Substitution for Indefinite Integrals

Learning Outcomes

  • Recognize the conditions under which substitution may be used to evaluate integrals
  • Use substitution to evaluate 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]

 

We can generalize the procedure in the following Problem-Solving Strategy.

Problem-Solving Strategy: Integration 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]

Example: Evaluating an inDefinite Integral Using Substitution

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

Watch the following video to see the worked solution to Example: Evaluating an Indefinite Integral Using Substitution.

Try It

Use substitution to find the antiderivative of [latex]\displaystyle\int 3{x}^{2}{({x}^{3}-3)}^{2}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.

Tip: 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! Watch how:

Example: Using Substitution with Alteration

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

Watch the following video to see the worked solution to Example: Using Substitution with Alteration.

Try It

Use substitution to find the antiderivative of [latex]\displaystyle\int {x}^{2}{({x}^{3}+5)}^{9}dx.[/latex]

Example: Using Substitution with Integrals of Trigonometric Functions

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

Try It

Use substitution to evaluate the integral [latex]\displaystyle\int \frac{ \cos t}{{ \sin }^{2}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.

Example: Finding an Antiderivative Using [latex]u[/latex]-Substitution

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

Try It

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

Try It