<s>[INST] does $\int_0^\infty x/(1+x^2 \sin^2x) \mathrm dx$ converge or diverge? $$\int_0^\infty x/(1+x^2\sin^2x) \mathrm dx$$
I'd be very happy if someone could help me out and tell me, whether the given integral converges or not (and why?). Thanks a lot.
 [/INST] Since $-1 \leq \sin{x} \leq 1$, you have $\sin^{2}{x} \leq 1$, and from this you can conclude that 
\begin{align*}
1+x^{2}\sin^{2}{x} &\leq 1+2x^{2} \\\ \Longrightarrow  \frac{1}{1+x^{2}\sin^{2}{x}} &\geq \frac{1}{1+2x^{2}} \\\ \Longrightarrow \frac{x}{1+x^{2}\sin^{2}{x}}&\geq \frac{x}{1+2x^{2}} \\\
 \Longrightarrow \int_{0}^{\infty} \frac{x}{1+x^{2}\sin^{2}{x}} \ dx &\geq \int_{0}^{\infty} \frac{x}{1+2x^{2}} \ dx
\end{align*} 
Now 
\begin{align*}
\int_{0}^{\infty} \frac{x}{1+2x^{2}} \ dx &= \frac{1}{4} \cdot \int_{0}^{\infty} \frac{4x}{1+2x^{2}} \ dx \\\ &= \frac{1}{4} \bigl(\log{x}\bigr)_{1}^{\infty} \to \infty
\end{align*}
</s>