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

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Answer 1 · 2015-08-23T01:55:37

I takes a couple of substitutions. See 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.)

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