In
Section 4.4, we learned the key role that antiderivatives play in the process of evaluating definite integrals exactly. The Fundamental Theorem of Calculus tells us that if
\(F\) is any antiderivative of
\(f\text{,}\) then
\begin{equation*}
\int_a^b f(x) \, dx = F(b) - F(a)\text{.}
\end{equation*}
Furthermore, we realized that each elementary derivative rule developed in
Chapter 2 leads to a corresponding elementary antiderivative, as summarized in
Table 4.4.5. Thus, if we wish to evaluate an integral such as
\begin{equation*}
\int_0^1 \left(x^3 - \sqrt{x} + 5^x \right) \,dx\text{,}
\end{equation*}
it is straightforward to do so, since we can easily antidifferentiate \(f(x) = x^3 - \sqrt{x} + 5^x\text{.}\) Because one antiderivative of \(f\) is \(F(x) = \frac{1}{4}x^4 - \frac{2}{3}x^{3/2} + \frac{1}{\ln(5)}5^x\text{,}\) the Fundamental Theorem of Calculus tells us that
\begin{align*}
\int_0^1 \left(x^3 - \sqrt{x} + 5^x\right) \,dx &= \left. \frac{1}{4}x^4 - \frac{2}{3}x^{3/2} + \frac{1}{\ln(5)}5^x\right|_0^1\\
&= \left( \frac{1}{4}(1)^4 - \frac{2}{3}(1)^{3/2} + \frac{1}{\ln(5)}5^1 \right) - \left( 0 - 0 + \frac{1}{\ln(5)}5^0 \right)\\
&= -\frac{5}{12} + \frac{4}{\ln(5)}\text{.}
\end{align*}
We see that we have a natural interest in being able to find such algebraic antiderivatives. We emphasize algebraic antiderivatives, as opposed to any antiderivative, since we know by the Second Fundamental Theorem of Calculus that \(G(x) = \int_a^x f(t) \, dt\) is indeed an antiderivative of the given function \(f\text{,}\) but one that still involves a definite integral. Our goal in this section is to “undo” the process of differentiation to find an algebraic antiderivative for a given function.
Preview Activity 5.3.1.
Subsection 5.3.1 Reversing the Chain Rule: First Steps
Whenever \(f\) is a familiar function whose antiderivative is known and \(u(x)\) is a linear function, it is straightforward to antidifferentiate a function of the form
\begin{equation*}
h(x) = f(u(x))\text{.}
\end{equation*}
Example 5.3.1.
Determine the general antiderivative of
\begin{equation*}
h(x) = (5x-3)^6\text{.}
\end{equation*}
Check the result by differentiating.
For this composite function, the outer function \(f\) is \(f(u) = u^6\text{,}\) while the inner function is \(u(x) = 5x - 3\text{.}\) Since the antiderivative of \(f\) is \(F(u) = \frac{1}{7}u^7+C\text{,}\) we see that the antiderivative of \(h\) is
\begin{equation*}
H(x) = \frac{1}{7} (5x-3)^7 \cdot \frac{1}{5} + C = \frac{1}{35} (5x-3)^7 + C\text{.}
\end{equation*}
The inclusion of the constant \(\frac{1}{5}\) is essential precisely because the derivative of the inner function is \(u'(x) = 5\text{.}\) Indeed, if we now compute \(H'(x)\text{,}\) we find by the Chain Rule (and Constant Multiple Rule) that
\begin{equation*}
H'(x) = \frac{1}{35} \cdot 7(5x-3)^6 \cdot 5 = (5x-3)^6 = h(x)\text{,}
\end{equation*}
and thus \(H\) is indeed the general antiderivative of \(h\text{.}\)
Hence, in the special case where the outer function is familiar and the inner function is linear, we can antidifferentiate composite functions according to the following rule.
