How do you find the integral of #(x^2 - 1)^(1/2)#?

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.