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Section 7.2 More on \(u\)-Substitution

We’re going to do some more thinking about an integration topic that we’ve already introduced in Section 5.6. We’re going to do a bit more with it now, and try to build some more flexibility, so definitely review that introduction section if you’d like!
Before moving on, we should work through these few examples, just to make sure we remember what we’re up to.

Activity 7.2.1. Recapping \(u\)-Substitution.

We’re going to consider a few integrals, and work through each of the questions for all integrals.
  1. \(\displaystyle \displaystyle \int \left(\frac{1}{\sqrt{4-3x}}\right)\;dx\)
  2. \(\displaystyle \displaystyle \int x^2(5+x^3)^7\;dx\)
  3. \(\displaystyle \displaystyle \int \left(\frac{\sin(x)}{\cos(x)}\right)\;dx\)
  4. \(\displaystyle \displaystyle \int \left(xe^{-x^2}\right)\;dx\)

(a)

For each integral, explain why \(u\)-substitution is a good choice. How can you tell, just by looking at the integral, that this strategy will be a reasonable thing to try?

(b)

For each integral, explain your choice of \(u\text{,}\) and what that means for how we define \(du\text{.}\)

(c)

For each integral, is your definition of \(du\) present in the integrand function? How do you go about making this substitution when the integrand function isn’t set up perfectly?

(d)

Finish the substitution and integration, and substitute back to the original variable.

Subsection Variable Substitution in Integrals

In \(u\)-substitution, we focus a lot on one specific kind of structure: composition in our integrand function and/or some function-derivative pairing present. We do this because we’re undoing the The Chain Rule. Variable substitutions can be much more general in their goal, but this is a good one to focus on because it solves a specific problem that we might run into while integrating.

Strategy for \(u\)-Substitution.

Goal: undo the Chain Rule, or antidifferentiate functions with composition.
Process: We’ll translate \(\int f(x)\;dx\) to \(\int g(u)\;du\) by labeling some “inside” function as \(u\text{,}\) and substituting its derivative, \(du=u'\;dx\text{.}\)
But we can use \(u\)-substitution more generally as a kind of grouping mechanism.

Activity 7.2.2. Turn Around Problems.

