Answer
Converges
Work Step by Step
Use a direct limit comparison test.
Since, $\sin x \leq 1$
This implies that $0 \leq \dfrac{1 +\sin x}{x^2} \leq \dfrac{2}{x^2}$ for all $x \geq \pi$
Now,
$$\int_{\pi}^{\infty} \dfrac{2 dx}{x^2}=\lim\limits_{a \to \infty}\int_{\pi}^{p} \dfrac{2dx}{x^2} \\= \lim\limits_{a \to \infty}\int_{\pi}^{a} \dfrac{2dx}{x^2} \\=\lim\limits_{a \to \infty}[\dfrac{-2}{x}]_{\pi}^{a} \\=\dfrac{-2}{\pi}$$
Thus, the integral converges by the direct comparison test.