How do you integrate #int dx/sqrt(x^2-4)# using trig substitutions?

2 Answers
Apr 6, 2018

#intdx/sqrt(x^2-4)=ln|(x+sqrt(x^2-4))|+C#

Explanation:

If our integral involves a root in the form of #sqrt(x^2-a^2),# we're best off using the substitution

#x=asectheta.# Here, we see #a^2=4, a=2,# so we use the substitution

#x=2sectheta#

#dx=2secthetatanthetad theta#

Apply to the integral #intdx/sqrt(x^2-4),# yielding

#int(2secthetatantheta)/sqrt(4sec^2theta-4)d theta#

Simplify.

#int(2secthetatantheta)/sqrt(4(sec^2theta-1)d theta#

#int(secthetatantheta)/sqrt(sec^2theta-1)d theta#

Recalling the identity #sec^2theta-1=tan^2theta,# we get

#int(secthetatantheta)/sqrt(tan^2theta)d theta=int(secthetacanceltantheta)/canceltanthetad theta#

So, we end up with the common integral (you should have this memorized)

#intsecthetad theta=ln|sectheta+tantheta|+C#

We need things in terms of #x.# Well, recalling that #x=2sectheta, sectheta=x/2.# We still need to find tangent.

Recalling that #sec^2theta-1=tan^2theta, x^2/4-4/4=tan^2theta, tan^2theta=(x^2-4)/4, tantheta=sqrt(x^2-4)/2#

Thus, we have

#intdx/sqrt(x^2-4)=ln|(x+sqrt(x^2-4))/2|+C#

#ln|(x+sqrt(x^4-4))/2|=ln|x+sqrt(x^2-4)|-ln|2|+C=ln|x+sqrt(x^2-4)|+C#, as we can absorb that logarithm into the constant of integration.

Finally,

#intdx/sqrt(x^2-4)=ln|(x+sqrt(x^2-4))|+C#

Apr 6, 2018

Ignore this answer for now.