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## anonymous 3 years ago \int _{ 0 }^{ pi }{ \frac { sin\theta }{ \sqrt { pi\quad -\quad \theta } } } using\quad \frac { 1 }{ \sqrt { pi\quad -\quad \theta } } \quad as\quad comparison

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1. anonymous

what???

2. anonymous

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

3. anonymous

sorry idk how to do it, good luck though!

4. anonymous

Hahaha, thanks.

5. anonymous

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

6. anonymous

Yeah

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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.

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