Chapter 4: The Integral
Section 4.4: Substitution
Learning Objectives
By the end of this section, the student should be able to:
- Use substitution to find indefinite integrals.
- Use limit of integration substitution for definite integrals.
Substitution and Indefinite Integrals
[latex]u[/latex]-substitution). This means [latex]\frac{du}{dx}=g'(x)[/latex], so [latex]du=g'(x)dx[/latex]. Making these substitutions, [latex]\int f'\left( g(x) \right)g'(x)\, dx[/latex] becomes [latex]\int f'(u)\, du[/latex], which will probably be easier to integrate. Try [latex]u[/latex]-substitution when you see a product in your integral, especially if you recognize one factor as the derivative of some part of the other factor.
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Key Takeaways
The [latex]u[/latex]-Substitution Method for Antiderivatives
The goal is to turn [latex]\int f\left( g(x) \right)\, dx[/latex] into [latex]\int f(u)\, du[/latex], where [latex]f(u)[/latex] is much less messy than [latex]f\left(g(x)\right)[/latex].
- Let [latex]u[/latex] be some part of the integrand. A good first choice is
one step inside the messiest bit.
- Compute [latex]du=\frac{du}{dx}dx[/latex].
- Translate all your [latex]x[/latex]‘s into [latex]u[/latex]‘s everywhere in the integral, including the [latex]dx[/latex]. When you’re done, you should have a new integral that is entirely in [latex]u[/latex]. If you have any [latex]x[/latex]‘s left, then that’s an indication that the substitution didn’t work or isn’t complete; you may need to go back to step 1 and try a different choice for [latex]u[/latex].
- Integrate the new [latex]u[/latex]-integral, if possible. If you still can’t integrate it, go back to step 1 and try a different choice for [latex]u[/latex].
- Finally, substitute back [latex]x[/latex]‘s for [latex]u[/latex]‘s everywhere in your answer.
Example 1
Evaluate [latex]\int \frac{x}{\sqrt{4-x^2}}\, dx[/latex].
This integrand is more complicated than anything in our list of basic integral formulas, so we’ll have to try something else. The only tool we have is substitution, so let’s try that!
- Let [latex]u[/latex] be some part of the integrand. A good first choice is
one step inside the messiest bit
: let [latex]u=4-x^2[/latex]. - Compute [latex]du=\frac{du}{dx}dx[/latex]: [latex]du=-2x\, dx[/latex]. There is [latex]x\, dx[/latex] in the integrand, so that’s a good sign; that will be [latex]-\frac{1}{2}du[/latex].
- Translate all your [latex]x[/latex]‘s into [latex]u[/latex]‘s everywhere in the integral, including the [latex]dx[/latex]: [latex]\begin{align*} \int\frac{x}{\sqrt{4-x^2}}\, dx=& \int\frac{1}{\sqrt{4-x^2}}(x\, dx) \\ =& \int\frac{1}{\sqrt{u}}\left(-\frac{1}{2}du\right) \\ =& -\frac{1}{2}\int\frac{1}{\sqrt{u}}\, du \\ =& -\frac{1}{2}\int u^{-1/2}\, du \end{align*}[/latex] Alternatively, we could have solved for dx and substituted that and simplified: [latex]dx=\frac{du}{-2x}[/latex], so
[latex]\begin{align*} \int\frac{x}{\sqrt{4-x^2}}\, dx=& \int\frac{x}{\sqrt{u}}\left(\frac{du}{-2x}\right) \\ =& \int\frac{1}{\sqrt{u}}\left(-\frac{1}{2}du\right) \\ =& -\frac{1}{2}\int\frac{1}{\sqrt{u}}\, du \\ =& -\frac{1}{2}\int u^{-1/2}\, du \end{align*}[/latex] - Integrate the new [latex]u[/latex]-integral, if possible: [latex]-\frac{1}{2}\int u^{-1/2}\, du = -\frac{1}{2}\frac{u^{1/2}}{1/2}+C=-u^{1/2}+C[/latex]
- Finally, substitute back [latex]x[/latex]‘s for [latex]u[/latex]‘s everywhere in the answer: undoing our [latex]u=4-x^2[/latex] substitution yields [latex]-u^{1/2}+C = -\sqrt{4-x^2}+C.[/latex]
Thus we have found [latex]\int\frac{x}{\sqrt{4-x^2}}\, dx= -\sqrt{4-x^2}+C[/latex]
How would we check this? By differentiating:
[latex]\begin{align*} \frac{d}{dx}\left(-\sqrt{4-x^2}+C\right)=& \frac{d}{dx}\left(-\left(4-x^2\right)^{1/2}+C\right) \\ =& -\frac{1}{2}\left(4-x^2\right)^{-1/2}(-2x) \\ =& x\left(4-x^2\right)^{-1/2} \\ =& \frac{x}{\sqrt{4-x^2}} \end{align*}[/latex]
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Example 2
Evaluate [latex]\int\frac{e^x\, dx}{\left(e^x+15\right)^3}[/latex].
