Comparison Test for Improper Integrals — Question 1

Question 1 – Solution

Step 1: Find an upper comparison. Since $\sin^2x\ge0$, $$x^2+\sin^2x\ge x^2.$$ Both sides are positive, so taking reciprocals reverses the inequality: $$0\le\frac1{x^2+\sin^2x}\le\frac1{x^2}.$$ Step 2: Test the comparison integral. $$\begin{align*} \int_1^\infty\frac{dx}{x^2} &=\lim_{b\to\infty}\left[-\frac1x\right]_1^b\\ &=\lim_{b\to\infty}\left(1-\frac1b\right)=1. \end{align*}$$ Step 3: Apply direct comparison. The larger integral converges, so the original integral also converges. $$\boxed{\text{converges}}$$