Question

How do you integrate `int 1/sqrt(x^2-a^2)` by trigonometric substitution?

Answer 1

`lnabs(x+sqrt(x^2-a^2))+C`

Explanation:

`intdx/sqrt(x^2-a^2)`

We will use the substitution `x=asectheta`. Thus `dx=asecthetatanthetad theta`. Substituting:

`=int(asecthetatanthetad theta)/sqrt(a^2sec^2theta-a^2)=int(asecthetatanthetad theta)/(asqrt(sec^2theta-1))`

Note that `tan^2theta=sec^2theta-1`:

`=int(secthetatanthetad theta)/tantheta=intsecthetad theta=lnabs(sectheta+tantheta)+C`

From `x=asectheta` we see that `sectheta=x/a`. Thus we have a right triangle where `x` is the hypotenuse and `a` is the adjacent side. Through the Pythagorean theorem we see that the opposite side is `sqrt(x^2-a^2)`.

So, `tantheta` would be opposite over adjacent, or `sqrt(x^2-a^2)/a`.

`=lnabs(x/a+sqrt(x^2-a^2)/a)+C`

Note that a `1/a` term can be factored from both of these, which can then be removed from the logarithm as a constant: `log(AB)=log(A)+log(B)`. The `ln(1/a)` constant will be absorbed into `C`.

`=lnabs(x+sqrt(x^2-a^2))+C`