# 5.3: Integration by Substitution

- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- How can we begin to find algebraic formulas for antiderivatives of more complicated algebraic functions?
- What is an indefinite integral and how is its notation used in discussing antiderivatives?
- How does the technique of \(u\)-substitution work to help us evaluate certain indefinite integrals, and how does this process rely on identifying function-derivative pairs?

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

\[ \int_a^b f(x) \, dx = F(b) - F(a)\text{.} \nonumber \]

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

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

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.

In Section 2.5, we learned the Chain Rule and how it can be applied to find the derivative of a composite function. In particular, if \(u\) is a differentiable function of \(x\text{,}\) and \(f\) is a differentiable function of \(u(x)\text{,}\) then

In words, we say that the derivative of a composite function \(c(x) = f(u(x))\text{,}\) where \(f\) is considered the “outer” function and \(u\) the “inner” function, is “the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function.”

- For each of the following functions, use the Chain Rule to find the function's derivative. Be sure to label each derivative by name (e.g., the derivative of \(g(x)\) should be labeled \(g'(x)\)).
- \(\displaystyle g(x) = e^{3x}\)
- \(\displaystyle h(x) = \sin(5x+1)\)
- \(\displaystyle p(x) = \arctan(2x)\)
- \(\displaystyle q(x) = (2-7x)^4\)
- \(\displaystyle r(x) = 3^{4-11x}\)

- For each of the following functions, use your work in (a) to help you determine the general antiderivative
^{ 1 }of the function. Label each antiderivative by name (e.g., the antiderivative of \(m\) should be called \(M\)). In addition, check your work by computing the derivative of each proposed antiderivative.- \(\displaystyle m(x) = e^{3x}\)
- \(\displaystyle n(x) = \cos(5x+1)\)
- \(\displaystyle s(x) = \frac{1}{1+4x^2}\)
- \(\displaystyle v(x) = (2-7x)^3\)
- \(\displaystyle w(x) = 3^{4-11x}\)

- Based on your experience in parts (a) and (b), conjecture an antiderivative for each of the following functions. Test your conjectures by computing the derivative of each proposed antiderivative.
- \(\displaystyle a(x) = \cos(\pi x)\)
- \(\displaystyle b(x) = (4x+7)^{11}\)
- \(\displaystyle c(x) = xe^{x^2}\)

Recall that the general antiderivative of a function includes “\(+C\)” to reflect the entire family of functions that share the same derivative.

## 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

Determine the general antiderivative of

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

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

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

\[ H(x) = \frac{1}{a} F(ax+b) + C\text{.} \nonumber \]

It is useful to have shorthand notation that indicates the instruction to find an antiderivative. Thus, in a similar way to how the notation

represents the derivative of \(f(x)\) with respect to \(x\text{,}\) we use the notation of the *indefinite integral*,

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

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 \(\Box\text{.}\)”

Evaluate each of the following indefinite integrals. Check each antiderivative that you find by differentiating.

- \(\displaystyle \int \sin(8-3x) \, dx\)
- \(\displaystyle \int \sec^2 (4x) \, dx\)
- \(\displaystyle \int \frac{1}{11x - 9} \, dx\)
- \(\displaystyle \int \csc(2x+1) \cot(2x+1) \, dx\)
- \(\displaystyle \int \frac{1}{\sqrt{1-16x^2}}\, dx\)
- \(\displaystyle \int 5^{-x}\, dx\)

## 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

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

we can equivalently write

Recall that the Chain Rule states that

Restating this relationship in terms of an indefinite integral,

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^{ 2 }, \(du = g'(x)\, dx\text{.}\) Now converting the indefinite integral to a new one in terms of \(u\text{,}\) we have

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 \(u = g(x)\text{,}\)

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

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

Evaluate the indefinite integral

and check the result by differentiating.

**Answer**-
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

\[ \int \sin (7x^4 + 3) \cdot x^3 \, dx\text{,} \nonumber \]and so by substituting \(u\) for \(7x^4 + 3\) and \(\frac{1}{28} \, du\) for \(x^3 \, dx\text{,}\) it follows that

\[ \int \sin (7x^4 + 3) \cdot x^3 \, dx = \int \sin(u) \cdot \frac{1}{28} \, du\text{.} \nonumber \]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\\[4pt] &= \frac{1}{28} \int \sin(u) \, du\\[4pt] &= \frac{1}{28} (-\cos(u)) + C\\[4pt] &= -\frac{1}{28} \cos(7x^4 + 3) + C\text{.} \end{align*}To check our work, we observe by the Chain Rule that

\[ \frac{d}{dx} \left[ -\frac{1}{28}\cos(7x^4 + 3) \right] = -\frac{1}{28} \cdot (-1)\sin(7x^4 + 3) \cdot 28x^3 = \sin(7x^4 + 3) \cdot x^3\text{,} \nonumber \]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

However, for the similar indefinite integral

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.

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.

- \(\displaystyle \int \frac{x^2}{5x^3+1} \, dx\)
- \(\displaystyle \int e^x \sin(e^x) \, dx\)
- \(\displaystyle \int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx\)

## 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

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

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

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

which is, of course, the same result.

Evaluate each of the following definite integrals exactly through an appropriate \(u\)-substitution.

- \(\displaystyle \int_1^2 \frac{x}{1 + 4x^2} \, dx\)
- \(\displaystyle \int_0^1 e^{-x} (2e^{-x}+3)^{9} \, dx\)
- \(\displaystyle \int_{2/\pi}^{4/\pi} \frac{\cos\left(\frac{1}{x}\right)}{x^{2}} \,dx\)

## Summary

- To find algebraic formulas for antiderivatives of more complicated algebraic functions, we need to think carefully about how we can reverse known differentiation rules. To that end, it is essential that we understand and recall known derivatives of basic functions, as well as the standard derivative rules.
- The indefinite integral provides notation for antiderivatives. When we write “\(\int f(x) \, dx\text{,}\)” we mean “the general antiderivative of \(f\text{.}\)” In particular, if we have functions \(f\) and \(F\) such that \(F' = f\text{,}\) the following two statements say the exact thing:
\[ \frac{d}{dx}[F(x)] = f(x) \ \text{and} \ \int f(x) \, dx = F(x) + C\text{.} \nonumber \]
That is, \(f\) is the derivative of \(F\text{,}\) and \(F\) is an antiderivative of \(f\text{.}\)

- The technique of \(u\)-substitution helps us to evaluate indefinite integrals of the form \(\int f(g(x))g'(x) \, dx\) through the substitutions \(u = g(x)\) and \(du = g'(x) \, dx\text{,}\) so that
\[ \int f(g(x))g'(x) \, dx = \int f(u) \, du\text{.} \nonumber \]
A key part of choosing the expression in \(x\) to be represented by \(u\) is the identification of a function-derivative pair. To do so, we often look for an “inner” function \(g(x)\) that is part of a composite function, while investigating whether \(g'(x)\) (or a constant multiple of \(g'(x)\)) is present as a multiplying factor of the integrand.