Another Integral

[NYC]

ok, I'm integrating dx divided by ( x^2 times the square root of the quantity (9+x^2) )
Reading the problem, it sounds like this: "dx over x squared times the square root of nine plus x squared"
I'm sorry, I don't know how to be much clearer, but I began by allowing x to equal 3 tangent (theta) and came to the integral of secant theta divided by the quantity (9 tangent squared of theta)
or Int of sec(@)/(9tan^2@)
(assuming @ is theta)
From here, I'm not sure where to go. Any pointers would be greatly appreciated.

 

[royhaas]

Write \(\displaystyle tan(\theta)=sin(\theta)sec(\theta)\).

 

[NYC]

hm

if I make that substitution, then I'm left with the integral of cos theta over sin squared theta. I can rewrite it as the integral of cos(@)/(1-cos^2(@)) but I still don't see where I can go from here

 

[galactus]

The derivative of \(\displaystyle \L\\tan({\theta})=sec^{2}({\theta})\)

\(\displaystyle \L\\x=3tan({\theta})\ and\ dx=3sec^{2}d{\theta}\)

Remember, \(\displaystyle \L\\1+tan^{2}{\theta}=sec^{2}{\theta}\)

 

[NYC]

I used those to come to the point I'm at now, and there doesn't seem to be another manipulation I can do to get an answer which I can integrate.

 

[NYC]

Ah

I rearranged it to become the integral of cot (@) csc (@)..which becomes -csc(@).. and then my final answer is negative sqrt of (9+x^2)divided by 9x times ..if this is incorrect, please let me know. Thanks for the help

 

[galactus]

Yes, that is correct. Very good!