Evaluate integral by making the given suvstitution

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

[latex]f\left[g(x)\right]^<\prime >(x)=<(^-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]^<\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 <\displaystyle\int f\left[g(x)\right]^<\prime >(x)dx>\hfill & = <\displaystyle\int f(u)du>\hfill \\ & =F(u)+C\hfill \\ & =F(g(x))+C.\hfill \end[/latex]

Proof

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

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

[latex]\displaystyle\int f\left[g(x)\right]^<\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 <\displaystyle\int f\left[g(x)\right]^<\prime >(x)dx>\hfill & = <\displaystyle\int f(u)du>\hfill \\ & =F(u)+C\hfill \\ & =F(g(x))+C.\hfill \end[/latex]

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

[latex]<\displaystyle\int \underset<\underbrace<(^-3)>>>^\underset<\underbrace<(2xdx)>>=\displaystyle\int ^du.[/latex]

Using the power rule for integrals, we have

[latex]\displaystyle\int ^du=\frac^>+C[/latex]

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

[latex]\dfrac<^>+C=\dfrac<<(^-3)>^>+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]^<\prime >(x)[/latex] is also part of the integrand.
  2. Substitute [latex]u=g(x)[/latex] and [latex]du=^<\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^+4)>^dx.[/latex]

Show Solution

The first step is to choose an expression for [latex]u[/latex]. We choose [latex]u=3^+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]:

[latex]\displaystyle\int 6x<(3^+4)>^dx=\displaystyle\int ^du.[/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]:

[latex]\begin <\displaystyle\int ^du>\hfill & =\frac^>+C\hfill \\ \\ \\ & =\frac<<(3^+4)>^>+C.\hfill \end[/latex]

Analysis

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<(3^+4)>^+1.[/latex] We have

[latex]y=\frac<1><(3^+4)>^+1,[/latex] [latex]\begin<>\\ \hfill ^<\prime >& =(\frac)5<(3^+4)>^6x\hfill \\ & =6x<(3^+4)>^.\hfill \end[/latex]

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.

Closed Captioning and Transcript Information for Video

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

Try It

Use substitution to find the antiderivative of [latex]\displaystyle\int 3^<(^-3)>^dx.[/latex]

Show Solution

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]^-5>[/latex], [latex]g'(x)=[/latex]. Although [latex]g'(x)=[/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^-5>dz.[/latex]

Show Solution

Rewrite the integral as [latex]\displaystyle\int z<(^-5)>^2>dz.[/latex] Let [latex]u=^-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.[/latex] we can solve this problem. Thus,

[latex]\begin<>\\ \hfill u& =^-5\hfill \\ \hfill du& =2zdz\hfill \\ \hfill \fracdu& =\frac(2z)dz=zdz.\hfill \end[/latex]

Write the integral in terms of [latex]u[/latex], but pull the [latex]\frac[/latex] outside the integration symbol:

[latex]\displaystyle\int z<(^-5)>^2>dz=\frac\displaystyle\int ^2>du.[/latex]

Integrate the expression in [latex]u[/latex]:

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

Closed Captioning and Transcript Information for Video

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

Try It

Use substitution to find the antiderivative of [latex]\displaystyle\int ^<(^+5)>^dx.[/latex]

Multiply the du equation by [latex]\frac.[/latex]