# Solving an improper integral using another

This was an old two part exam question that I was looking over. Essentially using the improper integral $$\displaystyle\int_1^9 \frac{1}{\sqrt{x-1}}dx$$ you are supposed to determine if the integral $$\displaystyle\int_1^9 \frac{\sin^2(x)}{\sqrt{x-1}} dx$$ converges or diverges. If it converges find its value. I determined that $$\displaystyle\lim_{t\to 1^+}\int_t^9 \frac{1}{\sqrt{x-1}}dx=6$$. Now I'm not sure which convergence test to use to prove that the integral diverges. Any help would be much appreciated.

• Comparison. Because $0 \leq \sin^2(x) \leq 1$. The integral with $0$ in the numerator is trivial. The integral with $1$ in the numerator is the one you did. So you know whether the sine integral converges or diverges and also (if it converges) an interval containing the value of its integral. Jan 14, 2022 at 5:29
• You write "$\lim_{t \rightarrow 1^+}$", but there is no "$t$" in the subsequent expression. Jan 14, 2022 at 5:30
• Now you have $\lim\limits_{x\to 1^+}$, but that too is wrong. Why? Jan 14, 2022 at 6:22 • The graph of the second function is squeezed between the graphs of the first function and the graph of $x=1$ and lies between the graphs of the first function and the graph of $y=0$, so I don't know what you mean by 'divergent' in this case. Jan 14, 2022 at 16:16