How do you integrate #int (e^x-1)/sqrt(e^(2x) -1)dx# using trigonometric substitution?

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Answer 1 · 2016-11-29T04:53:23

#ln(e^x+sqrt(e^(2x)-1))-sec^-1(e^x)+C#

#I=int(e^x-1)/sqrt(e^(2x)-1)dx#

Apply the substitution #e^x=sectheta#. This implies that #e^xdx=secthetatanthetad theta#. First rearranging, then substituting, we see that:

#I=int(e^x-1)/(e^xsqrt(e^(2x)-1))(e^xdx)#

#I=int(sectheta-1)/(secthetasqrt(sec^2theta-1))(secthetatanthetad theta)#

Using #sec^2theta-1=tan^2theta#:

#I=int(sectheta-1)d theta#

#I=lnabs(sectheta+tantheta)-theta#

#I=lnabs(sectheta+sqrt(sec^2theta-1))-theta#

Using #e^x=sectheta# and #theta=sec^-1(e^x)#:

#I=lnabs(e^x+sqrt(e^(2x)-1))-sec^-1(e^x)#

Since #e^x+sqrt(e^(2x)-1)>0# for all values of #x#, the absolute value bars aren't needed:

#I=ln(e^x+sqrt(e^(2x)-1))-sec^-1(e^x)+C#