Learning Outcomes
- Evaluate various integrals via the substitution method.
Section 13.1 Substitution Method (TI1)
Subsection 13.1.1 Activities
Activity 13.1.
Answer the following.
(a)
Using the chain rule, which of these is the derivative of \(e^{x^3}\) with respect to \(x\text{?}\)
- \(\displaystyle e^{3x^2}\)
- \(\displaystyle x^3e^{x^3-1}\)
- \(\displaystyle 3x^2e^{x^3}\)
- \(\displaystyle \dfrac{1}{4}e^{x^4}\)
(b)
Based on this result, which of these would you suspect to equal \(\displaystyle \int x^2e^{x^3}\,dx\text{?}\)
- \(\displaystyle e^{x^3+1}+C\)
- \(\displaystyle \dfrac{1}{3x}e^{x^3+1}+C\)
- \(\displaystyle 3e^{x^3}+C\)
- \(\displaystyle \dfrac{1}{3}e^{x^3}+C\)
Activity 13.2.
Recall that if \(u\) is a function of \(x\text{,}\) then \(\dfrac{d}{dx}[u^7]=7u^6 u'\) by the Chain Rule (Theorem 10.87).
For each question, choose from the following.
- \(\displaystyle \dfrac{1}{7}u^7+C\)
- \(\displaystyle u^7+C\)
- \(\displaystyle 7u^7+C\)
- \(\displaystyle \dfrac{6}{7}u^7+C\)
(a)
What is \(\displaystyle \int 7u^6 u'\,dx\text{?}\)
(b)
What is \(\displaystyle \int u^6 u'\,dx\text{?}\)
(c)
What is \(\displaystyle \int 6u^6 u'\,dx\text{?}\)
Activity 13.3.
Based on these activities, which of these choices seems to be a viable strategy for integration?
- Memorize an integration formula for every possible function.
- Attempt to rewrite the integral in the form \(\displaystyle \int g'(u)u'\,dx=g(u)+C\text{.}\)
- Keep differentiating functions until you come across the function you want to integrate.
Fact 13.4.
By the chain rule,
\begin{equation*}
\dfrac{d}{dx}[g(u)+C]=g'(u)u'\text{.}
\end{equation*}
There is a dual integration technique reversing this process, known as the substitution method.
This technique involves choosing an appropriate function \(u\) in terms of \(x\) to rewrite the integral as follows:
\begin{equation*}
\displaystyle \int f(x)\,dx=\dots=\displaystyle \int g'(u)u'\,dx=g(u)+C\text{.}
\end{equation*}
Observation 13.5.
Recall that \(\dfrac{du}{dx}=u'\text{,}\) and so \(du=u'\,dx\text{.}\) This allows for the following common notation:
\begin{equation*}
\displaystyle \int f(x)\,dx=\dots=\displaystyle \int g'(u)\, du=g(u)+C\text{.}
\end{equation*}
Therefore, rather than dealing with equations like \(u'=\dfrac{du}{dx}=x^2\text{,}\) we will prefer to write \(du=x^2\, dx\text{.}\)
Activity 13.6.
Consider \(\displaystyle \int x^2e^{x^3}\,dx\text{,}\) which we conjectured earlier to be \(\dfrac{1}{3}e^{x^3}+C\text{.}\)
Suppose we decided to let \(u=x^3\text{.}\)
(a)
Compute \(\dfrac{du}{dx}=\unknown\text{,}\) and rewrite it as \(du=\unknown\,dx\text{.}\)
(b)
This \(\unknown\,dx\) doesn’t appear in \(\displaystyle \int x^2e^{x^3}\,dx\) exactly, so use algebra to solve for \(x^2\,dx\) in terms of \(du\text{.}\)
(c)
Replace \(x^2\, dx\) and \(x^3\) with \(u\,du\) terms to rewrite \(\displaystyle \int x^2e^{x^3}\,dx\) as \(\displaystyle \int \dfrac{1}{3}e^u\,du\text{.}\)
(d)
Solve \(\displaystyle \int \dfrac{1}{3}e^u\,du\) in terms of \(u\text{,}\) then replace \(u\) with \(x^3\) to confirm our original conjecture.
Example 13.7.
Here is how one might write out the explanation of how to find \(\displaystyle \int x^2e^{x^3}\,dx\) from start to finish:
\begin{align*}
\displaystyle \int x^2e^{x^3}\,dx &&\text{Let }&u=x^3\\
&&& du = 3x^2\,dx\\
&&& \dfrac{1}{3}du = x^2\,dx\\
\displaystyle \int x^2e^{x^3}\,dx &= \displaystyle \int e^{(x^3)} (x^2\,dx)\\
&= \displaystyle \int e^{u} \dfrac{1}{3}\,du\\
&= \dfrac{1}{3}e^{u}+C\\
&= \dfrac{1}{3}e^{x^3}+C
\end{align*}
Activity 13.8.
Which step of the previous example do you think was the most important?
