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We motivate this section with an example. Let \(f(x) = (x^2+3x-5)^{10}\text{.}\) We can compute \(\fp(x)\) using the Chain Rule. It is: \begin{align*} \fp(x) \amp = 10(x^2+3x-5)^9\cdot(2x+3)\\ \amp = (20x+30)(x^2+3x-5)^9\text{.} \end{align*}

Now consider this: What is \(\int (20x+30)(x^2+3x-5)^9\ dx\text{?}\) We have the answer in front of us; \begin{equation*} \int (20x+30)(x^2+3x-5)^9\ dx = (x^2+3x-5)^{10}+C. \end{equation*}

How would we have evaluated this indefinite integral without starting with \(f(x)\) as we did?

This section explores integration by substitution. It allows us to “undo the Chain Rule.” Substitution allows us to evaluate the above integral without knowing the original function first.

The underlying principle is to rewrite a “complicated” integral of the form \(\int f(x)\ dx\) as a not–so–complicated integral \(\int h(u)\ du\text{.}\) We'll formally establish later how this is done. First, consider again our introductory indefinite integral, \(\int (20x+30)(x^2+3x-5)^9\ dx\text{.}\) Arguably the most “complicated” part of the integrand is \((x^2+3x-5)^9\text{.}\) We wish to make this simpler; we do so through a substitution. Let \(u=x^2+3x-5\text{.}\) Thus \begin{equation*} (x^2+3x-5)^9 = u^9. \end{equation*}

We have established \(u\) as a function of \(x\text{,}\) so now consider the differential of \(u\text{:}\) \begin{equation*} du = (2x+3)dx. \end{equation*}

Keep in mind that \((2x+3)\) and \(dx\) are multiplied; the \(dx\) is not “just sitting there.”

Return to the original integral and do some substitutions through algebra: \begin{align*} \int (20x+30)(x^2+3x-5)^9\ dx \amp = \int 10(2x+3)(x^2+3x-5)^9\ dx\\ \amp =\int 10(\underbrace{x^2+3x-5}_u)^9\underbrace{(2x+3)\ dx}_{du}\\ \amp =\int 10u^9\ du\\ \amp = u^{10} + C \quad \text{(replace \(u\) with \(x^2+3x-5\)) }\\ \amp = (x^2+3x-5)^{10} +C \end{align*}

One might well look at this and think “I (sort of) followed how that worked, but I could never come up with that on my own,” but the process is learnable. This section contains numerous examples through which the reader will gain understanding and mathematical maturity enabling them to regard substitution as a natural tool when evaluating integrals.

We stated before that integration by substitution “undoes” the Chain Rule. Specifically, let \(F(x)\) and \(g(x)\) be differentiable functions and consider the derivative of their composition: \begin{equation*} \frac{d}{dx}\Big(F\big(g(x)\big)\Big) = \Fp(g(x))g'(x). \end{equation*}

Thus \begin{equation*} \int \Fp(g(x))g'(x)\ dx = F(g(x)) + C. \end{equation*}

Integration by substitution works by recognizing the “inside” function \(g(x)\) and replacing it with a variable. By setting \(u=g(x)\text{,}\) we can rewrite the derivative as \begin{equation*} \frac{d}{dx}\Big(F\big(u\big)\Big) = \Fp(u)u'. \end{equation*}

Since \(du = g'(x)dx\text{,}\) we can rewrite the above integral as \begin{equation*} \int \Fp(g(x))g'(x)\ dx = \int \Fp(u) du = F(u)+C = F(g(x))+ C. \end{equation*}

This concept is important so we restate it in the context of a theorem.

The point of substitution is to make the integration step easy. Indeed, the step \(\int \Fp(u)\ du = F(u) + C\) looks easy, as the antiderivative of the derivative of \(F\) is just \(F\text{,}\) plus a constant. The “work” involved is making the proper substitution. There is not a step–by–step process that one can memorize; rather, experience will be one's guide. To gain experience, we now embark on many examples.

Example6.1.2Integrating by substitution

Evaluate \(\ds \int x\sin(x^2+5)\ dx\text{.}\)

Example6.1.3Integrating by substitution

Evaluate \(\ds \int \cos(5x)\ dx\text{.}\)


The previous example exhibited a common, and simple, type of substitution. The “inside” function was a linear function (in this case, \(y = 5x\)). When the inside function is linear, the resulting integration is very predictable, outlined here.

Key Idea6.1.4Substitution With A Linear Function

Consider \(\int \Fp(ax+b)\ dx\text{,}\) where \(a\neq 0\) and \(b\) are constants. Letting \(u = ax+b\) gives \(du = a\cdot dx\text{,}\) leading to the result \begin{equation*} \int \Fp(ax+b)\ dx = \frac{1}{a}F(ax+b) + C. \end{equation*}

Thus \(\int \sin(7x-4)\ dx = -\frac17\cos(7x-4)+C\text{.}\) Our next example can use Key Idea 6.1.4, but we will only employ it after going through all of the steps.

Example6.1.5Integrating by substituting a linear function

Evaluate \(\ds \int \frac{7}{-3x+1}\ dx\text{.}\)


Not all integrals that benefit from substitution have a clear “inside” function. Several of the following examples will demonstrate ways in which this occurs.

Example6.1.6Integrating by substitution

Evaluate \(\ds \int \sin(x) \cos(x) \ dx\text{.}\)


Our examples so far have required “basic substitution.” The next example demonstrates how substitutions can be made that often strike the new learner as being “nonstandard.”

