Answer
Converges
Work Step by Step
Apply a direct limit comparison test.
We have $\sin x \leq 1$
This suggests that $0 \leq \dfrac{1 +\sin x}{x^2} \leq \dfrac{2}{x^2}$ for all $x \geq \pi$
We also know:
$\int_{\pi}^{\infty} \dfrac{2 dx}{x^2}=\lim\limits_{p \to \infty}\int_{\pi}^{p} \dfrac{2dx}{x^2}$
and $\lim\limits_{p \to \infty}\int_{\pi}^{p} \dfrac{2dx}{x^2}=\lim\limits_{p \to \infty}[\dfrac{-2}{x}]_{\pi}^{p}=\dfrac{-2}{\pi}$
Hence, the given integral converges by the direct comparison test.