Convergence of $\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx$ I need to find out whether the following improper integral converges:
$$\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx$$
I tried two comparison tests that failed, any ideas?
 A: Note that we can write
$$\begin{align}
\int_0^{N\pi/5}\frac{1}{1+x^2\sin^2(5x)}\,dx&=\frac15\int_0^{N\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\
&=\frac15\int_0^{\pi/2}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\
&+\frac15\sum_{n=1}^{N-1}\int_{(n-1/2)\pi}^{(n+1/2)\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\\\\
&+\frac15\int_{(N-1/2)\pi}^{N\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx\tag 1
\end{align}$$
The first on the right-hand side of $(1)$ creates no issue and the third integral is easily seen to converge as $N\to \infty$.
We focus attention on the second integral on the right-hand side of $(1)$ and write
$$\begin{align}
\int_{(n-1/2)\pi}^{(n+1/2)\pi}\frac{1}{1+(x/5)^2\sin^2(x)}\,dx&\ge 2\int_{0}^{\pi/2}\frac{1}{1+((n+1/2)\pi)/5)^2\sin^2(x)}\,dx\\\\
&\ge 2\int_{0}^{\pi/2}\frac{1}{1+((n+1/2)\pi)/5)^2 x^2}\,dx\\\\
&=\frac{20}{(2n+1)\pi}\arctan((n+1/2)\pi^2/10)\\\\
&\ge \frac{20}{(2n+1)\pi}\left(\frac{\pi}{2}-\frac{1}{(n+1/2)\pi^2/5}\right)
\end{align}$$
Inasmuch as the harmonic series diverges, the integral of interest diverges likewise.
