Showing an improper integral converges

Question

Im trying to evaluate

$$ \int\limits_0^{\pi} \frac{ \sin^2 x}{\sqrt{x} } dx $$

This seems like a convergent integral. I know we can do it by brute force: that is: Use that $\sin^2 x = \frac{ 1 - \cos 2x }{2}$ and then use integration by parts. Is there another way to solve this integral?

Answer 1

For $0\le x\le \pi$, $\sin x \le x$. Thus $0\le \frac{\sin^2 x}{\sqrt{x}} \le x^{\frac{3}{2}}$ for $0<x<\pi$. and $\int_0^{\pi} x^{\frac{3}{2}} dx=\frac{2}{5}\pi^{\frac{5}{2}}$ and so it converges. Therefore given integral converges. It is called comparison test for integrals.

Answer 2

$$\int_0^\pi\dfrac{\sin^2x}{\sqrt x} = 2\int_0^\pi\sin^2x\,d\sqrt x = 2\int_0^{\sqrt\pi}\sin^2u^2du\in(0,2\sqrt\pi),$$ so converges.