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

Calculus1
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what???
\[\int\limits _{ 0 }^{ \pi }{ \frac { \sin\theta }{ \sqrt { \pi\quad -\quad \theta } } } using\quad \frac { 1 }{ \sqrt { \pi\quad -\quad \theta } } \quad as\quad comparison\]
sorry idk how to do it, good luck though!

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Hahaha, thanks.
you're asking if the integral is convergent or divergent?
Yeah
if \[\int_0^{\pi}\frac{d\theta}{\sqrt{\pi-\theta}}\] is convergent, you can conclude that the integral in question is also convergent.
How do I know that integral you stated above is convergent? That's the part that is troubling me.
why don't you evaluate the definite integral? if it is a value, then the integral is convergent.
Oh okay, I thought it was improper. I just realized that it was definite. Thanks!
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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