- Choosing \(u=x^3\text{.}\)
- Finding \(du=3x^2\,dx\) and \(\dfrac{1}{3}du=x^2\,dx\text{.}\)
- Substituting \(\displaystyle \int x^2e^{x^3}\,dx\) with \(\displaystyle \int\dfrac{1}{3}e^u\,du\text{.}\)
- Integrating \(\displaystyle \int\dfrac{1}{3}e^u\,du=\dfrac{1}{3}e^u+C\text{.}\)
- Unsubstituting \(\dfrac{1}{3}e^u+C\) to get \(\dfrac{1}{3}e^{x^3}+C\text{.}\)
Activity 13.9.
Suppose we wanted to try the substitution method to find \(\displaystyle \int e^x\cos(e^x+3)\,dx\text{.}\) Which of these choices for \(u\) appears to be most useful?
- \(u=x\text{,}\) so \(du=dx\)
- \(u=e^x\text{,}\) so \(du=e^x\,dx\)
- \(u=e^x+3\text{,}\) so \(du=e^x\,dx\)
- \(u=\cos(x)\text{,}\) so \(du=-\sin(x)\,dx\)
- \(u=\cos(e^x+3)\text{,}\) so \(du=e^x\sin(e^x+3)\,dx\)
Activity 13.10.
Complete the following solution using your choice from the previous activity to find \(\displaystyle \int e^x\cos(e^x+3)\,dx\text{.}\)
\begin{align*}
\displaystyle \int e^x\cos(e^x+3)\,dx &&\text{Let }&u=\unknown\\
&&& du = \unknown\,dx\\
\displaystyle \int e^x\cos(e^x+3)\,dx &= \displaystyle \int \unknown\, du\\
&= \cdots\\
&= \sin(e^x+3)+C
\end{align*}
Activity 13.11.
Complete the following integration by substitution to find \(\displaystyle \int \dfrac{x^3}{x^4+4}\,dx\text{.}\)
\begin{align*}
\displaystyle \int \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=\unknown\\
&&& du = \unknown\,dx\\
&&& \unknown \, du = \unknown\,dx\\
\displaystyle \int \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int \dfrac{\unknown}{\unknown}\, du\\
&= \cdots\\
&= \dfrac{1}{4}\ln|x^4+4|+C
\end{align*}
Activity 13.12.
Given that \(\displaystyle \int \dfrac{x^3}{x^4+4}\,dx
= \dfrac{1}{4}\ln|x^4+4|+C
\text{,}\) what is the value of \(\displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx
\text{?}\)
- \(\displaystyle \dfrac{8}{20}\)
- \(\displaystyle -\dfrac{8}{20}\)
- \(\displaystyle \dfrac{1}{4}\ln(20)-\dfrac{1}{4}\ln(4)\)
- \(\displaystyle \dfrac{1}{4}\ln(4)-\dfrac{1}{4}\ln(20)\)
Activity 13.13.
What’s wrong with the following computation?
\begin{align*}
\displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4\\
&&& du = 4x^3\,dx\\
&&& \dfrac{1}{4} du = x^3\,dx\\
\displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int_0^2 \dfrac{1/4}{u}\, du\\
&= \left[\dfrac{1}{4}\ln|u|\right]_0^2\\
&= \dfrac{1}{4}\ln 2-\dfrac{1}{4}\ln 0
\end{align*}
- The wrong \(u\) substitution was made.
- The antiderivative of \(\dfrac{1/4}{u}\) was wrong.
- The \(x\) values \(0,2\) were plugged in for the variable \(u\text{.}\)
Example 13.14.
Here’s one way to show the computation of this definite integral by tracking \(x\) values in the bounds.
\begin{align*}
\displaystyle \int_0^2 \dfrac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4\\
&&& du = 4x^3\,dx\\
&&& \dfrac{1}{4} du = x^3\,dx\\
\displaystyle \int_{x=0}^{x=2} \dfrac{x^3}{x^4+4}\,dx &= \displaystyle \int_{x=0}^{x=2} \dfrac{1/4}{u}\, du\\
&= \left[ \dfrac{1}{4}\ln|u|\right]_{x=0}^{x=2}\\
&= \left[ \dfrac{1}{4}\ln|x^4+4|\right]_{x=0}^{x=2}\\
&= \dfrac{1}{4}\ln(20)-\dfrac{1}{4}\ln(4)
\end{align*}
Example 13.15.
Instead of unsubstituting \(u\) values for \(x\) values, definite integrals may be computed by also substituting \(x\) values in the bounds with \(u\) values. Use this idea to complete the following solution:
\begin{align*}
\displaystyle \int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=\unknown\\
&&&du = 3x^2\,dx\\
&&&\dfrac{1}{3}du = x^2\,dx\\
\displaystyle \int_1^3 x^2e^{x^3}\,dx &= \displaystyle \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx)\\
&= \displaystyle \int_{u=\unknown}^{u=\unknown} e^{u} \dfrac{1}{3}\, du\\
&= \left[\dfrac{1}{3}e^{u}\right]_{\unknown}^{\unknown}\\
&= \unknown
\end{align*}
Example 13.16.
