How to integrate $\sqrt{x^2-1}$ How do you integrate the following integral? I have tried u-substitution, but it doesn't work out. 
$$\int(x^2-1)^{0.5}dx$$
 A: HINT(My approach): We can perform trigonometric substitution by substituting $x =\sec u$ and thus our integral transforms to $\displaystyle\int \sec u\sqrt{\sec^2 u-1} \tan u du = \displaystyle\int \sec u \tan^2 u du =\displaystyle \int \sec^3 u du-\displaystyle \int \sec u du$. Now using reduction formula for $\sec u$ in the first integral, the entire integral can be easily solved.  

Hope it helps.
A: $I=\int 1\cdot\sqrt{x^2-1}dx$
Put $x = \sec y$
then $dx = \sec y.\tan y dy$
$= \int \sqrt{\sec^2 y-1}. \sec y.\tan y dy$
$= \int \sqrt{\tan^2 y}. \sec y.\tan y dy$
$= \int \tan y. \sec y. \tan y dy$
$= \int \tan^2 y. \sec y dy$
$= \int (\sec^2 y - 1). \sec y dy$
$= \int \sec^3 y dy - \int \sec y dy$
$= \int \sec^3 y dy + log|\sec y + \tan y|$  .......(1)
Now,
$\int \sec^3 y dy=\int \sec y (\sec^2 y dy)$
Put $\tan y = t$
$\sec^2 y dy = dt$ 
and $\sec y = \sqrt{1+t^2}$
$\int \sqrt{1 + t^2}dt$ 
$\frac{t\sqrt{1+t^2}}{2} + \frac{\log(t+\sqrt{1+t^2})}{2}$
$t = \tan y$ 
$\frac{\tan y\sqrt{1 + \tan^2 y}}{2} + \frac{\log(\tan y + \sqrt{1 + \tan^2 y})}{2}$
Now from equation (1)
$= \frac{\tan y \sqrt{1 + \tan^2 y}}{2} + \frac{\log(\tan y + \sqrt{1 + \tan^2 y})}{2} + log|\sec y + \tan y|$ 
And replace y = $\sec^{-1}x$ and you got answer.
A: Hint

In $(1,+\infty)$

Put $x=\cosh(t)$.
then, we compute
$$\int \sinh^2(t) dt=\int \frac{1-\cosh(2t)}{2}dt$$
$$=\frac{t}{2}-\frac{\sinh(2t)}{4}.$$

In $(-\infty,-1)$

You put $x=-\cosh(u)$.
