}$ for $k>1$, the integral on $I_k$ is greater than $\frac{1}{2k+1}$. Therefore, the integral diverges.
