Answer
Converges
Work Step by Step
$\displaystyle \int_0^\pi \frac{\sin^2x}{\sqrt{x}}\ dx$
Find a greater-valued function:
$\displaystyle \frac{\sin^2x}{\sqrt{x}} < \frac{2}{\sqrt{x}}$
(Recall: $\displaystyle 0 \le \sin^2x \le 1$, so we can substitute with a higher value in numerator to get a greater-valued function)
$\displaystyle \int_0^\pi \frac{2}{\sqrt{x}}\ dx = 2\lim\limits_{t \to 0} \int_t^\pi \frac{1}{\sqrt{x}}\ dx$
$= \displaystyle 2\lim\limits_{t \to 0} \left[2\sqrt{x}\right]_t^\pi = 4\lim\limits_{t \to 0}(\sqrt{\pi}-\sqrt{t}) = 4\sqrt{\pi}$
We know that the greater-valued function $\displaystyle \frac{2}{\sqrt{x}}$ converges to a value, so therefore by the convergence theorem, $\displaystyle \int_0^\pi \frac{\sin^2x}{\sqrt{x}}\ dx$ converges.
