How do you integrate #int sqrt(e^(8x)-9)# using trig substitutions?
1 Answer
Explanation:
#intsqrt(e^(8x)-9)dx#
Apply the substitution
Note that
#=intsqrt((e^(4x))^2-9)dx#
#=intsqrt(9sec^2theta-9)(1/4tanthetad theta)#
#=3/4intsqrt(sec^2theta-1)(tanthetad theta)#
Note that
#=3/4inttan^2thetad theta#
#=3/4int(sec^2theta-1)d theta#
#=3/4intsec^2thetad theta-3/4intd theta#
#=3/4tantheta-3/4theta+C#
From
#=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#