Here is how one might write out the explanation of how to find \(\displaystyle \int_1^3 x^2e^{x^3}\,dx\) from start to finish by leaving bounds in terms of \(x\) instead:
\begin{align*}
\displaystyle \int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=x^3\\
&&& du = 3x^2\,dx\\
&&& \dfrac{1}{3}du = x^2\,dx\\
\displaystyle \int_1^3 x^2e^{x^3}\,dx &= \displaystyle \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx)\\
&= \displaystyle \int_{x=1}^{x=3} e^{u} \dfrac{1}{3}\, du\\
&= \left[\dfrac{1}{3}e^{u}\right]_{x=1}^{x=3}\\
&= \left[\dfrac{1}{3}e^{x^3}\right]_{x=1}^{x=3}\\
&= \dfrac{1}{3}e^{3^3} - \dfrac{1}{3}e^{1^3}\\
&= \dfrac{1}{3}e^{27} - \dfrac{1}{3}e
\end{align*}
Activity 13.17.
Use substitution to show that
\begin{equation*}
\displaystyle \int_1^4 \dfrac{e^{\sqrt x}}{\sqrt x}\,dx=2e^2-2e\text{.}
\end{equation*}
Activity 13.18.
Use substitution to show that
\begin{equation*}
\displaystyle \int_0^{\pi/4} \sin(2\theta)\,d\theta=\dfrac{1}{2}\text{.}
\end{equation*}
Activity 13.19.
Consider \(\displaystyle \int (3x-5)^2\,dx\text{.}\)
(a)
Solve this integral using substitution.
(b)
Replace \((3x-5)^2\) with \((9x^2-30x+25)\) in the original integral, the solve using the reverse power rule.
(c)
Which method did you prefer?
Activity 13.20.
Consider \(\displaystyle \int \tan(x)\,dx\text{.}\)
(a)
Replace \(\tan(x)\) in the integral with a fraction involving sine and cosine.
(b)
Use substitution to solve the integral.
Remark 13.21.
As you’ve noticed by now, sometimes it’s not so easy to see the substitution that will “work”. It doesn’t jump out at us. It’s important to look for deeper possibilities for substitution, subexpressions that may result in a tractable integral, and explore what happens when we try them out.
Sometimes the “right” choice of \(u\) is quite unexpected. It’s important to think about unusual possibilities in times of desperation. And of course, there is often more than one way to do a problem.
Activity 13.22.
Below are two possible solutions to the same integral, using two different choices for \(u\text{.}\) Read each one carefully and answer the question that follows.
(a)
\begin{align*}
\int x\sqrt{4x+4}\,dx &\phantom{==}\text{Let }u=x+1\\
&\phantom{==} 4u=4x+4\\
&\phantom{==} x=u-1\\
&\phantom{==} du = dx\\
&\\
\int x\sqrt{4x+4}\,dx &= \int (u-1)\sqrt{4u}\,du\\
&= \int (2u^{3/2}-2u^{1/2})\,du\\
&= \dfrac{4}{5}u^{5/2}-\dfrac{4}{3}u^{3/2}+C\\
&= \dfrac{4}{5}(x+1)^{5/2}\\
&\phantom{==}-\dfrac{4}{3}(x+1)^{3/2}+C
\end{align*}
Is this calculation valid?
- Yes
- No
- More information is needed
(b)
\begin{align*}
\int x\sqrt{4x+4}\,dx &\phantom{==}\text{Let }u=\sqrt{4x+4}\\
&\phantom{==} u^2=4x+4\\
&\phantom{==} x=\dfrac{1}{4}u^2-1\\
&\phantom{==} dx=\dfrac{1}{2}u\,du\\
&\\
\int x\sqrt{4x+4}\,dx &= \int \left(\dfrac{1}{4}u^2-1\right)(u)\left(\dfrac{1}{2}u\,du\right)\\
&= \int \left(\dfrac{1}{8}u^4-\dfrac{1}{2}u^2\right)\,du\\
&= \dfrac{1}{40}u^5-\dfrac{1}{6}u^3+C\\
&= \dfrac{1}{40}(4x+4)^{5/2}\\
&\phantom{==}-\dfrac{1}{6}(4x+4)^{3/2}+C
\end{align*}
Is this calculation valid?
- Yes
- No
- More information is needed
Activity 13.23.
Use substitution to show that
\begin{equation*}
\displaystyle \int t^5(t^3+1)^{1/3}\,dt=
\dfrac{1}{7}(t^3+1)^{7/3}-
\dfrac{1}{4}(t^3+1)^{4/3}+C\text{.}
\end{equation*}
Activity 13.24.
Use substitution to evaluate the integrals.
(a)
\(\int x^2 \sqrt{x-1} \, dx\)(b)
\(\int x \sqrt{x^2 - 1} \, dx\)Subsection 13.1.2 Videos
Note: a \(1/6\) was accidentally forgotten in the last example shown in the video above.
Subsection 13.1.3 Exercises
Exercises available at
https://tbil.org/preview/calculus/exercises/#/bank/TI1/
