3.4 Integration by Substitution
3.4 Integration by Substitution
We don’t have many integration rules. For quite a few of the problems we see, the rules won’t directly apply; we’ll have to do some algebraic manipulation first. In practice, it is much harder to write down the antiderivative of a function than it is to find a derivative. (In fact, it’s very easy to write a function that doesn’t have any antiderivative you can find with algebra, although proving that it doesn’t have an antiderivative is much more difficult.)
The Substitution Method (also sometimes called [latex]u[/latex]-Substitution) is one way of algebraically manipulating an integrand so that the rules apply. This is a way to unwind or undo the Chain Rule for derivatives. When you find the derivative of a function using the Chain Rule, you end up with a product of something like the original function times a derivative. We can reverse this to write an integral:
[latex]\frac{d}{dx} f\left( g(x) \right) = f'\left( g(x) \right)g'(x)[/latex]
so
[latex]f\left( g(x) \right) =\int f'\left( g(x) \right)g'(x)\, dx[/latex]
With substitution, we will substitute [latex]u=g(x)[/latex] (hence the name [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 the substitution method when you see a composition of functions in your integral, usually multiplied by another function, and especially if you recognize the latter as the derivative of the inner function of the composition.
Video Demonstration
u-Substitution
© 2014 Eric Bancroft
Integration by 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 an inner function of some part of the integrand. A good first choice is one step inside the messiest bit.
- Identify the other factors of the integrand that involve the original variable [latex]x[/latex] – you will have to replace them with factors that involve [latex]u[/latex] only, i.e., [latex]x[/latex] will have to disappear.
- Compute [latex]\frac{du}{dx}[/latex] and use the equation to isolate the factors with [latex]x[/latex] – hopefully they are those you identified in step 2.
- 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] (or a different integration method).
- 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 the [latex]x[/latex] expression for [latex]u[/latex]‘s everywhere in your answer.
Example 1
Evaluate [latex]\displaystyle \int \frac{x}{\sqrt{4-x^2}}\, dx[/latex].
Answer: [latex]\displaystyle \int \frac{x}{\sqrt{4-x^2}}\, dx=?[/latex]
Note that we can rewrite
[latex]\displaystyle \int \frac{x}{\sqrt{4-x^2}}\, dx=\displaystyle \int x\left(4-x^2\right)^{-1/2}\, dx[/latex]
We can see that there is a composition of functions [latex]\left(4-x^2\right)^{-1/2}[/latex] and the remaining part involving [latex]x[/latex] is [latex]x\,dx[/latex]. Let’s try substitution by setting the inner function of the composition to be some new variable, say [latex]u[/latex]. Then we have:
[latex]u=4-x^2\Rightarrow \frac{du}{dx}=-2x\Rightarrow -\frac{1}{2}du=x\,dx[/latex]
Now 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 x\left(4-x^2\right)^{-1/2}\, dx\\ &=\int\left(4-x^2\right)^{-1/2}x\, dx \\ & = \int u^{-1/2}\left(-\frac{1}{2}du\right) \\ & = -\frac{1}{2}\int u^{-1/2}\, du \\ & = -\frac{1}{2}\frac{1}{-\frac{1}{2}+1}u^{-1/2+1}+C\\ &=-u^{1/2}+C\\ &=-\left(4-x^2\right)^{1/2}+C\\ &=-\sqrt{4-x^2}+C \end{align*}[/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]
Video Demonstration
Examples
© 2014 Eric Bancroft
Example 2
Evaluate [latex]\displaystyle \int\frac{e^x\, dx}{\left(e^x+15\right)^3}[/latex].
Answer: [latex]\displaystyle \int\frac{e^x\, dx}{\left(e^x+15\right)^3}=?[/latex]
Note that we can rewrite
[latex]\displaystyle \int \frac{e^x\, dx}{\left(e^x+15\right)^3}=\displaystyle \int e^x\left(e^x+15\right)^{-3}\, dx[/latex]
We can see that there is a composition of functions [latex]\left(e^x+15\right)^3[/latex] and the remaining part involving [latex]x[/latex] is [latex]e^x\,dx[/latex]. Let’s try substitution by setting the inner function of the composition to be some new variable, say [latex]u[/latex]. Then we have:
[latex]u=e^x+15\Rightarrow \frac{du}{dx}=e^x\Rightarrow du=e^x\,dx[/latex]
