Answer
The integral $\int_{1}^{\infty} \dfrac{2}{x^2} dx$ converges to $2$.
Hence, the integral $\int_{1}^\infty \dfrac{1-\sin x}{x^2} dx $ converges as well by the comparison test.
Work Step by Step
We can see that $\dfrac{2}{x^2}\geq \dfrac{1-\sin x}{x^2}\geq \dfrac{-1}{x^2}$
So, we will compute the integral as follows:
$\int_{1}^{\infty} \dfrac{2}{x^2} dx =\int_{1}^{\infty} 2x^{-2} \ dx \\=[-2x^{-1}]_1^\infty \\=[-\dfrac{2}{x}]_1^{\infty}\\=2$
So, the integral $\int_{1}^{\infty} \dfrac{2}{x^2} dx$ converges to $2$.
Hence, the integral $\int_{1}^\infty \dfrac{1-\sin x}{x^2} dx $ converges as well by the comparison test.
