Show that the integral is divergent $\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}$ 
Show that the integral is divergent
   $$\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}$$

It has no point of discontinuity in range of integration. Also, I have found $\int_0^\infty\frac{\ dx}{1+x^2\sin^2x}>\frac{\pi}{2}$, but that seems to be of no consequence. I don't how to move ahead with this.
 A: We have that
\begin{align*}\int_0^\infty\frac{dx}{1+x^2\sin^2x}
&=\sum_{n=0}^{\infty}
\int_0^{\pi}\frac{dx}{1+(x+n\pi)^2\sin^2(x+n\pi)}
\\&\geq  \sum_{n=0}^{\infty}
\int_0^{\pi/2}\frac{2dx}{1+\pi^2(1+n)^2\sin^2(x)}\\
&=\sum_{n=0}^{\infty}\frac{\pi}{\sqrt{1+\pi^2(n+1)^2}}\\
&\geq \frac{\pi}{\sqrt{1+\pi^2}}\sum_{n=0}^{\infty}\frac{1}{n+1}=+\infty
\end{align*}
where we used the fact that
$$\int\frac{\ dx}{1+a^2\sin^2(x)}=\frac{\arctan(\sqrt{1+a^2}\tan(x))}{\sqrt{1+a^2}}+C.$$
P.S. We can also use the inequality $\sin^2(x)\leq x^2$ and evaluate 
$$\int_0^{\pi/2}\frac{2dx}{1+\pi^2(1+n)^2x^2}=\frac{2\arctan\left((n+1)\pi^2/2\right)}{\pi(n+1)}.$$
A: As an alternative to Robert Z' fine answer, you may consider that for any $n\in\mathbb{N}^+$ the function $x^2 \sin^2 x$ is bounded by $2\pi^2$ on the interval $I_n=\left[\pi n-\frac{1}{n},\pi n+\frac{1}{n}\right]$. Since the integrand function is non-negative and these intervals are disjoint,
$$ \int_{0}^{+\infty}\frac{dx}{1+x^2 \sin^2 x}\geq \int_{\bigcup I_n}\frac{dx}{1+2\pi^2} \geq \frac{1}{1+2\pi^2}\sum_{n\geq 1}\left|I_n\right|=+\infty $$
since $|I_n|=\frac{2}{n}$ and the harmonic series is divergent.
