Question

Use the Comparison Theorem to determine whether the integral is convergent or divergent.

Answer 1

First, examine what happens over the integration domain.

$\displaystyle\frac{\sin^2x}{\sqrt x}=x^{3/2}\frac{\sin^2x}{x^2}\implies\lim_{x\to0^+}x^{3/2}\frac{\sin^2x}{x^2}=0\times1=0$

$\dfrac{\sin^2x}{\sqrt x}$ is continuous for $x\neq0$, so we don't need to worry about the right endpoint.

Now, note that

$\dfrac{\sin^2x}{\sqrt x}\le\dfrac1{\sqrt x}$

due to the fact that $|\sin x|\le1\implies0\le\sin^2x\le1$. So, if the larger integral converges, so must the smaller one. You have

$\displaystyle\int_0^\pi\frac{\mathrm dx}{\sqrt x}=\lim_{t\to0^+}\int_t^\pi x^{-1/2}\,\mathrm dx$
$=\displaystyle\lim_{t\to0^+}2x^{1/2}\bigg|_{x=t}^{x=\pi}=2\left(\sqrt\pi-\lim_{t\to0^+}\sqrt t\right)=2\sqrt\pi$

so the larger integral converges. Therefore the smaller one does, too.