How do you integrate #int sqrt(e^(8x)-9)# using trig substitutions?

1 Answer
mason m
Sep 15, 2016

#(sqrt(e^(8x)-9)-3"arcsec"(e^(4x)/3))/4+C#

Explanation:

#intsqrt(e^(8x)-9)dx#

Apply the substitution #e^(4x)=3sectheta#. Thus #4e^(4x)dx=3secthetatanthetad theta#.

Note that #dx=(3secthetatanthetad theta)/(4e^(4x))=(3secthetatanthetad theta)/(4(3sectheta))=1/4tanthetad theta#.

#=intsqrt((e^(4x))^2-9)dx#

#=intsqrt(9sec^2theta-9)(1/4tanthetad theta)#

#=3/4intsqrt(sec^2theta-1)(tanthetad theta)#

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

#=3/4inttan^2thetad theta#

#=3/4int(sec^2theta-1)d theta#

#=3/4intsec^2thetad theta-3/4intd theta#

#=3/4tantheta-3/4theta+C#

From #e^(4x)=3sectheta#, we see that #theta="arcsec"(e^(4x)/3)#. Also, since #sectheta=e^(4x)/3#, we see that the hypotenuse is #e^(4x)#, the adjacent side is #3#, and the opposite side is #sqrt(e^(8x)-9)#. Thus #tantheta=sqrt(e^(8x)-9)/3#.

#=3/4(sqrt(e^(8x)-9)/3)-3/4"arcsec"(e^(4x)/3)+C#

#=(sqrt(e^(8x)-9)-3"arcsec"(e^(4x)/3))/4+C#

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