# 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

• Hint: $\sin x=\sin(\pi-x)$ and $\sin x\approx x$ for small $x$. So $\frac{\sqrt x}{\sin x}\approx \frac\pi{\pi-x}$ for $x$ close to $\pi$. Can you fill in the details and find the conclusion? – Benjamin Sep 11 '20 at 16:27

## 1 Answer

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.

• Thank you @ Angina Seng, one question, how did you know that you need to change variable like that ? – simon Sep 13 '20 at 5:47