4.4: The Substitution Rule

4.4 The Substitution Rule

Learning Objectives

• Apply the substitution rule to evaluate both definite and indefinite integrals by making an appropriate substitution to simplify the integrand and compute the integral.

Reversing the Chain Rule: \(u\)-substitution

How can we find the antiderivative of the function

\[ g(x) = x e^{x^2} \ \text{and} \ h(x) = e^{x^2}? \nonumber \]

It is important to remember that differentiation and antidifferentiation are almost inverse processes (that they are not is due to the \(+C\) that arises when antidifferentiating). This almost-inverse relationship enables us to take any known derivative rule and rewrite it as a corresponding rule for an indefinite integral. For example, since

\[ \frac{d}{dx} \left[x^5\right] = 5x^4\text{,} \nonumber \]

we can equivalently write

\[ \int 5x^4 \, dx = x^5 + C\text{.} \nonumber \]

Recall that the Chain Rule states that

\[ \frac{d}{dx} \left[ f(g(x)) \right] = f'(g(x)) \cdot g'(x)\text{.} \nonumber \]

Restating this relationship in terms of an indefinite integral,

\[ \int f'(g(x)) g'(x) \, dx = f(g(x))+C\text{.}\label{iLY}\tag{\(\PageIndex{1}\)} \]

Equation \(\PageIndex{1}\) tells us that if we can view a given function as \(f'(g(x)) g'(x)\) for some appropriate choices of \(f\) and \(g\text{,}\) then we can antidifferentiate the function by reversing the Chain Rule. Note that both \(g(x)\) and \(g'(x)\) appear in the form of \(f'(g(x)) g'(x)\text{;}\) we will sometimes say that we seek to identify a function-derivative pair (\(g(x)\) and \(g'(x)\)) when trying to apply the rule in Equation \(\PageIndex{1}\).

If we can identify a function-derivative pair, we will introduce a new variable \(u\) to represent the function \(g(x)\text{.}\) With \(u = g(x)\text{,}\) it follows in Leibniz notation that \(\frac{du}{dx} = g'(x)\text{,}\) so that in terms of differentials, \(du = g'(x)\, dx\text{.}\) Now converting the indefinite integral to a new one in terms of \(u\text{,}\) we have

\[ \int f'(g(x)) g'(x) \, dx = \int f'(u) \,du\text{.} \nonumber \]

If we recall from the definition of the derivative that \(\dfrac{du}{dx} \approx \dfrac{\Delta{u}}{\Delta{x}}\) and use the fact that \(\dfrac{du}{dx} = g'(x)\text{,}\) then we see that \(g'(x) \approx \dfrac{\Delta{u}}{\Delta{x}}\text{.}\) Solving for \(\Delta u\text{,}\) \(\Delta u \approx g'(x) \Delta x\text{.}\) It is this last relationship that, when expressed in “differential” notation enables us to write \(du = g'(x) \, dx\) in the change of variable formula.

Provided that \(f'\) is an elementary function whose antiderivative is known, we can easily evaluate the indefinite integral in \(u\text{,}\) and then go on to determine the desired overall antiderivative of \(f'(g(x)) g'(x)\text{.}\) We call this process \(u\)-substitution, and summarize the rule as follows:

Note

With the substitution \(u = g(x)\text{,}\)

\[ \int f'(g(x)) g'(x) \, dx = \int f'(u) \,du = f(u) + C = f(g(x)) + C\text{.} \nonumber \]

To see \(u\)-substitution at work, we consider the following example.

Example \(\PageIndex{9}\)

Evaluate the indefinite integral

\[ \int 5x^4 \cdot \sin (x^5 + 1) \, dx \nonumber \]

and check the result by differentiating.

Evaluating Definite Integrals via \(u\)-substitution

We have introduced \(u\)-substitution as a means to evaluate indefinite integrals of functions that can be written, up to a constant multiple, in the form \(f(g(x))g'(x)\text{.}\) This same technique can be used to evaluate definite integrals involving such functions, though we need to be careful with the corresponding limits of integration. Consider, for instance, the definite integral

\[ \int_2^5 xe^{x^2} \, dx\text{.} \nonumber \]

Whenever we write a definite integral, it is implicit that the limits of integration correspond to the variable of integration. To be more explicit, observe that

\[ \int_2^5 xe^{x^2} \, dx = \int_{x=2}^{x=5} xe^{x^2} \, dx\text{.} \nonumber \]

When we execute a \(u\)-substitution, we change the variable of integration; it is essential to note that this also changes the limits of integration. For instance, with the substitution \(u = x^2\) and \(du = 2x \, dx\text{,}\) it also follows that when \(x = 2\text{,}\) \(u = 2^2 = 4\text{,}\) and when \(x = 5\text{,}\) \(u = 5^2 = 25\text{.}\) Thus, under the change of variables of \(u\)-substitution, we now have

\begin{align*} \int_{x=2}^{x=5} xe^{x^2} \, dx &= \int_{u=4}^{u=25} e^{u} \cdot \frac{1}{2} \, du\\[4pt] &= \left. \frac{1}{2}e^u \right|_{u=4}^{u=25}\\[4pt] &= \frac{1}{2}e^{25} - \frac{1}{2}e^4\text{.} \end{align*}

Alternatively, we could consider the related indefinite integral \(\int xe^{x^2} \, dx\text{,}\) find the antiderivative \(\frac{1}{2}e^{x^2}\) through \(u\)-substitution, and then evaluate the original definite integral. With that method, we'd have

\begin{align*} \int_{2}^{5} xe^{x^2} \, dx &= \left. \frac{1}{2}e^{x^2} \right|_{2}^{5}\\[4pt] &= \frac{1}{2}e^{25} - \frac{1}{2}e^4\text{,} \end{align*}

which is, of course, the same result.

Example \(\PageIndex{10}\)

Evaluate the following integrals:

1. \( \displaystyle \int 6x(3x^2 + 4)^4 \, dx \)
2. \( \displaystyle \int x\sqrt{x^2 - 5} \, dx \)
3. \( \displaystyle \int \frac{\sin t}{\cos^3 x} \, dx \)
4. \( \displaystyle \int \frac{x}{\sqrt{x - 1}} \, dx \)
5. \( \displaystyle \int_0^1 x^2 (1 + 2x^3)^5 \, dx \)
6. \( \displaystyle \int_0^1 x e^{4x^2 + 3} \, dx \)
7. \( \displaystyle \int_0^1 x^2 \cos\left(\frac{\pi}{2} x^3\right) \, dx \)
8. \( \displaystyle \int_0^{\pi/2} \cos^2 x \, dx \)