1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$ – Benjamin Sep 11 '20 at 16:27
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you @ Angina Seng, one question, how did you know that you need to change variable like that ? $\endgroup$ – simon Sep 13 '20 at 5:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.