- #1
TomasRiker
- 4
- 0
Hello,
I found something surprising (at least to me) while looking at the following integral:
[tex]\int \sqrt{\frac{e^x-1}{e^x+1}} dx[/tex]
Wolfram Alpha suggests the following substitution as the first step:
[tex]u = \frac{1}{e^x+1}[/tex]
Which leads to the following integral:
[tex]\int \frac{\sqrt{1-2u}}{(u-1)u} du[/tex]
The next substitution is:
[tex]s = \sqrt{1-2u}[/tex]
From there, the solution is just a partial fraction decomposition away. Thinking that one should be able to do both substitutions in one step, I plugged [itex]u[/itex] into [itex]s[/itex] and got:
[tex]s = \sqrt{1-\frac{2}{e^x+1}} = \sqrt{\frac{e^x-1}{e^x+1}}[/tex]
So, this is the original integrand that was substituted here! I don't remember having learned this technique. Does this approach of substituting the whole integrand have a name?
Thank you!David
I found something surprising (at least to me) while looking at the following integral:
[tex]\int \sqrt{\frac{e^x-1}{e^x+1}} dx[/tex]
Wolfram Alpha suggests the following substitution as the first step:
[tex]u = \frac{1}{e^x+1}[/tex]
Which leads to the following integral:
[tex]\int \frac{\sqrt{1-2u}}{(u-1)u} du[/tex]
The next substitution is:
[tex]s = \sqrt{1-2u}[/tex]
From there, the solution is just a partial fraction decomposition away. Thinking that one should be able to do both substitutions in one step, I plugged [itex]u[/itex] into [itex]s[/itex] and got:
[tex]s = \sqrt{1-\frac{2}{e^x+1}} = \sqrt{\frac{e^x-1}{e^x+1}}[/tex]
So, this is the original integrand that was substituted here! I don't remember having learned this technique. Does this approach of substituting the whole integrand have a name?
Thank you!David