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,
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,
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]:
We can generalize the procedure in the following Problem-Solving Strategy.
Problem-Solving Strategy: Integration by Substitution
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.
Substitute [latex]u=g(x)[/latex] and [latex]du={g}^{\prime }(x)dx.[/latex] into the integral.
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].
Evaluate the integral in terms of [latex]u[/latex].
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]
Show Solution
The first step is to choose an expression for [latex]u[/latex]. We choose [latex]u=3{x}^{2}+4.[/latex] because then [latex]du=6xdx.,[/latex] and we already have du in the integrand. Write the integral in terms of [latex]u[/latex]:
Remember that du is the derivative of the expression chosen for [latex]u[/latex], regardless of what is inside the integrand. Now we can evaluate the integral with respect to [latex]u[/latex]:
We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for C of 1, we let [latex]y=\frac{1}{5}{(3{x}^{2}+4)}^{5}+1.[/latex] We have
This is exactly the expression we started with inside the integrand.
Watch the following video to see the worked solution to Example: Evaluating an Indefinite Integral Using Substitution.
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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]
Show Solution
Rewrite the integral as [latex]\displaystyle\int z{({z}^{2}-5)}^{1\text{/}2}dz.[/latex] Let [latex]u={z}^{2}-5[/latex] and [latex]du=2zdz.[/latex] Now we have a problem because [latex]du=2zdz[/latex] and the original expression has only [latex]zdz.[/latex] We have to alter our expression for du or the integral in [latex]u[/latex] will be twice as large as it should be. If we multiply both sides of the du equation by [latex]\frac{1}{2}.[/latex] we can solve this problem. Thus,
Watch the following video to see the worked solution to Example: Using Substitution with Alteration.
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Use substitution to find the antiderivative of [latex]\displaystyle\int {x}^{2}{({x}^{3}+5)}^{9}dx.[/latex]
Hint
Multiply the du equation by [latex]\frac{1}{3}.[/latex]
Show Solution
[latex]\frac{{({x}^{3}+5)}^{10}}{30}+C[/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]
Show Solution
We know the derivative of [latex] \cos t[/latex] is [latex]\text{−} \sin t,[/latex] so we set [latex]u= \cos t.[/latex] Then [latex]du=\text{−} \sin tdt.[/latex] Substituting into the integral, we have
Use substitution to evaluate the integral [latex]\displaystyle\int \frac{ \cos t}{{ \sin }^{2}t}dt.[/latex]
Show Solution
[latex]-\dfrac{1}{ \sin t}+C[/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]
Show Solution
If we let [latex]u=x-1,[/latex] then [latex]du=dx.[/latex] But this does not account for the [latex]x[/latex] in the numerator of the integrand. We need to express [latex]x[/latex] in terms of [latex]u[/latex]. If [latex]u=x-1,[/latex] then [latex]x=u+1.[/latex] Now we can rewrite the integral in terms of [latex]u[/latex]:
Let [latex]u={x}^{3}-3.[/latex]
[latex]\displaystyle\int 3{x}^{2}{({x}^{3}-3)}^{2}dx=\frac{1}{3}{({x}^{3}-3)}^{3}+C[/latex]