<s>[INST] Integrate $\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ via trig substitution $x = a\sec\theta, dx = \sec\theta \tan\theta$
$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{(\tan^2\theta)^{\frac{3}{2}}}$ = 
$ \int \frac{dx}{\tan^{\frac{7}{2}}\theta}$ = 
$\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}$ = 
$\int \tan^{\frac{-5}{2}}\theta \sec\theta$
Here is where I get stuck...I tried converting $\tan\theta$ and $\sec\theta$ in terms of $\cos\theta$ and $\sin\theta$, but that didn't seem to get me anywhere...What is my next move from here? Did I even start this problem correctly? I can't tell :(

Update with more work after initial answers:
$\int \frac{\cos\theta}{\sin^2\theta}$ $u = \sin\theta, du = \cos\theta d\theta$
I found $\sin^{-1}\theta = \frac{\sqrt{x^2-1}}{x}$ 
$= \int \frac{du}{u^2} = \frac{1}{ \frac{1}{3}u^3} = 
\frac{1}{3\sin^3\theta}
= 3 \bigg( \frac{x}{\sqrt{x^2-1}} \bigg)^3$
 [/INST] If you do $x=\sec\theta\,\mathrm d\theta$ and $\mathrm dx=\sec(\theta)\tan(\theta)\,\mathrm d\theta$, what you get is$$\int\frac{\sec(\theta)\tan(\theta)}{\tan^3\theta}\,\mathrm d\theta=\int\frac{\cos^2\theta}{\sin^3\theta}\,\mathrm d\theta=\int\frac{\cos^2(\theta)\sin(\theta)}{\bigl(1-\cos^2(\theta)\bigr)^2}\,\mathrm d\theta.$$Now, do $\cos\theta=u$ and $\sin\theta\,\mathrm d\theta=\mathrm du$.
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