Question

How do you integrate `intdx/ sqrt(x^2 - a^2)`?

Answer 1

`intdx/sqrt(x^2-a^2)="arcosh"(x/a)+"c"`

Explanation:

`intdx/sqrt(x^2-a^2)=intdx/(asqrt(x^2/a^2-1))=int1/(asqrt((x/a)^2-1))dx`

Now, let `u=x/a` and `du=1/adx`

Then

`int1/(asqrt((x/a)^2-1))dx=int1/sqrt(u^2-1)dx="arcosh"(u)+"c"="arcosh"(x/a)+"c"`

Answer 2

`int dx/sqrt(x^2-a^2) = ln abs (x +sqrt(x^2-a^2)) +C`

Explanation:

The integrand is defined for `x in (-oo,-a) uu (a,+oo)`. Let us focus first on `x in (a,+oo)` and substitute:

`x = a sect`

`dx = asect tant`

with `t in (0,pi/2)`

so:

`int dx/sqrt(x^2-a^2) = int (asect tant dt)/sqrt(a^2sec^2t-a^2)`

`int dx/sqrt(x^2-a^2) = int (sect tant dt)/sqrt(sec^2t-1)`

Use now the trigonometric identity:

`sec^2t-1 = tan^2t`

and as for `t in (0,pi/2)` the tangent is positive:

`sqrt(sec^2t-1) = tan^`

Then:

`int dx/sqrt(x^2-a^2) = int (sect tant dt)/tant`

`int dx/sqrt(x^2-a^2) = int sect dt`

`int dx/sqrt(x^2-a^2) = ln abs (sect +tant) +C`

Undoing the substitution:

`int dx/sqrt(x^2-a^2) = ln abs (x/a +sqrt(x^2/a^2-1)) +C`

`int dx/sqrt(x^2-a^2) = ln abs (x +sqrt(x^2-a^2)) +C`

By differentiating we can see that the solution is valid also for `x in (-oo,-3)`