4.5: The Substitution Rule

Motivation

Integrals require antiderivatives. Antiderivatives require undoing derivative rules.

This section shows you how to undo the chain rule. The method is called U-Substitution!

For Indefinite Integrals

The Substitution Rule
When integrating something with a composition, like $f(g(x))$, set $u = g(x)$, or the inside of the composition. Then \[\int f(g(x)) g'(x) \ dx = \int f(u) \ du\]

Recall from Section 2.9 the definition of a differential:

If $u = f(x)$, then the differential $du$ is equivalent to $du = f'(x) \ dx$.

After picking your $u$, you need to replace the $dx$ in your integral! The differential $du$ allows you to do this replacement.

Integrate \[\int x^3 \cos(x^4 + 2) \ dx\]

First, we will rewrite the integral: \[\int x^3 \cos(x^4 + 2) \ dx = \int \cos(x^4 + 2) \cdot x^3 \ dx\]

Let $u = x^4 + 2$, the inside of the composition.

Then $du = 4x^3 \ dx$.

Look at your integral. You need to replace $x^3 \ dx$. Solving for this in the previous equation: $x^3 \ dx = \dfrac{1}{4} \ du$.

Now substitute:

\begin{align} \int \cos(x^4 + 2) \cdot x^3 \ dx &= \int cos(u) \dfrac{1}{4} \ du \\&= \dfrac{1}{4} \int \cos(u) \ du \\&= \dfrac{1}{4} \left(\sin(u) + C\right) \\&= \dfrac{\sin{u}}{4} + \dfrac{1}{4}C \\&= \dfrac{\sin(x^4 + 2)}{4} + C \end{align}

Notice that I replaced $\dfrac{1}{4}C$ with $C$. You can do this because it's just a constant, it doesn't matter if $C$ is divided by four or not.

Integrate $\displaystyle \int \sqrt{2x + 1} \ dx$.
Integrate $\displaystyle \int \sqrt{1 + x^2} \ x^5 \ dx$.

Insight

Recall in Section 4.2 we said the $dx$ is symbolically equivalent to the width of one rectangle.

In this section we are manipulating $dx$ as part of a differential.

We are allowed to do this because of the following reasons:

  1. First, the differential $du = f'(x) \ dx$ is only valid if $dx \neq 0$.
  2. For the definite integral, which represents the area underneath the curve, we have the limit sum formula from the bottom of Section 4.1.
  3. If you think about it, the $\displaystyle \lim_{n\rightarrow \infty}$ will never take $n$ to be infinity. $n$ is always a number.
  4. Because $n$ represents the number of rectangles, a finite number of rectangles will always have a width of a rectangle $dx$ that is nonzero.
  5. Therefore, we are allowed to manipulate the $dx$ in the integral as a differential!

For Definite Integrals

Using U-substition on definite integrals is the same method as indefinite integrals.

The only difference is the top and bottom bound of integration must be changed from $x$ to $u$'s.

Integrate \[\int^4_0 \sqrt{2x + 1} \ dx \]
Integrate \[\int^2_1 \dfrac{1}{(3 - 5x)^2} \ dx\]