improper integral convergent , divergent can someone please explain how should i proceed from here?
$$\int_{0}^{\pi} \frac{\sqrt{x}}{\sin x} dx$$
I did these steps
$$\int_{0}^{1} \frac{\sqrt{x}}{\sin x} dx+\int_{1}^{\pi} \frac{\sqrt{x}}{\sin x} dx$$
and then $$\lim_{x\to\ 0^+}\frac{\frac{\sqrt{x}}{\sin x}}{\frac{1}{\sqrt{x}}}=\lim_{x\to\ 0^+}\frac{x}{\sin x}=1$$ and therefor first part converge but i stuck here, how should i evaluate second integral? i mean for limit comparison test i don't know to which simpler integral should i compare the second one?
Any help or suggestion would be great

In the second integral, write $y=\pi-x$. It becomes
$$\int_0^{\pi-1}\frac{\sqrt{\pi-y}}{\sin y}\,dy$$
and the integrand here is approximately $\sqrt\pi/y$ for $y$ near zero.
