Integration using Substitution: Learn It 1

  • Identify when to use substitution to simplify and solve integrals
  • Apply substitution methods to find indefinite integrals
  • Apply substitution methods to find definite integrals

The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.

Substitution for Indefinite Integrals

At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—that is, the integrand shows you what to do; it is a matter of recognizing the form of the function.

So, what are we supposed to see? We are looking for an integrand of the form [latex]f\left[g(x)\right]{g}^{\prime }(x)dx.[/latex]

For example, in the integral [latex]\displaystyle\int {({x}^{2}-3)}^{3}2xdx,[/latex] we have [latex]f(x)={x}^{3},g(x)={x}^{2}-3,[/latex] and [latex]g\text{‘}(x)=2x.[/latex] Then,

[latex]f\left[g(x)\right]{g}^{\prime }(x)={({x}^{2}-3)}^{3}(2x),[/latex]

and we see that our integrand is in the correct form.

The method is called substitution because we substitute part of the integrand with the variable [latex]u[/latex] and part of the integrand with du. It is also referred to as change of variables because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.

substitution with indefinite integrals

Let [latex]u=g(x),,[/latex] where [latex]{g}^{\prime }(x)[/latex] is continuous over an interval, let [latex]f(x)[/latex] be continuous over the corresponding range of [latex]g[/latex], and let [latex]F(x)[/latex] be an antiderivative of [latex]f(x).[/latex] Then,

[latex]\begin{array}{cc} {\displaystyle\int f\left[g(x)\right]{g}^{\prime }(x)dx}\hfill & = {\displaystyle\int f(u)du}\hfill \\ & =F(u)+C\hfill \\ & =F(g(x))+C.\hfill \end{array}[/latex]

Proof

Let [latex]f[/latex], [latex]g[/latex], [latex]u[/latex], and F be as specified in the theorem. Then

[latex]\begin{array}{cc}\frac{d}{dx}F(g(x))\hfill & ={F}^{\prime }(g(x)){g}^{\prime }(x)\hfill \\ & =f\left[g(x)\right]{g}^{\prime }(x).\hfill \end{array}[/latex]

 

Integrating both sides with respect to [latex]x[/latex], we see that

[latex]\displaystyle\int f\left[g(x)\right]{g}^{\prime }(x)dx=F(g(x))+C.[/latex]

 

If we now substitute [latex]u=g(x),[/latex] and [latex]du=g\text{‘}(x)dx,[/latex] we get

[latex]\begin{array}{cc} {\displaystyle\int f\left[g(x)\right]{g}^{\prime }(x)dx}\hfill & = {\displaystyle\int f(u)du}\hfill \\ & =F(u)+C\hfill \\ & =F(g(x))+C.\hfill \end{array}[/latex]

[latex]_\blacksquare[/latex]

Returning to the problem we looked at originally, we let [latex]u={x}^{2}-3[/latex] and then [latex]du=2xdx.[/latex] Rewrite the integral in terms of [latex]u[/latex]:

[latex]{\displaystyle\int \underset{u}{\underbrace{({x}^{2}-3)}}}^{3}\underset{du}{\underbrace{(2xdx)}}=\displaystyle\int {u}^{3}du.[/latex]

Using the power rule for integrals, we have

[latex]\displaystyle\int {u}^{3}du=\frac{{u}^{4}}{4}+C[/latex]

 

Substitute the original expression for [latex]x[/latex] back into the solution:

[latex]\dfrac{{u}^{4}}{4}+C=\dfrac{{({x}^{2}-3)}^{4}}{4}+C[/latex]

How To: Integrate by Substitution

  1. Look carefully at the integrand and select an expression [latex]g(x)[/latex] within the integrand to set equal to [latex]u[/latex]. Let’s select [latex]g(x).[/latex] such that [latex]{g}^{\prime }(x)[/latex] is also part of the integrand.
  2. Substitute [latex]u=g(x)[/latex] and [latex]du={g}^{\prime }(x)dx[/latex] into the integral.
  3. We should now be able to evaluate the integral with respect to [latex]u[/latex]. If the integral can’t be evaluated we need to go back and select a different expression to use as [latex]u[/latex].
  4. Evaluate the integral in terms of [latex]u[/latex].
  5. Write the result in terms of [latex]x[/latex] and the expression [latex]g(x).[/latex]

Use substitution to find the antiderivative of [latex]\displaystyle\int 6x{(3{x}^{2}+4)}^{4}dx.[/latex]

Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.

As long as you select a [latex]g(x)[/latex] for [latex]u[/latex] such that a multiple of [latex]g'(x)[/latex] exists in the integrand, it will work! In other words, make sure the exponents work – don’t worry about the constants.

For instance, in the example below, if we select [latex]{u={z}^{2}-5}[/latex], [latex]g'(x)={2z}[/latex]. Although [latex]g'(x)={2z}[/latex] doesn’t appear in the integrand, [latex]z[/latex] does. Substitution can work here! 

Use substitution to find the antiderivative of [latex]\displaystyle\int z\sqrt{{z}^{2}-5}dz.[/latex]

Use substitution to evaluate the integral [latex]\displaystyle\int \frac{ \sin t}{{ \cos }^{3}t}dt.[/latex]

Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, [latex]u[/latex] should be the only variable in the integrand.

In some cases, this means solving for the original variable in terms of [latex]u[/latex]. This technique should become clear in the next example.

Use substitution to find the antiderivative of [latex]\displaystyle\int \frac{x}{\sqrt{x-1}}dx.[/latex]