## University Calculus: Early Transcendentals (3rd Edition)

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.