<s>[INST] What did I get wrong when solving $\int\frac{\sqrt{x^2-1}}{x^4}dx$? I'm not sure that this is the problem, but I think I may not know how to find the $\theta$ value when solving an integral problem with trigonometric substitution. 
I got $\frac{\sin^3(\sec^{-1}(x))}{3}+C$ for the answer, but the answer should be, $\frac{1}{3}\frac{(x^2-1)^{3/2}}{x^3}+C$

$$\int\frac{\sqrt{x^2-1}}{x^4}dx$$

Let $x=\sec\theta$
Then $dx=\sec\theta\tan\theta d\theta$
$$\int\frac{\sqrt{\sec^2\theta-1}}{\sec^4\theta}\sec\theta\tan\theta d\theta$$
$$=\int\frac{\sec\theta}{\sec^4\theta}\sqrt{\tan^2\theta}\tan\theta d\theta$$
$$=\int\frac{1}{\sec^3\theta} \tan^2\theta d\theta$$
$$=\int\frac{1}{\sec^3\theta}\frac{\sec^2\theta}{\csc^2\theta}d\theta$$
$$=\int\frac{1}{\sec\theta}\frac{1}{\csc^2\theta}d\theta$$
$$=\int \cos\theta\sin^2\theta d \theta$$
Using $u$-substition, let $u=\sin\theta$
Then $du=\cos\theta d\theta$ and $dx = \frac{1}{\cos\theta}du$
$$\int\cos\theta u^2 \frac{1}{\cos\theta}du$$
$$=\int u^2 du$$
$$=\frac{u^3}{3}+C$$
$$=\frac{\sin^3\theta}{3}+C$$
Since $x=\sec\theta$, $\sec^{-1}(x)=\theta$
$$=\frac{\sin^3(\sec^{-1}(x))}{3}+C$$
What am I doing wrong?
 [/INST] Taking your final answer we can sub back in the original subs
$$
\cos (\theta) =\frac{1}{x}
$$
We can then use the relationship 
$$
\sin^2 \theta + \cos^2\theta = 1 = \sin^2 \theta +\frac{1}{x^2}
$$
The rest is straight forward. 
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