How do you integrate #int 1/sqrt(x^2-a^2)# by trigonometric substitution?

1 Answer
Sep 12, 2016

Answer:

#lnabs(x+sqrt(x^2-a^2))+C#

Explanation:

#intdx/sqrt(x^2-a^2)#

We will use the substitution #x=asectheta#. Thus #dx=asecthetatanthetad theta#. Substituting:

#=int(asecthetatanthetad theta)/sqrt(a^2sec^2theta-a^2)=int(asecthetatanthetad theta)/(asqrt(sec^2theta-1))#

Note that #tan^2theta=sec^2theta-1#:

#=int(secthetatanthetad theta)/tantheta=intsecthetad theta=lnabs(sectheta+tantheta)+C#

From #x=asectheta# we see that #sectheta=x/a#. Thus we have a right triangle where #x# is the hypotenuse and #a# is the adjacent side. Through the Pythagorean theorem we see that the opposite side is #sqrt(x^2-a^2)#.

So, #tantheta# would be opposite over adjacent, or #sqrt(x^2-a^2)/a#.

#=lnabs(x/a+sqrt(x^2-a^2)/a)+C#

Note that a #1/a# term can be factored from both of these, which can then be removed from the logarithm as a constant: #log(AB)=log(A)+log(B)#. The #ln(1/a)# constant will be absorbed into #C#.

#=lnabs(x+sqrt(x^2-a^2))+C#