If \(h(x) = f(ax + b)\) and \(F\) is a known algebraic antiderivative of \(f\text{,}\) then the general antiderivative of \(h\) is given by
\begin{equation*}
H(x) = \frac{1}{a} F(ax+b) + C\text{.}
\end{equation*}
It is useful to have shorthand notation that indicates the instruction to find an antiderivative. Thus, in a similar way to how the notation
\begin{equation*}
\frac{d}{dx} \left[ f(x) \right]
\end{equation*}
represents the derivative of \(f(x)\) with respect to \(x\text{,}\) we use the notation of the indefinite integral,
\begin{equation*}
\int f(x) \, dx
\end{equation*}
to represent the general antiderivative of \(f\) with respect to \(x\text{.}\) Returning to the earlier example with \(h(x) = (5x-3)^6\text{,}\) we can rephrase the relationship between \(h\) and its antiderivative \(H\) through the notation
\begin{equation*}
\int (5x-3)^6 \, dx = \frac{1}{35} (5x-6)^7 + C\text{.}
\end{equation*}
When we find an antiderivative, we will often say that we evaluate an indefinite integral. Just as the notation \(\frac{d}{dx} [ \Box ]\) means “find the derivative with respect to \(x\) of \(\Box\text{,}\)” the notation \(\int \Box \, dx\) means “find a function of \(x\) whose derivative is \(\Box\text{.}\)”
Activity 5.3.2.
Evaluate each of the following indefinite integrals. Check each antiderivative that you find by differentiating.
\(\displaystyle \int \sin(8-3x) \, dx\)
\(\displaystyle \int \sec^2 (4x) \, dx\)
\(\displaystyle \int \frac{1}{11x - 9} \, dx\)
\(\displaystyle \int \csc(2x+1) \cot(2x+1) \, dx\)
\(\displaystyle \int \frac{1}{\sqrt{1-16x^2}}\, dx\)
\(\displaystyle \int 5^{-x}\, dx\)
Subsection 5.3.2 Reversing the Chain Rule: \(u\)-substitution
A natural question arises from our recent work: what happens when the inner function is not linear? For example, can we find antiderivatives of such functions as
\begin{equation*}
g(x) = x e^{x^2} \ \text{and} \ h(x) = e^{x^2}?
\end{equation*}
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
\begin{equation*}
\frac{d}{dx} \left[x^5\right] = 5x^4\text{,}
\end{equation*}
we can equivalently write
\begin{equation*}
\int 5x^4 \, dx = x^5 + C\text{.}
\end{equation*}
Recall that the Chain Rule states that
\begin{equation*}
\frac{d}{dx} \left[ f(g(x)) \right] = f'(g(x)) \cdot g'(x)\text{.}
\end{equation*}
Restating this relationship in terms of an indefinite integral,
\begin{equation}
\int f'(g(x)) g'(x) \, dx = f(g(x))+C\text{.}\tag{5.3.1}
\end{equation}
Equation (5.3.1) 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 (5.3.1).
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
1 ,
\(du = g'(x)\, dx\text{.}\) Now converting the indefinite integral to a new one in terms of
\(u\text{,}\) we have
\begin{equation*}
\int f'(g(x)) g'(x) \, dx = \int f'(u) \,du\text{.}
\end{equation*}
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:
With the substitution \(u = g(x)\text{,}\)
\begin{equation*}
\int f'(g(x)) g'(x) \, dx = \int f'(u) \,du = f(u) + C = f(g(x)) + C\text{.}
\end{equation*}
To see \(u\)-substitution at work, we consider the following example.
Example 5.3.2.
Evaluate the indefinite integral
\begin{equation*}
\int x^3 \cdot \sin (7x^4 + 3) \, dx
\end{equation*}
and check the result by differentiating.
Solution.
We can make two algebraic observations regarding the integrand, \(x^3 \cdot \sin (7x^4 + 3)\text{.}\) First, \(\sin (7x^4 + 3)\) is a composite function; as such, we know we’ll need a more sophisticated approach to antidifferentiating. Second, \(x^3\) is almost the derivative of \((7x^4 + 3)\text{;}\) the only issue is a missing constant. Thus, \(x^3\) and \((7x^4 + 3)\) are nearly a function-derivative pair. Furthermore, we know the antiderivative of \(f(u) = \sin(u)\text{.}\) The combination of these observations suggests that we can evaluate the given indefinite integral by reversing the chain rule through \(u\)-substitution.
