Does the integral $\int\limits^{{\pi}}_{0} \frac{3\sin^2\left(2x\right)}{\sqrt{x}}\,\mathrm{d}x$ converge or diverge? Does $$\displaystyle\int\limits^{{\pi}}_{0} \dfrac{3\sin^2\left(2x\right)}{\sqrt{x}}\,\mathrm{d}x$$
converge or diverge?
I'm not even going to evaluate the integral, it's too ugly. I also don't know of a "simpler" function, that can tell me if it will converge or diverge? How can I do this question? I first have to figure out if it converges or diverges, and then find a appropriate function, but I need help figuring that out.
 A: Hint. The integrand is continuous over $(0,\pi]$ and one has
$$
\left|\int\limits^{{\pi}}_{0} \dfrac{3\sin^2\left(2x\right)}{\sqrt{x}}\,\mathrm{d}x\right|\le\int\limits^{{\pi}}_{0} \left|\dfrac{3\sin^2\left(2x\right)}{\sqrt{x}}\right|\,\mathrm{d}x\le3\int\limits^{{\pi}}_{0} \dfrac{1}{\sqrt{x}}\,\mathrm{d}x<\infty
$$
A: In particular $$I=3\int_{0}^{\pi}\frac{\sin^{2}\left(2x\right)}{\sqrt{x}}dx=\frac{3}{2}\int_{0}^{\pi}\frac{1-\cos\left(4x\right)}{\sqrt{x}}dx$$ $$=3\sqrt{\pi}-\frac{3}{2}\int_{0}^{\pi}\frac{\cos\left(4x\right)}{\sqrt{x}}dx\stackrel{x=\pi t^{2}/8}{=}3\sqrt{\pi}-\frac{3\sqrt{\pi}}{2\sqrt{2}}\int_{0}^{2\sqrt{2}}\cos\left(\pi t^{2}/2\right)dt$$ $$=\color{red}{3\sqrt{\pi}-\frac{3\sqrt{\pi}}{2\sqrt{2}}C\left(2\sqrt{2}\right)\approx4.3856}$$ where $C\left(x\right)$ is the Fresnel $C$ integral.
A: It converges: the only problem can be at $x=0$, and $\sin^2{2x} = 4x^2+o(x^2) $ there, so the whole integrand looks like $x^{3/2}$ near $x=0$, which obviously has a convergent integral since it is bounded.
A: Ok, Here is a standard approach: We don't know whether or not it converges, so hence my intial step to do a u-sub, which is straight forward: $\sqrt{x}=t$ which gives a new integral of the form $\int{6sin^2{2t}}dt$ with boundaries $0$ and $\sqrt{\pi}$. At this point you can hopefully see that the integral went from improper to proper, and it is going to be convergent. Now you state that the integral is "ugly" to evaluate, but I think this one is quite cute? Set $2t=v$ and you get a $sin^2v$ integral, which is standard. You can tag the limits along. Note, the answer won't be pretty as you are dealing with "ugly" angles. But nonetheless, convergent.
