First of all, notice how this equation has this relationship:
#sqrt(x^2 - 1) prop sqrt(a^2x^2 - a^2) prop sqrt(a^2sec^2theta - a^2)#
where #prop# means "proportional to", and #a = 1#. Thus, let:
#a = 1#
#x = asectheta = sectheta#
#dx = secthetatanthetad theta#
#sqrt(x^2 - 1) = sqrt(sec^2theta - 1) = sqrt(tan^2theta) = tantheta#
Now you can write this as:
#int sqrt(x^2 - 1)dx#
#= int tanthetasecthetatanthetad theta#
#= intsecthetatan^2thetad theta#
Something you can do to simplify this (and it may seem odd at first) is to rewrite this as:
#= int sectheta(sec^2theta - 1)d theta#
since #tan^2theta = sec^2theta - 1#. Now we only have #sectheta# to deal with.
#= intsec^3theta - secthetad theta#
These two integrals require special tricks, but they do have real answers. Let's take them separately for simplicity.
#int secthetad theta#
#= int (sectheta)((sectheta + tantheta)/(sectheta + tantheta))d theta#
#= int (sec^2theta + secthetatantheta)/(sectheta + tantheta)d theta#
Now notice that if you take the derivative of the denominator, you get the numerator times #d theta#. i.e.:
#d[sectheta + tantheta] = (sec^2theta + secthetatantheta)d theta#
Thus, let:
#u = sectheta + tantheta#
#du = secthetatantheta + sec^2thetad theta#
#=> int 1/u du#
#= ln|u| = ln|sectheta + tantheta|#
Now let's take the other integral.
#int sec^3thetad theta#
With this, the best trick one can try is Integration by Parts. Typically, you let #u = sectheta#, so let:
#u = sectheta#
#du = secthetatanthetad theta#
#dv = sec^2thetad theta#
#v = tantheta#
#= secthetatantheta - int secthetatan^2thetad theta#
Look at that, we got back the original integral (didn't even need an identity for this step)!
The whole thing now is:
#int secthetatan^2thetad theta#
#= secthetatantheta - color(darkgreen)(int secthetatan^2thetad theta) - int secthetad theta#
#2int secthetatan^2thetad theta = secthetatantheta - int secthetad theta#
Technically, we're done. Now we just have:
#int secthetatan^2thetad theta = 1/2[secthetatantheta - int secthetad theta]#
#= 1/2[secthetatantheta - ln|sectheta + tantheta|]#
Now if you look all the way at the top:
#x = sectheta#
#sqrt(x^2 - 1) = tantheta#
Thus:
#= color(blue)(1/2[xsqrt(x^2 - 1) - ln|x + sqrt(x^2 - 1)|] + C#
You can also see it here.
