Integration question: $\int \sqrt{x^2-4}\, dx $

Question

I am looking for a way to integrate $$\int \sqrt{x^2-4}\ dx $$ using trigonometric substitutions.

All my attempts so far lead to complicated solutions that were uncomputable.

Answer 1

Hint: Let $x = 2 \cosh \theta \Rightarrow dx = 2 \sinh \theta\ d\theta $.

Edit: The OP said that he hasn't seen hyperbolic functions just yet. Then what about $x = 2\sec \theta \Rightarrow dx = 2\sec \theta \tan\theta\ d\theta$

So

$$2\int \sqrt{4\sec^2\theta - 4}\sec\theta\tan\theta \ d\theta = 4\int \sec\theta \tan^2\theta d\theta$$

Answer 2

For a trigonometric substitution, $ x = 2\sec \theta $ will work if you can integrate certain other trig functions.

Answer 3

I'll focus on $\displaystyle \int \sqrt{x^2-1}\,\mathrm dx$ for simplicity. This is how I think about this sort of problem when I don't remember the appropriate substitution.

If you suspect that it's useful to use a trigonometric substitution, since the fundamental property $$\forall \theta \in\mathbb R\left((\sin(\theta))^2+(\cos(\theta))^2=1\right)$$ involves only squared numbers, the substitution $x=\varphi(\theta)$ has to be such that $(\varphi(\theta))^2-1$ is a square as opposed to the negative of a square (because of the square root).

So now the goal is to obtain something that looks like $(\varphi(\theta))^2-1$ from $(\sin(\theta))^2+(\cos(\theta))^2=1$.

Clearly something like $(\sin(\theta))^2-1=-(\cos(\theta))^2$ won't help because the RHS is the negative of a square.

The next usual trick with this trigonometric identities is to divide by $\cos$ yielding $(\tan(\theta))^2+1=(\sec(\theta))^2$ from where one gets $(\tan(\theta))^2=(\sec(\theta))^2-1$ which satisfies the criteria above.

Consider the substitution $x=\sec(\theta)$. It will result, after a few calculations, in known antiderivatives.

Answer 4

Following up on Aaron Maroja's (second) hint, note that

$$\sec\theta\tan^2\theta\ d\theta={\sin^2\theta\ d\theta\over\cos^3\theta}={\sin^2\theta\cos\theta\ d\theta\over\cos^4\theta}={s^2\ ds\over(1-s^2)^2}$$

where $s=\sin\theta$. Some tedious partial fractions can wrap things up:

$${s^2\over(1-s^2)^2}={A\over1-s}+{B\over(1-s)^2}+{C\over1+s}+{D\over(1+s)^2}$$

Answer 5

Using the substitution $x = \sec \theta$, we end up with

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

This can be done using integration by parts

$$ \begin{align} \int \sec \theta \tan^2 \theta \,d\theta &= \int \tan \theta \,d(\sec \theta) \\&= \tan \theta \sec \theta - \int \sec\theta \,d(\tan\theta) \\&= \tan \theta \sec \theta - \int \sec\theta \,(\tan^2 \theta + 1) \,d\theta \\&= \tan \theta \sec \theta - \int \sec \theta \tan^2 \theta \,d\theta - \int \sec\theta \,d\theta \\ 2\int \sec \theta \tan^2 \theta \,d\theta &= \tan \theta \sec \theta - \int \sec\theta\,d\theta \end{align}$$ For reference: Integral of secant cubed