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. sirm3d

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

6. anonymous

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. anonymous

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. anonymous

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.

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