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

Question

Aurora Alvarado · Accepted Answer

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

Christopher Agresta · Answer

undefined To integrate ∫√(e^(8x) - 9) dx using trigonometric substitution, perform the following steps:

1. Let $ e^{4x} = 9\sec^2(\theta) $, then $ e^{8x} = 81\sec^4(\theta) $.
2. Find $ dx $ in terms of $ d\theta $ using $ x = \frac{1}{8}\ln\left(\frac{81\sec^4(\theta)}{9}\right) $.
3. Substitute $ e^{8x} = 81\sec^4(\theta) $ and $ dx $ into the integral.
4. Simplify the integral in terms of $ \theta $.
5. Integrate the simplified expression with respect to $ \theta $.
6. Express the result in terms of $ x $.

The integration process involves trigonometric identities and integration techniques.