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

Question

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

anonymous · Answer

what???

anonymous · Answer

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

anonymous · Answer

sorry idk how to do it, good luck though!

anonymous · Answer

Hahaha, thanks.

sirm3d · Answer

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

anonymous · Answer

Yeah

sirm3d · Answer

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

anonymous · Answer

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

sirm3d · Answer

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

anonymous · Answer

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

sirm3d · Answer

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.