# Inverse Hyperbolic Funcs. or Trigonometric Substitution?

So, in integrals that lead to inverse hyperbolics and can be solved with trigonometric substitution i just get lost. I know how to use both of them but i don't know which to use.

For the sake of simplicity i'll just go with this one

∫ dx/sqrt(4+x^2)

We know this equals arcsinh(x/2) ,right?

or we can just write x=2tant and find a completely different answer?

Which one of these solutions are "more true" , assuming you are in the final exam?
Or just they aren't true and i'm missing something?

Any solution you find will be equally valid. In fact, you will only find one solution, however, it may happen that the solutions look very, very different. The reason for this is because the solutions vary "up to the constant".

For example, you may find log|2x|+C as solution or log|x|+C. But these two are the same since log|2x|+X=log[x|+log|2|+C. And we just put log|2| in the constant (which is arbitrary anyway).

Thus every solution you find are actually equal (up to a constant). Hence all solutions you find will be equally valid!

Any solution you find will be equally valid. In fact, you will only find one solution, however, it may happen that the solutions look very, very different.

Indeed! But even if they look very different, they can be proven to be the same, with different constants added.

At the specific OPs case, the two different results one gets are:
$$\int\frac{dx}{\sqrt{4+x^2}}=arcsinh\left(\frac{x}{2}\right)+C_1$$
and (if one uses the trigonometric substituition x=tan(t) ):
$$\int\frac{dx}{\sqrt{4+x^2}}=ln\left(\frac{x}{2}+ \sqrt{ x^2/4+1 } \right)+C_2$$
For proving that both are the same, we start as:
$$y=arcsinh\left(\frac{x}{2}\right)$$
$$sinh(y)=\frac{x}{2}$$
$$\frac{e^y-e^{-y}}{2}=\frac{x}{2}$$
$$u=e^y$$
$$u-\frac{1}{u}=x$$
$$u^2-ux-1=0$$
$$u=\frac{x\pm\sqrt{x^2+4}}{2}$$
$$y=ln(u) = ln\left(\frac{x+\sqrt{x^2+4}}{2}\right)$$
Where we choose the + at the sqrt because the - would lead to a negative argument to the log. So:
$$y= ln\left(\frac{x}{2}+\sqrt{x^2/4+1}\right)$$
which is the same result we got when we integrated using ths trigonometric substituition x=tan t. In this specific case, the two solutions were exactly the same, but usually they differ by a constant, as pointed by micromass.