How to show the divergence of the improper integral $\int_{0}^{\infty} \frac{dx}{1+x^{2}\sin^{2}(x)}$? How can I show the given improper integral $\int_{0}^{\infty} \frac{dx}{1+x^{2}\sin^{2}(x)}$  is divergent?
Approach:
\begin{align*}
x^{2}\sin^{2}(x) \le x^2 \\
1+x^{2}\sin^{2}(x) \le 1+ x^2 \\ 
\frac{1}{1+x^{2}\sin^{2}(x)} \ge \frac{1}{1+x^{2}}
\end{align*}
but this inequality is not useful as $\int_{0}^{\infty}\frac{1}{1+x^{2}}dx$ is convergent.
 A: There are intervals either side of $n\pi$ where $(x\sin x)^2\lt 1/2$.   Add up their lengths.
A: Let $ n $ be a positive integer, we have : \begin{aligned} \int_{0}^{n\pi}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}&=\sum_{k=0}^{n-1}{\int_{k\pi}^{\left(k+1\right)\pi}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}}\\ &=\sum_{k=0}^{n-1}{\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+\left(x+k\pi\right)^{2}\sin^{2}{x}}}}\\ &\geq\sum_{k=0}^{n-1}{\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+\left(\pi+k\pi\right)^{2}\sin^{2}{x}}}}=\sum_{k=1}^{n}{\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}} \end{aligned}
Now, notice that $ \left(\forall k\in\mathbb{N}^{*}\right) : $ \begin{aligned}\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}=\int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}+\int_{\frac{\pi}{2}}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}\end{aligned}
Let's substitute $ \small\left\lbrace\begin{aligned}y&=\pi -x\\ \mathrm{d}x&=-\,\mathrm{d}x\end{aligned}\right. $, to get : $$ \int_{0}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}=2\int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}} $$
Let's use another substitution $ \small\left\lbrace\begin{aligned}u&=\tan{x}\\ \mathrm{d}x&=\frac{\mathrm{d}u}{1+u^{2}}\end{aligned}\right. $, to get : \begin{aligned}\int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}&=\int_{0}^{+\infty}{\frac{\mathrm{d}u}{\left(1+k^{2}\pi^{2}\frac{u^{2}}{1+u^{2}}\right)\left(1+u^{2}\right)}}\\ &=\int_{0}^{+\infty}{\frac{\mathrm{d}u}{1+\left(1+k^{2}\pi^{2}\right)u^{2}}}\\ &=\frac{1}{\sqrt{1+k^{2}\pi^{2}}}\int_{0}^{+\infty}{\frac{\sqrt{1+k^{2}\pi^{2}}}{1+\left(\sqrt{1+k^{2}\pi^{2}}u\right)^{2}}\,\mathrm{d}u}\\ &=\frac{1}{\sqrt{1+k^{2}\pi^{2}}}\left[\arctan{\left(\sqrt{1+k^{2}\pi^{2}}u\right)}\right]_{0}^{+\infty}\\ &=\frac{\pi}{2\sqrt{1+k^{2}\pi^{2}}}\end{aligned}
Thus, $$ \int_{0}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}=\frac{\pi}{\sqrt{1+k^{2}\pi^{2}}} $$
Hence, $$ \int_{0}^{n\pi}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}\geq\pi\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k^{2}\pi^{2}}}}=\sum_{k=1}^{n}{\frac{1}{\sqrt{\frac{1}{\pi^{2}}+k^{2}}}}\geq\sum_{k=1}^{n}{\frac{1}{k}} $$
Which means : $$ \lim_{n\to +\infty}{\int_{0}^{n\pi}}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}=+\infty $$
In other words : $$ \int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}\textrm{ DV} $$
