<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] Note that $$(\tan^2 x)^{3/2}=\tan^3 x$$
Thus, yous should instead obtain
$$\int\frac1{(x^2-1)^{3/2}}dx
=\int\frac{\sec t\tan t}{\tan^3 t}dt
$$
which simplifies to
$$\int\frac{\cos t}{\sin^2 t}=\int\frac1{\sin ^2 t}d(\sin t)=-\frac1{\sin t}+C=-\csc t+C$$
Now reverse the substitution by the identity $\csc^2 t=\frac1{1-\cos^2 t}=\frac1{1-x^{-2}}$.
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