Now 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{e^x\, dx}{\left(e^x+15\right)^3}& = \int e^x\left(e^x+15\right)^{-3}\, dx\\ &=\int\left(e^x+15\right)^{-3}e^x\, dx \\ & = \int u^{-3}\,du \\ & = \frac{1}{-3+1}u^{-3+1}+C\\ &=-\frac{1}{2}u^{-2}+C\\ &=-\frac{1}{2}\left(e^x+15\right)^{-2}+C\\ &=-\frac{1}{2\left(e^x+15\right)^{2}}+C \end{align*}[/latex]
Example 3
Evaluate
a. [latex]\displaystyle \int\frac{x^2}{x^3+5}\, dx[/latex]
b. [latex]\displaystyle \int\frac{x^3+5}{x^2}\, dx[/latex]
Answer:
a. [latex]\displaystyle \int\frac{x^2}{x^3+5}\, dx=?[/latex]
Note that we can rewrite
[latex]\displaystyle \int\frac{x^2}{x^3+5}\, dx=\displaystyle \int x^2\left(x^3+5\right)^{-1}\, dx[/latex]
We can see that there is a composition of functions [latex]\left(x^3+5\right)^{-1}[/latex] and the remaining part involving [latex]x[/latex] is [latex]x^2\,dx[/latex]. Let’s try substitution by setting the inner function of the composition to be some new variable, say [latex]u[/latex]. Then we have:
[latex]u=x^3+5\Rightarrow \frac{du}{dx}=3x^2\Rightarrow \frac{1}{3}du=x^2\,dx[/latex]
Now 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^2}{x^3+5}\, dx& =\int x^2\left(x^3+5\right)^{-1}\, dx\\ &=\int\left(x^3+5\right)^{-1}x^2\, dx \\ & = \int u^{-1}\left(\frac{1}{3}\,du\right) \\ &=\frac{1}{3} \int u^{-1}\,du\\ & =\frac{1}{3}\ln |u|+C\\ &=\frac{1}{3}\ln\left|x^3+5\right|+C \end{align*}[/latex]
b. [latex]\displaystyle \int\frac{x^3+5}{x^2}\, dx=?[/latex]
It is tempting to start this problem the same way we did the last, but note that there is no composition of functions in this integrand. 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\\ &=\frac{1}{1+1}x^{1+1}+\frac{5}{-2+1}x^{-2+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]
Answer: [latex]\int\limits_0^1 (3x-1)^4\, dx=?[/latex]
We can see that there is a composition of functions [latex](3x-1)^4[/latex] and the remaining part involving [latex]x[/latex] is [latex]dx[/latex]. Let’s try substitution by setting the inner function of the composition to be some new variable, say [latex]u[/latex]. Then we have:
[latex]u=3x-1\Rightarrow \frac{du}{dx}=3\Rightarrow \frac{1}{3}du=dx[/latex]
[latex]@\ x=0, u=3(0)-1=-1[/latex]
[latex]@\ x=1, u=3(1)-1=2[/latex]
Now translate all your [latex]x[/latex]‘s into [latex]u[/latex]‘s everywhere in the integral, including the [latex]dx[/latex] and the change of integral boundaries:
[latex]\begin{align*} \int\limits_0^1 (3x-1)^4\, dx& = \int\limits_{-1}^2 u^{4}\left(\frac{1}{3}\,du\right)\\ &=\frac{1}{3} \int\limits_{-1}^2 u^{4}\,du\\ & =\left.\frac{1}{3}\frac{1}{1+4}u^{1+4}\right|_{-1}^2\\ &=\left.\frac{1}{15}u^5\right|_{-1}^2\\ &=\frac{1}{15}\left(2^5\right)-\frac{1}{15}\left(-1\right)^5\\ & = \frac{32}{15}+\frac{1}{15} \\ & = \frac{33}{15} \end{align*}[/latex]
Video Demonstration
Definite Integrals and u-Substitution
© 2014 Eric Bancroft
Example 5
Evaluate [latex]\int\limits_2^{10} \frac{\left(\ln(x)\right)^6}{x}\, dx[/latex].
Answer: [latex]\int\limits_2^{10} \frac{\left(\ln(x)\right)^6}{x}\, dx=?[/latex]
Note that we can rewrite
[latex]\int\limits_2^{10} \frac{\left(\ln(x)\right)^6}{x}\, dx=\displaystyle \int \left(\ln(x)\right)^6x^{-1}\, dx[/latex]
We can see that there is a composition of functions [latex]\left(\ln(x)\right)^6[/latex] and the remaining part involving [latex]x[/latex] is [latex]x^{-1}\, dx[/latex]. Let’s try substitution by setting the inner function of the composition to be some new variable, say [latex]u[/latex]. Then we have:
[latex]u=\ln(x)\Rightarrow \frac{du}{dx}=\frac{1}{x}\Rightarrow du=x^{-1}dx[/latex]
[latex]@\ x=2, u=\ln(2)[/latex]
[latex]@\ x=10, u=\ln(10)[/latex]
Now translate all your [latex]x[/latex]‘s into [latex]u[/latex]‘s everywhere in the integral, including the [latex]dx[/latex] and the change of integral boundaries:
[latex]\begin{align*} \int\limits_2^{10}\frac{\left(\ln(x)\right)^6}{x}\, dx& = \int\limits_2^{10} \left(\ln(x)\right)^6x^{-1}\, dx\\ &=\int\limits_{\ln(2)}^{\ln(10)} u^{6}\,du\\ & =\left.\frac{1}{1+6}u^{1+6}\right|_{\ln(2)}^{\ln(10)}\\ & =\left.\frac{1}{7}u^{7}\right|_{\ln(2)}^{\ln(10)}\\ &=\frac{1}{7}\left(\ln(2)\right)^7-\frac{1}{7}\left(\ln(10)\right)^7\\ & \approx 49.01 \end{align*}[/latex]
