Learning Outcomes
- Evaluate various integrals via the substitution method.
Section 5.1 Substitution Method (TI1)
Subsection 5.1.1 Activities
Activity 5.1.1.
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 5.1.2.
Recall that if \(u\) is a function of \(x\text{,}\) then \(\dfrac{d}{dx}[u^7]=7u^6 u'\) by the Chain Rule (Theorem 2.5.5).
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 5.1.3.
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 5.1.4.
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 5.1.5.
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 5.1.6.
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 5.1.7.
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 5.1.8.
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 5.1.9.
Below are two correct solutions to the same integral, using two different choices for \(u\text{.}\) Which method would you prefer to use yourself?
\begin{align*}
\displaystyle \int x\sqrt{4x+4}\,dx &\phantom{=}\text{Let }u=x+1\\
&\phantom{=} 4u=4x+4\\
&\phantom{=} x=u-1\\
&\phantom{=} du = dx\\
\displaystyle \int x\sqrt{4x+4}\,dx &= \displaystyle \int (u-1)\sqrt{4u}\,du\\
&= \displaystyle \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*}
\begin{align*}
\displaystyle \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\\
\displaystyle \int x\sqrt{4x+4}\,dx &= \displaystyle \int \left(\dfrac{1}{4}u^2-1\right)(u)\left(\dfrac{1}{2}u\,du\right)\\
&= \displaystyle \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*}
Activity 5.1.10.
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 5.1.11.
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 5.1.12.
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 5.1.13.
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 5.1.14.
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 5.1.15.
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 5.1.16.
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 5.1.17.
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 5.1.18.
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 5.1.19.
Use substitution to show that
\begin{equation*}
\displaystyle \int_0^{\pi/4} \sin(2\theta)\,d\theta=\dfrac{1}{2}\text{.}
\end{equation*}
Activity 5.1.20.
Use substitution to show that
\begin{equation*}
\displaystyle \int u^5(u^3+1)^{1/3}\,du=
\dfrac{1}{7}(u^3+1)^{7/3}-
\dfrac{1}{4}(u^3+1)^{4/3}+C\text{.}
\end{equation*}
Activity 5.1.21.
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 5.1.22.
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.
Subsection 5.1.2 Videos
Note: a \(1/6\) was accidentally forgotten in the last example shown in the video above.
Subsection 5.1.3 Exercises
Exercises available at
https://tbil.org/calculus/preview/exercises/#/bank/TI1/
