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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.

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

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  • $\begingroup$ Thanks, that was exactly what I was looking for. Not sure why that was so hard for me to see, but that breakdown helps me visualize it. Now I must simplify further and sub out for theta, which i think I can manage. Ty again. $\endgroup$ Jun 11, 2016 at 21:05

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