Question

Use the comparison theorem to determine whether the integral is convergent or divergent integral sin^2x/sqrt(x) from 0 to pi

Answer 1

`-1\le\sin x\le1\implies0\le\sin^2x\le1\implies0<(\sin^2x)/(\sqrt x)\le\frac1{\sqrt x}`

Does


`\displaystyle\int_0^\pi(\mathrm dx)/(\sqrt x)`

converge?