# Substituting for the entire integrand

• TomasRiker
In summary, the conversation discusses a surprising substitution method used for an integral, where the entire original integrand is substituted for in one step. This approach is possible because of the relationship between the derivative of e^x and e^x itself.
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

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

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.

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:
Thank you! It is clear to me now.

## 1. What does it mean to substitute for the entire integrand?

Substituting for the entire integrand means replacing the entire expression being integrated with a new variable or expression, in order to simplify the integration process.

## 2. When should I substitute for the entire integrand?

You should substitute for the entire integrand when the integrand is a complicated expression that cannot be easily integrated using traditional methods. This technique can make the integration process more manageable and efficient.

## 3. How do I substitute for the entire integrand?

To substitute for the entire integrand, you can start by letting u equal a part of the integrand that can be simplified, and then use the substitution formula to replace all instances of that part with u. From there, you can continue to simplify the integral using algebraic manipulations.

## 4. What is the substitution formula for the entire integrand?

The substitution formula for the entire integrand is u = g(x), where u is the new variable and g(x) is the part of the integrand being substituted. This formula allows you to rewrite the integral in terms of u and then solve for u in terms of x.

## 5. Are there any limitations to substituting for the entire integrand?

While substituting for the entire integrand can be a helpful technique, it is not always possible or appropriate. Some integrals may not lend themselves to this method, and in some cases, it may actually make the integration more complicated. It is important to assess each integral individually and determine if substitution is the best approach.

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