I was looking at some integrals to do with trigonometric substitutions and I stumbled across this one
$$\int\frac{1}{\sqrt{x^2-1}}dx$$
I know you can do it with a regular trigonometric substitution or just use a hyperbolic substitution but I was wondering if you can do it the following way.
$$ \int \frac{1}{\sqrt{x^2-1}} dx = \int \frac{\cos \theta}{\sqrt{-\cos^2\theta}}d\theta = \int \frac{1}{i}d\theta = \frac{1}{i}\arcsin x, $$ where the $x=\sin \theta$ substitution was used. Could anybody please explain to me why I don't get the same result as one would get if a hyperbolic or other trigonometric substitution was used?
Thanks in advance.
To keep you original integral real you require $$ x^2 \gt 1 $$ If you let $$ x=\sin \theta$$ then for real $\theta$ it must be true that $$ -1 \le x \le 1 \Rightarrow x^2 \le 1 $$