Letting \(u\) represent the inner function of the composite function \(\sin (7x^4 + 3)\text{,}\) we have \(u = 7x^4 + 3\text{,}\) and thus \(\frac{du}{dx} = 28x^3\text{.}\) In differential notation, it follows that \(du = 28x^3 \, dx\text{,}\) and thus \(x^3 \, dx = \frac{1}{28} \, du\text{.}\) The original indefinite integral may be slightly rewritten as
\begin{equation*}
\int \sin (7x^4 + 3) \cdot x^3 \, dx\text{,}
\end{equation*}
and so by substituting \(u\) for \(7x^4 + 3\) and \(\frac{1}{28} \, du\) for \(x^3 \, dx\text{,}\) it follows that
\begin{equation*}
\int \sin (7x^4 + 3) \cdot x^3 \, dx = \int \sin(u) \cdot \frac{1}{28} \, du\text{.}
\end{equation*}
Now we may evaluate the easier integral in \(u\text{,}\) and then replace \(u\) by the expression \(7x^4 + 3\text{.}\) Doing so, we find
\begin{align*}
\int \sin (7x^4 + 3) \cdot x^3 \, dx &= \int \sin(u) \cdot \frac{1}{28} \, du\\
&= \frac{1}{28} \int \sin(u) \, du\\
&= \frac{1}{28} (-\cos(u)) + C\\
&= -\frac{1}{28} \cos(7x^4 + 3) + C\text{.}
\end{align*}
To check our work, we observe by the Chain Rule that
\begin{equation*}
\frac{d}{dx} \left[ -\frac{1}{28}\cos(7x^4 + 3) \right] = -\frac{1}{28} \cdot (-1)\sin(7x^4 + 3) \cdot 28x^3 = \sin(7x^4 + 3) \cdot x^3\text{,}
\end{equation*}
which is indeed the original integrand.
The \(u\)-substitution worked because the function multiplying \(\sin (7x^4 + 3)\) was \(x^3\text{.}\) If instead that function was \(x^2\) or \(x^4\text{,}\) the substitution process would not have worked. This is one of the primary challenges of antidifferentiation: slight changes in the integrand make tremendous differences. For instance, we can use \(u\)-substitution with \(u = x^2\) and \(du = 2xdx\) to find that
\begin{align*}
\int xe^{x^2} \, dx &= \int e^u \cdot \frac{1}{2} \, du\\
&= \frac{1}{2} \int e^u \, du\\
&= \frac{1}{2} e^u + C\\
&= \frac{1}{2} e^{x^2} + C\text{.}
\end{align*}
However, for the similar indefinite integral
\begin{equation*}
\int e^{x^2} \, dx\text{,}
\end{equation*}
the \(u\)-substitution \(u = x^2\) is no longer possible because the factor of \(x\) is missing. Hence, part of the lesson of \(u\)-substitution is just how specialized the process is: it only applies to situations where, up to a missing constant, the integrand is the result of applying the Chain Rule to a different, related function.
Activity 5.3.3.
Evaluate each of the following indefinite integrals by using these steps:
Find two functions within the integrand that form (up to a possible missing constant) a function-derivative pair;
Make a substitution and convert the integral to one involving \(u\) and \(du\text{;}\)
Evaluate the new integral in \(u\text{;}\)
Convert the resulting function of \(u\) back to a function of \(x\) by using your earlier substitution;
Check your work by differentiating the function of \(x\text{.}\) You should come up with the integrand originally given.
\(\displaystyle \int \frac{x^2}{5x^3+1} \, dx\)
\(\displaystyle \int e^x \sin(e^x) \, dx\)
\(\displaystyle \int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx\)
Subsection 5.3.3 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
\begin{equation*}
\int_2^5 xe^{x^2} \, dx\text{.}
\end{equation*}
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
\begin{equation*}
\int_2^5 xe^{x^2} \, dx = \int_{x=2}^{x=5} xe^{x^2} \, dx\text{.}
\end{equation*}
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\\
&= \left. \frac{1}{2}e^u \right|_{u=4}^{u=25}\\
&= \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}\\
&= \frac{1}{2}e^{25} - \frac{1}{2}e^4\text{,}
\end{align*}
which is, of course, the same result.
Activity 5.3.4.
Evaluate each of the following definite integrals exactly through an appropriate \(u\)-substitution.
\(\displaystyle \int_1^2 \frac{x}{1 + 4x^2} \, dx\)
\(\displaystyle \int_0^1 e^{-x} (2e^{-x}+3)^{9} \, dx\)
\(\displaystyle \int_{2/\pi}^{4/\pi} \frac{\cos\left(\frac{1}{x}\right)}{x^{2}} \,dx\)
