# Using Integration Techniques, how do you manipulate specific trig identities?

I've been working on this problem for the last couple hours, and I simply cannot figure out how to solve it. I've scoured the internet, checked the answer using symbolab and wolfram alpha, and yet I still cannot figure out how to make the connection between the refinements. The original problem is as follows:

$$\int\frac{1}{x^2\sqrt{x^2+9}}dx$$

I identify that this is a tangent substitution, where $x=3\tan\theta$ and $dx = 3\sec^2\theta$. After some manipulation and cancelation, I am left with:

$$\frac{1}{9}\int\frac{\sec\theta}{\tan^2\theta}\;d\theta$$

I know I am at the right spot at this point from checking symbolab, but I simply cannot figure out how to carry on from here. I've tried using the identity $\tan^2\theta = \sec^2\theta-1$, and i've also tried using a power reducing formula on $\tan^2\theta$, to no avail. Symbolab shows a refinement from this point:

$$\frac{1}{9}\int\frac{\sec\theta}{\tan^2\theta}\;d\theta = \frac{1}{9}\int\frac{1}{sin\theta}\cot\theta\;d\theta$$ but this makes absolutely no sense to me. I've spent about an hour messing around with identities but I simply cannot figure out how they came to this conclusion. My trig is a bit rusty but i'm working hard to get better at it, but after much trial and error I simply cannot figure it out. How is this conclusion reached? Should I have tackled this problem in a different manner? Any help would be appreciated as I would like to know how to solve this for future reference. Asking you folks has been my last resort. Thanks in advance.

If your problem is the last step, note that: $$\frac{\sec \theta}{\tan^2 \theta}=\frac{\frac{1}{\cos \theta}}{\frac{\sin^2 \theta}{\cos^2 \theta}}=\frac{1}{\frac{\sin^2 \theta}{\cos \theta}}=\frac{1}{\sin \theta}\frac{\cos \theta}{\sin\theta}=\frac{\cot \theta}{\sin \theta}$$