How do you integrate #int dx/(4x^2-1)^(3/2)# using trig substitutions?

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Answer 1 · 2018-06-20T03:07:44

Use the substitution #2x=sectheta#.

Let

#I=intdx/(4x^2-1)^(3/2)#

Apply the substitution #2x=sectheta#:

#I=int(1/2secthetatanthetad theta)/(tan^3theta)#

Simplify:

#I=1/2intcscthetacotthetad theta#

Integrate directly:

#I=-1/2csctheta+C#

Rewrite in terms of #sectheta# and #tantheta#:

#I=-1/2 sectheta/tantheta+C#

Reverse the substitution:

#I=-x/sqrt(4x^2-1)+C#