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

1 Answer
Aug 23, 2015

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.)