This integral is not in our list of building blocks. But notice that the derivative of [latex]e^x+15 [/latex] (which we see in the denominator) is just [latex]e^x [/latex] (which we see in the numerator), so substitution will be a good choice for this.
Let [latex]u=e^x+15[/latex]. Then [latex]du=e^x\, dx[/latex], and this integral becomes [latex]\int\frac{du}{u^3} = \int u^{-3}\, du[/latex].
Luckily, that is on our list of building block formulas: [latex]\int\frac{du}{u^3} = \frac{u^{-2}}{-2}+C = -\frac{1}{2u^2}+C[/latex].
Finally, translating back: [latex]\int\frac{e^x\, dx}{\left(e^x+15\right)^3} = -\frac{1}{2\left(e^x+15\right)^2} +C.[/latex]
Example 3
Evaluate
- [latex]\int\frac{x^2}{x^3+5}\, dx[/latex]
- [latex]\int\frac{x^3+5}{x^2}\, dx[/latex]
Solutions
- This is not a basic integral, but the composition is less obvious. Here, we can treat the denominator as the inside of the [latex]\frac{1}{x}[/latex] function. Let [latex]u=x^3+5[/latex]. Then [latex]du=3x^2\, dx[/latex]. Solving for [latex]dx[/latex], [latex]dx=\frac{du}{3x^2}[/latex]. Substituting, [latex]\int\frac{x^2}{x^3+5}\, dx = \int\frac{x^2}{u}\frac{du}{3x^2} = \int \frac{1}{u}\frac{du}{3} = \frac{1}{3}\int \frac{1}{u}\, du.[/latex] Using our basic formulas, [latex]\frac{1}{3}\int \frac{1}{u}\, du = \frac{1}{3}\ln|u| +C.[/latex] Undoing the substitution, [latex]\int\frac{x^2}{x^3+5}\, dx = \frac{1}{3}\ln\left|x^3+5\right| +C.[/latex]
- It is tempting to start this problem the same way we did the last, but if we try it will not work, since the numerator of this fraction is not the derivative of the denominator. Instead, we need to try a different approach. For this problem, we can use some basic algebra:
[latex]\begin{align*} \int\frac{x^3+5}{x^2}\, dx =& \int\left(\frac{x^3}{x^2}+\frac{5}{x^2}\right)\, dx \\ =& \int\left(x+5x^{-2}\right)\, dx. \end{align*}[/latex] We can integrate this using our basic rules, without needing substitution:
[latex]\begin{align*} \int\left(x+5x^{-2}\right)\, dx=& \frac{x^2}{2}+5\frac{x^{-1}}{-1}+C \\ =& \frac{1}{2}x^2-\frac{5}{x}+C. \end{align*}[/latex]
Substitution and Definite Integrals
When you use substitution to help evaluate a definite integral, you have a choice for how to handle the limits of integration. You can do either of these, whichever seems better to you. The important thing to remember is that the original limits of integration were values of the original variable (say, [latex]x[/latex]), not values of the new variable (say, [latex]u[/latex]).
