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

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anonymous
  • anonymous
Hahaha, thanks.
sirm3d
  • sirm3d
you're asking if the integral is convergent or divergent?
anonymous
  • anonymous
Yeah
sirm3d
  • sirm3d
if \[\int_0^{\pi}\frac{d\theta}{\sqrt{\pi-\theta}}\] is convergent, you can conclude that the integral in question is also convergent.
anonymous
  • anonymous
How do I know that integral you stated above is convergent? That's the part that is troubling me.
sirm3d
  • sirm3d
why don't you evaluate the definite integral? if it is a value, then the integral is convergent.
anonymous
  • anonymous
Oh okay, I thought it was improper. I just realized that it was definite. Thanks!
sirm3d
  • 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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