# 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 $4\int \tan^2\theta \sec\theta d\theta$

• Are you sure it should be a four factored out? – Quality Mar 23 '15 at 19:24
• @LearningMath The 4 looks right to me. 2 from the square root and 2 from dx. – Mike Mar 23 '15 at 19:33

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.

$$\int tan^2 \theta \sec \theta d\theta = \int \sec^3 \theta d\theta - \int \sec \theta d\theta$$

Now, just follow: http://en.wikipedia.org/wiki/Integral_of_secant_cubed

$$\int{\tan^2\theta \sec\theta} \ d\theta = \int{\frac{\sin^2\theta}{\cos^3\theta}}d\theta = \int{\frac{d\theta}{\cos^3\theta}} - \int{\frac{d\theta}{\cos\theta}}$$

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.

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.