The two integrals that we’re going to look at are “just” some \(u\)-substitution problems, but I like to call integrals like these turn-around problems. We’ll see why!
(a)
Consider the integral:
\begin{equation*} \int x\sqrt[3]{x+5}\;dx\text{.} \end{equation*}
First, explain why \(u\)-substitution is reasonable here.
Hint.
Do you see composition? A function inside of something?
(b)
Identify \(du\) for your chosen \(u\)-substitution. When you substitute, you should notice that there are some extra bits in this integrand function that have not been assigned to be translated over to be written in terms of \(u\text{.}\) Which parts?
Solution.
If we let \(u=x+5\text{,}\) then \(du=dx\text{.}\)
\begin{equation*} \int x\sqrt[3]{x+5}\;dx = \int {\color{red}{x}} \sqrt[3]{u}\;du \end{equation*}
(c)
We need to think about how to write \(x\) in terms of \(u\text{.}\) Luckily, we already have everything we need! We have defined a link between the \(x\) variable and the \(u\) variable. We defined it as \(u\) being written as some function of \(x\text{,}\) but can we “turn around” that link to write \(x\) in terms of \(u\text{?}\)
Hint.
We let \(u=x+5\text{.}\) Solve this for \(x\)
(d)
Substitute the integral to be fully written in terms of \(u\text{.}\)
Solution.
\begin{align*} u = x+5 \amp \longrightarrow x=u-5\\ du \amp = dx \\ \int x\sqrt[3]{x+5}\;dx \amp = \int (u-5)\sqrt[3]{u}\;du\\ \amp = \int (u-5)u^{1/3}\;du \end{align*}
(e)
Before antidifferentiating, compare this integral with the original one. Specifically thinking about how we might multiply, describe the differences between the integrals with regard to composition and rewriting our integrand.
Then, go ahead and use this nicely rewritten version to antidifferentiate and substitute back to \(x\text{.}\)
Solution.
\begin{align*} \int x\sqrt[3]{x+5}\;dx \amp = \int (u-5)u^{1/3}\;du\\ \amp = \int u^{4/3}-5^{1/3}\;du\\ \amp = \frac{3u^{7/3}}{7}-\frac{15u^{4/3}}{4}+C\\ \amp = \frac{3(x+5)^{7/3}}{7}-\frac{15(x+5)^{4/3}}{4}+C \end{align*}
(f)
Apply this same strategy to the following integral:
\begin{equation*} \int \frac{x}{x+5}\;dx\text{.} \end{equation*}
This integral might be a bit trickier to find the composition in order to identify the \(u\)-substitution! Give some things a try!
Solution.
\begin{align*} u = x+5 \amp \longrightarrow x=u-5\\ du \amp = dx \\ \int \frac{x}{x+5}\;dx \amp = \int \frac{u-5}{u}\;du\\ \amp = \int \frac{u}{u}-\frac{5}{u}\;du \\ \amp \int 1 - \frac{5}{u}\;du\\ \amp = u-5\ln|u|+C\\ \amp = (x+5) - 5\ln|x+5|+C \end{align*}
Since this antiderivative has an extra constant (the \(5\) being added to \(x\)), we can write a smaller version of this by combining the \(5\) with the constant of integration:
\begin{equation*} \int \frac{x}{x+5}\;dx = x-5\ln|x+5|+C\text{.} \end{equation*}
(g)
Compare your integral in terms of \(x\) with the substituted version, in terms of \(u\text{.}\) Why was the second one so much easier to think about or rewrite?
In both of these examples, we got around not being able to multiply (using the distributive property) or divide (by splitting up our fraction into two with common denominators) by grouping some terms together with our substitution. Once we wrote \(x+5\) as \(u\) in both of these, we were able to distribute \(u^{1/3}\) across the two other terms, and we were able to divide \(u-5\) by \(u\) through splitting the single fraction into two fractions.
The term turn-around problem is a good one because we’re turning around two things:
  1. The substitution itself, by solving for \(u\) instead of \(x\text{.}\)
  2. The structure of the integral, by grouping \(x+5\) into one term, \(u\text{,}\) and expanding \(x\) into two terms, \(u-5\text{.}\) This allowed us to change how the algebra would work, making it much friendlier!

Example 7.2.1.

Find the following indefinite integral:
\begin{equation*} \int \frac{x^2+3x-1}{x-1}\;dx \end{equation*}
Hint.
Try letting \(u=x-1\) so that \(du=dx\text{.}\) Then we can say that \(x=u+1\text{.}\)
Solution.
\begin{align*} u=x-1 \amp \longrightarrow x=u+1\\ du \amp = dx \\ \int \frac{x^2+3x-1}{x-1}\;dx\amp = \int \frac{(u+1)^2+3(u+1)-1}{u}\;du\\ \amp = \int \frac{u^2+5u+3}{u}\;du\\ \amp = \int u + 5 + \frac{3}{u}\;du\\ \amp = \frac{u^2}{2}+5u+3\ln|u|+C\\ \amp = \frac{(x-1)^2}{2}+5(x-1)+3\ln|x-1|+C \end{align*}
There are some ways of rewriting this antiderivative family: we could try to group up all of the constant terms by multiplying everything out. Feel free to do this, but it is completely unnecessary.
This specific example is an interesting one, because we actually have a couple of different options as to how we approach it. This is true in a lot of cases: there is very rarely only a single approach to an integral that will eventually work out. Sometimes there are approaches or more techniques that are more obvious to some people, and sometimes there are approaches that seem more easy/difficult for some people. But even still, we are often presented with many choices we could make in how we approach our integration.
Moving forward in this chapter, we’ll present a whole host of strategies for how we might integrate different types of functions and how we might approach different structures that we see in the integrals we’ll look at. We’ll try to balance a difficult duality:
  • There is rarely no single “right” way to do things! We can’t summarize things with strongly worded rules like “if your function looks like this, then you have to do this to antidifferentiate.”
  • We would like to build some good intuition, and so having some tried-and-true strategies to fall back on will help! We can try to identify some intuitive strategies, even if they’re not the only ones that will work.
All of this to simply say: we are going to present a lot of problems with a lot of solutions, and there simply isn’t enough space to write out alternative approaches for each one. We will try to revisit some integrals to think about alternative strategies when we are able to, though!