Simple integral calculation: $\int \sqrt{x^2-4}\,dx$ Calculate $$\int \sqrt{x^2-4}\,dx$$
I tried using substitution for $x = 2\sec\theta $ but I got stuck at $4\int \tan^2\theta \sec\theta d\theta$
 A: Write $$ \int \tan^2 \theta \sec \theta \, d\theta = \int \tan \theta (\sec \theta \tan \theta) \, d\theta$$
and integrate by parts. You should (after some algebra) be able to express the integral using the integral of the secant.
A: $$\int tan^2 \theta \sec \theta d\theta = \int \sec^3 \theta d\theta - \int \sec \theta d\theta$$
Now, just follow:
http://en.wikipedia.org/wiki/Integral_of_secant_cubed
A: $$\int{\tan^2\theta \sec\theta} \ d\theta = \int{\frac{\sin^2\theta}{\cos^3\theta}}d\theta = \int{\frac{d\theta}{\cos^3\theta}} - \int{\frac{d\theta}{\cos\theta}}$$
For both integrals, multiply the numerator and denominator by $\cos\theta$ and use the change of variable $t = \sin\theta$.
Things should be easy from here.
A: Alternatively, one may recall that
$$
\cosh^2 u -\sinh^2 u=1 
$$ or 
$$
\cosh^2 u -1=\sinh^2 u
$$ then you may try the change of variable $\displaystyle x=2\cosh u$, giving $\displaystyle dx=2\sinh u\: du\,$ and
$$
\int\sqrt{x^2-4}\:dx =4\int\sinh^2 u \:du
$$ Hoping you can take it from here.
