Integrate 1/(9-x^2)dx - Basic Integral Help

[jspek9]

the problem is integrate 1/(9-x^2)dx.

For some reason I'm just drawing a blank. I've tried to fit arccot, ln, and trig sub and get nowhere. Is it part frac decomp? I don't know i think I'm just making it harder than it is.
any insight is appreciated.

 

[quantumdude]

Let's see what you've tried for a trig sub (that's how I'd do it).

 

[Mark44]

Partial fractions would also work.

 

[quantumdude]

Yes it would, but I didn't want to torture him by demanding that he type out 2 attempts. So I picked my favorite method. :D

 

[jspek9]

I'm using u= asinθ so sinθ= x/3 & dx = 3cosθ dθ & 3cosθ = 9-x²

so subing in I get the integral 3cosθ/3cosθ dθ which reduces to one and equates to x after integration.

The thing is this is really an improper integral problem
Integral from 0 to 3 of 1/(9-x²)
so if I get x when I integrate I have

lim [x]^a₀
a-->3^-

so the lim is 3 but this doesn't seem right

 

[quantumdude]

I'm using u= asinθ so sinθ= x/3 & dx = 3cosθ dθ

Sure, that's just what I would have done.

& 3cosθ = 9-x²

No, that's not right. It should be the following:

$$3\cos(\theta)=\sqrt{9-x^2}$$

This implies the following.

$$9\cos^2(\theta)=9-x^2$$

Do you see why?

 

[jspek9]

oooooooohhhhh it doesn't fit the form the way i was trying to use it. i probably should have looked up the form before i got so frustrated. oh well ill know next time thanks for the help:)