Substituting for the entire integrand

[TomasRiker]

Hello,

I found something surprising (at least to me) while looking at the following integral:
\int \sqrt{\frac{e^x-1}{e^x+1}} dx
Wolfram Alpha suggests the following substitution as the first step:
u = \frac{1}{e^x+1}
Which leads to the following integral:
\int \frac{\sqrt{1-2u}}{(u-1)u} du
The next substitution is:
s = \sqrt{1-2u}
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 u into s and got:
s = \sqrt{1-\frac{2}{e^x+1}} = \sqrt{\frac{e^x-1}{e^x+1}}
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

 

[ShayanJ]

This is just another change of variables. It was done in two steps just to reduce the complexity of calculations.

 

[TomasRiker]

I understand this - maybe I wasn't clear enough.
What surprises me here is that, using those substitutions, we effectively substituted for the entire original integrand.
I don't remember having seen such a thing before, and I am wondering whether this is common and in which cases it works.

 

[HallsofIvy]

Any time you make a substitution, u= u(x), you have to also substitute for the derivative, dx. Here, with u= \left(\frac{e^x- 1}{e^x+ 1}\right) we have u'= \frac{1}{2}\left(\frac{e^x- 1}{e^x+ 1}\right)^{-1/2}\left(\frac{e^u- 1}{e^u+ 1}\right)dx= \frac{1}{2}\left(\frac{e^x- 1}{e^x+ 1}\right)^{1/2} dx so that 2du= \left(\frac{e^x- 1}{e^x+ 1}\right)^{1/2} dx, what you are referring to as "substituting for the entire integral".

That is, basically, a result of the fact that e^x is its own derivative. A much simpler example would be \int e^x dx. Making the substitution u= e^x which give e^x dx= du and the integral becomes \int du.

 

[TomasRiker]

Thank you! It is clear to me now.