- You can solve the antiderivative as a side problem, translating back to [latex]x[/latex]’s, and then use the antiderivative with the original limits of integration. Or…
- You can substitute for the limits of integration at the same time as you’re substituting for everything inside the integral, and then skip the
translate back into [latex]x[/latex]
step. If the original integral had endpoints [latex]x =a[/latex] and [latex]x =b[/latex], and we make the substitution [latex]u = g(x)[/latex] and [latex]du = g'(x )\, dx[/latex], then the new integral will have endpoints [latex]u= g(a)[/latex] and [latex]u=g(b)[/latex] and [latex]\int_{x=a}^{x=b}\text{(original integrand)}\, dx[/latex] becomes [latex]\int_{u=g(a)}^{u=g(b)} \text{(new integrand)}\, du.[/latex]
Method 1 seems more straightforward for most students, but it can involve some messy algebra. Method 2 is often neater and usually involves fewer steps.
Example 4
Evaluate [latex]\int\limits_0^1 (3x-1)^4\, dx[/latex].
We’ll need substitution to find an antiderivative, so we’ll need to handle the limits of integration carefully. Let’s solve this example both ways.
- Doing the antiderivative as a side problem. Step One: Find the antiderivative using substitution. Let [latex]u=3x-1[/latex]. Then [latex]du=3\, dx[/latex] and [latex]\int(3x-1)^4\, dx = \int u^4\left(\frac{1}{3}\, du\right) = \frac{1}{3}\frac{u^5}{5}+C.[/latex] Translating back to x: [latex]\frac{1}{3}\frac{u^5}{5}+C = \frac{(3x-1)^5}{15}+C.[/latex] Step Two: Evaluate the definite integral. [latex]\int\limits_0^1 (3x-1)^4\, dx = \left. \frac{(3x-1)^5}{15}\right]_0^1 = \frac{\left(3(1)-1\right)^5}{15} - \frac{\left(3(0)-1\right)^5}{15}[/latex] [latex]= \frac{32}{15}-\frac{-1}{15}=\frac{33}{15}.[/latex]
- Substituting for the limits of integration. Let [latex]u=3x-1[/latex]. Then [latex]du=3\, dx[/latex] and, substituting for the limits of integration, when [latex]x = 0[/latex], [latex]u = -1[/latex], and when [latex]x = 1[/latex], [latex]u = 2[/latex]. So
[latex]\begin{align*} \int_{x=0}^{x=1} (3x-1)^4\, dx =& \int_{u=-1}^{u=2} u^4\left(\frac{1}{3}\, du\right) \\ =& \left.\frac{u^5}{15}\right]_{u=-1}^{u=2} \\ =& \frac{(2)^5}{15}-\frac{(-1)^5}{15} \\ =& \frac{32}{15}-\frac{-1}{15} \\ =& \frac{33}{15} \end{align*}[/latex]
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Example 5
Evaluate [latex]\int\limits_2^{10} \frac{\left(\ln(x)\right)^6}{x}\, dx[/latex].
I can see the derivative of [latex]\ln(x) [/latex] in the integrand, so I can tell that substitution is a good choice.
Let [latex]u=\ln(x)[/latex]. Then [latex]du=\frac{1}{x}\, dx[/latex]. When [latex]x=2 , u=\ln(2)[/latex]. When [latex]x=10, u=\ln(10)[/latex]. So the new definite integral is
[latex]\begin{align*} \int\limits_{x=2}^{x=10}\frac{\left(\ln(x)\right)^6}{x}\, dx =& \int\limits_{u=\ln(2)}^{u=\ln(10)} u^6\, du \\ =& \left.\frac{u^7}{7}\right]_{u=\ln(2)}^{u=\ln(10)} \\ =& \frac{1}{7}\left(\left(\ln(10)\right)^7-\left(\ln(2)\right)^7\right) \\ \approx & 49.01. \end{align*}[/latex]
