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

Question

Paisley Johnson · Accepted Answer

#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#

Aurora Alvarado · Answer

undefined To integrate $ \int \frac{e^x - 1}{\sqrt{e^{2x} - 1}} \, dx $ using trigonometric substitution, let $ e^x = \sec(\theta) $. Then, $ dx = \sec(\theta) \tan(\theta) \, d\theta $.

Substitute these into the integral, and simplify the expression in terms of $ \theta $:

$ \int \frac{e^x - 1}{\sqrt{e^{2x} - 1}} \, dx = \int \frac{\sec(\theta) - 1}{\sqrt{\sec^2(\theta) - 1}} \sec(\theta) \tan(\theta) \, d\theta $

Now, simplify the expression under the square root:

$ \sqrt{\sec^2(\theta) - 1} = \sqrt{\tan^2(\theta)} = \tan(\theta) $

The integral becomes:

$ \int (\sec(\theta) - 1) \, d\theta $

Integrate term by term:

$ \int \sec(\theta) \, d\theta - \int d\theta $

$ = \ln|\sec(\theta) + \tan(\theta)| - \theta + C $

Finally, substitute back $ e^x $ for $ \sec(\theta) $ and simplify the result:

$ = \ln|e^x + \sqrt{e^{2x} - 1}| - \arccos(e^x) + C $