## David.Butler Group Title \int _{ 0 }^{ pi }{ \frac { sin\theta }{ \sqrt { pi\quad -\quad \theta } } } using\quad \frac { 1 }{ \sqrt { pi\quad -\quad \theta } } \quad as\quad comparison one year ago one year ago

1. ChelseaSweets

what???

2. David.Butler

$\int\limits _{ 0 }^{ \pi }{ \frac { \sin\theta }{ \sqrt { \pi\quad -\quad \theta } } } using\quad \frac { 1 }{ \sqrt { \pi\quad -\quad \theta } } \quad as\quad comparison$

3. ChelseaSweets

sorry idk how to do it, good luck though!

4. David.Butler

Hahaha, thanks.

5. sirm3d

you're asking if the integral is convergent or divergent?

6. David.Butler

Yeah

7. sirm3d

if $\int_0^{\pi}\frac{d\theta}{\sqrt{\pi-\theta}}$ is convergent, you can conclude that the integral in question is also convergent.

8. David.Butler

How do I know that integral you stated above is convergent? That's the part that is troubling me.

9. sirm3d

why don't you evaluate the definite integral? if it is a value, then the integral is convergent.

10. David.Butler

Oh okay, I thought it was improper. I just realized that it was definite. Thanks!

11. sirm3d

it is an improper integral, but after integration, the function is continuous in that interval, so you can treat it like an ordinary definite integral.