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?