<s>[INST] integrate $\int\sqrt{x^2-1}dx$ unproven step In order to find $\int\sqrt{x^2-1}dx$ one makes substitutions $x=\sec(\theta)$ , $dx=\sec(\theta)\tan(\theta)d\theta$ and $\sqrt{x^2-1}$ = $\sqrt{\tan^2(\theta)}$. Then you find $\int{\tan^2(\theta) \sec(\theta)}d\theta$ directly as easy as $1+1=2$ it seems from many online sources and even WolframAlpha. But I tried for hours but every way I try, $\sqrt{\tan^2(\theta)}\neq\tan(\theta)$. I already taken into account $\theta$ is a substitution for $\sec^{-1}(x)$, does not make a difference. What do I think wrong? (edit: and also $\sqrt{\tan^2(\theta)} * \tan(\theta)$ != $\tan^2(\theta)$ )
 [/INST] $\int\sqrt (x^2-1)dx$
(1)Use a substitution of $x=cosh \theta$, $dx=sinh\theta$
=$\int\sqrt(cosh^2-1).(sinh\theta)d\theta$
=$\int\sqrt sinh^2\theta.(sinh\theta)d\theta$
=$\int\sinh^2\theta d\theta$
Using double angle formulae: $cosh 2\theta=1+2sinh^2\theta$
Hence, $sinh^2\theta=\frac{1}{2}(cosh 2\theta-1)$
=$\frac{1}{2}\int (cosh2\theta-1)d\theta$
=$\frac{1}{2}[\frac{sinh 2\theta}{2}-\theta]$
Need to otain final answer in terms of $x$,
so from (1), substitute $cosh\theta=x$ 
in $cosh^2\theta-sinh^2\theta=1$. Hence $sinh\theta=\sqrt(x^2-1)$
Therefore $sinh 2\theta=2cosh\theta sinh\theta=2x\sqrt(x^2-1)$ 
$\frac{1}{2}[\frac{2x\sqrt(x^2-1)}{2}-cosh^{-1}x]$+C
$\frac{x\sqrt(x^2-1)}{2}-\frac{cosh^{-1}x}{2}$+C
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