# Substituting for the entire integrand

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1. Jul 7, 2015

### 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!

Best regards,
David

2. Jul 8, 2015

### ShayanJ

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

3. Jul 8, 2015

### 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.

4. Jul 8, 2015

### HallsofIvy

Staff Emeritus
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 $\int du$.

Last edited by a moderator: Jul 8, 2015
5. Jul 9, 2015

### TomasRiker

Thank you! It is clear to me now.