Example6.1.7Integrating by substitution

Evaluate \(\ds\int x\sqrt{x+3}\ dx\text{.}\)

Example6.1.8Integrating by substitution

Evaluate \(\ds \int \frac{1}{x\ln(x) }\ dx\text{.}\)


Subsection6.1.1Integrals Involving Trigonometric Functions

Section 6.3 delves deeper into integrals of a variety of trigonometric functions; here we use substitution to establish a foundation that we will build upon.

The next three examples will help fill in some missing pieces of our antiderivative knowledge. We know the antiderivatives of the sine and cosine functions; what about the other standard functions tangent, cotangent, secant and cosecant? We discover these next.

Example6.1.9Integration by substitution: antiderivatives of \(\tan(x)\)

Evaluate \(\ds \int \tan(x) \ dx\text{.}\)

Example6.1.10Integrating by substitution: antiderivatives of \(\sec(x)\)

Evaluate \(\ds\int \sec(x) \ dx\text{.}\)


We can use similar techniques to those used in Examples 6.1.9 and 6.1.10 to find antiderivatives of \(\cot(x)\) and \(\csc(x)\) (which the reader can explore in the exercises.) We summarize our results here.

We explore one more common trigonometric integral.

Example6.1.12Integration by substitution: powers of \(\cos(x)\) and \(\sin(x)\)

Evaluate \(\ds \int \cos^2(x) \ dx\text{.}\)


Subsection6.1.2Simplifying the Integrand

It is common to be reluctant to manipulate the integrand of an integral; at first, our grasp of integration is tenuous and one may think that working with the integrand will improperly change the results. Integration by substitution works using a different logic: as long as equality is maintained, the integrand can be manipulated so that its form is easier to deal with. The next two examples demonstrate common ways in which using algebra first makes the integration easier to perform.

Example6.1.13Integration by substitution: simplifying first

Evaluate \(\ds\int \frac{x^3+4x^2+8x+5}{x^2+2x+1}\ dx\text{.}\)

Example6.1.14Integration by alternate methods

Evaluate \(\ds\int \frac{x^2+2x+3}{\sqrt{x}}\ dx\) with, and without, substitution.


Subsection6.1.3Substitution and Inverse Trigonometric Functions

When studying derivatives of inverse functions, we learned that \begin{equation*} \frac{d}{dx}\big(\tan^{-1}(x) \big) = \frac{1}{1+x^2}. \end{equation*}

Applying the Chain Rule to this is not difficult; for instance, \begin{equation*} \frac{d}{dx}\big(\tan^{-1}(5x) \big) = \frac{5}{1+25x^2}. \end{equation*}

We now explore how Substitution can be used to “undo” certain derivatives that are the result of the Chain Rule applied to Inverse Trigonometric functions. We begin with an example.

Example6.1.15Integrating by substitution: inverse trigonometric functions

Evaluate \(\ds \int \frac{1}{25+x^2}\ dx\text{.}\)


Example 6.1.15 demonstrates a general technique that can be applied to other integrands that result in inverse trigonometric functions. The results are summarized here.

Let's practice using Theorem 6.1.16.

Example6.1.17Integrating by substitution: inverse trigonometric functions

Evaluate the given indefinite integrals. \begin{equation*} \int \frac{1}{9+x^2}\ dx, \int \frac{1}{x\sqrt{x^2-\frac{1}{100}}}\ dx \text{ and } \int \frac{1}{\sqrt{5-x^2}}\ dx. \end{equation*}


Most applications of Theorem 6.1.16 are not as straightforward. The next examples show some common integrals that can still be approached with this theorem.

Example6.1.18Integrating by substitution: completing the square

Evaluate \(\ds \int\frac{1}{x^2-4x+13}\ dx\text{.}\)

Example6.1.19Integrals requiring multiple methods

Evaluate \(\ds \int \frac{4-x}{\sqrt{16-x^2}}\ dx\text{.}\)


Subsection6.1.4Substitution and Definite Integration

This section has focused on evaluating indefinite integrals as we are learning a new technique for finding antiderivatives. However, much of the time integration is used in the context of a definite integral. Definite integrals that require substitution can be calculated using the following workflow:

  1. Start with a definite integral \(\ds \int_a^b f(x)\ dx\) that requires substitution.

  2. Ignore the bounds; use substitution to evaluate \(\ds \int f(x)\ dx\) and find an antiderivative \(F(x)\text{.}\)

  3. Evaluate \(F(x)\) at the bounds; that is, evaluate \(F(x)\Big|_a^b = F(b) - F(a)\text{.}\)

This workflow works fine, but substitution offers an alternative that is powerful and amazing (and a little time saving).

At its heart, (using the notation of Theorem 6.1.1) substitution converts integrals of the form \(\int \Fp(g(x))g'(x)\ dx\) into an integral of the form \(\int \Fp(u)\ du\) with the substitution of \(u = g(x)\text{.}\) The following theorem states how the bounds of a definite integral can be changed as the substitution is performed.

In effect, Theorem 6.1.20 states that once you convert to integrating with respect to \(u\text{,}\) you do not need to switch back to evaluating with respect to \(x\text{.}\) A few examples will help one understand.

Example6.1.21Definite integrals and substitution: changing the bounds

Evaluate \(\ds\int_0^2 \cos(3x-1)\ dx\) using Theorem 6.1.20.

Example6.1.23Definite integrals and substitution: changing the bounds

Evaluate \(\ds \int_0^{\pi/2} \sin(x) \cos(x) \ dx\) using Theorem 6.1.20.


Integration by substitution is a powerful and useful integration technique. The next section introduces another technique, called Integration by Parts. As substitution “undoes” the Chain Rule, integration by parts “undoes” the Product Rule. Together, these two techniques provide a strong foundation on which most other integration techniques are based.


Terms and Concepts

In the following exercises, evaluate the indefinite integral to develop an understanding of Substitution.