Question

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

Answer 1

I takes a couple of substitutions. See explanation.

Explanation:

Use a trigonometric substitution: `x = asectheta`

so `dx = asectheta tantheta d theta`

With a bit of work you can simplify `int dx/ sqrt(x^2 - a^2)` to

`int sectheta" " d theta`

If you know this integral, you can skip the next section.

To get that integral multiply by `1` in the form
`(sectheta+tantheta)/(sectheta+tantheta)`

This gets us:

`int (sec^2theta+sec theta tantheta)/(sectheta+tantheta) d theta`

And the numerator is the derivative of the denominator so we get an `ln`

We get
`int sectheta d theta = ln(sec theta + tan theta)+C`

Now that we have integrated the secant, note that due to the first substitution, `sec theta = x/a`.

Our trigonometry then gets us
`tan theta = sqrt(sec^2 theta -1) = sqrt(x^2/a^2 - 1)`

So our answer is:

`ln(sec theta + tan theta) +C = ln(x/a + sqrt(x^2/a^2 - 1)) +C`

We can rewrite in several ways. Perhaps the simplest is to write:

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

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

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

But assuming that `a` is a constant, `lna` is a constant so we can let it be absorbed by the `C`.

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

Checking the answer by differentiating is left as an exercise. (It's not very tedious.)