How do you find the integral of #dx/ sqrt(x^2 - a^2)#?

1 Answer
Jul 22, 2015

With this general case, notice how #sqrt(x^2 - a^2) prop sqrt(sec^2theta - 1)#. So, we can use the following substitution:

#x = asectheta#
#dx = asecthetatanthetad theta#
#sqrt(x^2 - a^2) = sqrt(a^2sec^2theta - a^2) = atantheta#

Thus, we have:

#= int 1/(cancel(atantheta))*cancel(a)secthetacancel(tantheta)d theta#

#= int secthetad theta#

Then just a little trick:
#= int sectheta((sectheta + tantheta)/(sectheta + tantheta))d theta#

#= int (sec^2theta + secthetatantheta)/(sectheta + tantheta)d theta#

Now, let:
#u = sectheta + tantheta#
#du = secthetatantheta + sec^2thetad theta#

Therefore:

#= int1/udu#

#= ln|u|#

#= ln|sectheta + tantheta|#

Recall that:
#sectheta = x/a#
#tantheta = sqrt(x^2 - a^2)/a#

Thus we have:

#= color(green)(ln|x/a + sqrt(x^2 - a^2)/a| + C)#

...which is perfectly acceptable. But, you could simplify this more and be a little sneaky.

#= ln|(1/a)[x + sqrt(x^2 - a^2)]| + C#

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

but since #a# is a constant... it gets embedded into #C#.

#= color(blue)(ln|x + sqrt(x^2 - a^2)| + C)#

So if you see Wolfram Alpha give you this answer, that's why.