Question

How do you integrate `int 2/sqrt(x^2-4)dx` using trigonometric substitution?

Answer 1

Use x = 2*sec(u) as a first substitution.

Explanation:

With `x = 2*sec(u), therefore dx = 2*sec(u)*tan(u)`

`sqrt(x^2 - 4) = sqrt(4*sec^2(u) - 4) = 2*sqrt(sec^2(u) - 1)`

From `sin^2(a) + cos^2(a) = 1`, divide through by `cos^2(a)` to obtain

`tan^2(a) + 1 = sec^2(a)`

So, we have:

`int 2/(sqrt(x^2-4) dx` `= int (4*sec(u)*tan(u))/(2sqrt(tan^2(u))` `du`

So, integral of `2*int sec(u) du`

Multiply numerator and denominator by `sec(u) + tan(u)` to obtain

`2*int (sec^2(u) + sec(u)*tan(u))/(sec(u) + tan(u)) du`

Use substitution `r = sec(u) + tan(u)` which gives

`2*int (dr)/r`

`= 2*ln(r) + C`

`= 2*ln(sec(u) + tan(u)) + C`

`= 2*ln(sec(arcsec(x/2) + tan(arcsec(x/2)) + C`

`= 2*ln(1/2*(x + sqrt(x^2 - 4))) + C`

And there we have it, let me know if you need any help with the last bit but it's fairly easy to figure out if you look at a few triangles.