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Section 5.1 Substitution method (TI1)

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 \(x\text{?}\)

  1. \(\displaystyle e^{3x^2}\)

  2. \(\displaystyle x^3e^{x^3-1}\)

  3. \(\displaystyle 3x^2e^{x^3}\)

  4. \(\displaystyle \frac{1}{4}e^{x^4}\)

(b)

Based on this result, which of these would you suspect to equal \(\int x^2e^{x^3}\,dx\text{?}\)

  1. \(\displaystyle e^{x^3+1}+C\)

  2. \(\displaystyle \frac{1}{3x}e^{x^3+1}+C\)

  3. \(\displaystyle 3e^{x^3}+C\)

  4. \(\displaystyle \frac{1}{3}e^{x^3}+C\)

Activity 5.1.2.

Recall that if \(u\) is a function of \(x\text{,}\) then \(\frac{d}{dx}[u^7]=7u^6 u'\) by the Chain Rule.

For each question, choose from the following.

  1. \(\displaystyle \frac{1}{7}u^7+C\)

  2. \(\displaystyle u^7+C\)

  3. \(\displaystyle 7u^7+C\)

  4. \(\displaystyle \frac{6}{7}u^7+C\)

(a)

What is \(\int 7u^6 u'\,dx\text{?}\)

(b)

What is \(\int u^6 u'\,dx\text{?}\)

(c)

What is \(\int 6u^6 u'\,dx\text{?}\)

Activity 5.1.3.

Based on these activities, which of these choices seems to be a viable strategy for integration?

  1. Memorize an integration formula for every possible function.

  2. Attempt to rewrite the integral in the form \(\int g'(u)u'dx=g(u)+C\text{.}\)

  3. Keep differentiating functions until you come across the function you want to integrate.

Observation 5.1.5.

Recall that \(u'=\frac{du}{dx}\text{.}\) This allows for the following:

\begin{equation*} \int g'(u)u' dx=\int g'(u)\frac{du}{dx}\,dx=\int g'(u)du=g(u)+C\text{.} \end{equation*}

Therefore, rather than dealing with equations like \(u'=\frac{du}{dx}=x^2\text{,}\) we will prefer to write \(du=x^2\,dx\text{.}\)

Activity 5.1.6.

Consider \(\int x^2e^{x^3}\,dx\text{,}\) which we conjectured earlier to be \(\frac{1}{3}e^{x^3}+C\text{.}\)

Suppose we decided to let \(u=x^3\text{.}\)

(a)

Compute \(\frac{du}{dx}=\unknown\text{,}\) and rewrite it as \(du=\unknown\,dx\text{.}\)

(b)

Multiply both sides of the equation \(du=\unknown\,dx\) so that its right-hand side appears in \(\int x^2e^{x^3}\,dx\text{.}\)

(c)

Show why \(\int x^2e^{x^3}\,dx\) may now be rewritten as \(\int \frac{1}{3}e^u\,du\text{.}\)

(d)

Solve \(\int \frac{1}{3}e^u\,du\) 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 \(\int x^2e^{x^3}\,dx\) from start to finish:

\begin{align*} \int x^2e^{x^3}\,dx &&\text{Let }&u=x^3\\ &&& du = 3x^2\,dx\\ &&& \frac{1}{3}du = x^2\,dx\\ \int x^2e^{x^3}\,dx &= \int e^{(x^3)} (x^2\,dx)\\ &= \int e^{u} \frac{1}{3}du\\ &= \frac{1}{3}e^{u}+C\\ &= \frac{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?

  1. Choosing \(u=x^3\text{.}\)

  2. Finding \(du=3x^2\,dx\) and \(\frac{1}{3}du=x^2\,dx\text{.}\)

  3. Substituting \(\int x^2e^{x^3}\,dx\) with \(\int\frac{1}{3}e^u\,du\text{.}\)

  4. Integrating \(\int\frac{1}{3}e^u\,du=\frac{1}{3}e^u+C\text{.}\)

  5. Unsubstituting \(\frac{1}{3}e^u+C\) to get \(\frac{1}{3}e^{x^3}+C\text{.}\)

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*} \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\\ &= \frac{4}{5}u^{5/2}-\frac{4}{3}u^{3/2}+C\\ &= \frac{4}{5}(x+1)^{5/2}\\ &\phantom{=}-\frac{4}{3}(x+1)^{3/2}+C \end{align*}
\begin{align*} \int x\sqrt{4x+4}\,dx &\phantom{=}\text{Let }u=\sqrt{4x+4}\\ &\phantom{=} u^2=4x+4\\ &\phantom{=} x=\frac{1}{4}u^2-1\\ &\phantom{=} dx=\frac{1}{2}u\,du\\ \int x\sqrt{4x+4}\,dx &= \int \left(\frac{1}{4}u^2-1\right)(u)\left(\frac{1}{2}u\,du\right)\\ &= \int \left(\frac{1}{8}u^4-\frac{1}{2}u^2\right)\,du\\ &= \frac{1}{40}u^5-\frac{1}{6}u^3+C\\ &= \frac{1}{40}(4x+4)^{5/2}\\ &\phantom{=}-\frac{1}{6}(4x+4)^{3/2}+C \end{align*}

Activity 5.1.10.

Suppose we wanted to try the substitution method to find \(\int e^x\cos(e^x+3)\,dx\text{.}\) Which of these choices for \(u\) appears to be most useful?

  1. \(u=x\text{,}\) so \(du=dx\)

  2. \(u=e^x\text{,}\) so \(du=e^x\,dx\)

  3. \(u=e^x+3\text{,}\) so \(du=e^x\,dx\)

  4. \(u=\cos(x)\text{,}\) so \(du=-\sin(x)\,dx\)

  5. \(u=\cos(e^x+3)\text{,}\) 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 \(\int e^x\cos(e^x+3)\,dx\text{.}\)

\begin{align*} \int e^x\cos(e^x+3)\,dx &&\text{Let }&u=\unknown\\ &&& du = \unknown\,dx\\ \int e^x\cos(e^x+3)\,dx &= \int \unknown\, du\\ &= \cdots\\ &= \sin(e^x+3)+C \end{align*}

Activity 5.1.12.

Complete the following integration by substitution to find \(\int \frac{x^3}{x^4+4}\,dx\text{.}\)

\begin{align*} \int \frac{x^3}{x^4+4}\,dx &&\text{Let }&u=\unknown\\ &&& du = \unknown\,dx\\ &&& \unknown du = \unknown\,dx\\ \int \frac{x^3}{x^4+4}\,dx &= \int \frac{\unknown}{\unknown}\, du\\ &= \cdots\\ &= \frac{1}{4}\ln|x^4+4|+C \end{align*}

Activity 5.1.13.

Given that \(\int \frac{x^3}{x^4+4}\,dx = \frac{1}{4}\ln|x^4+4|+C \text{,}\) what is the value of \(\int_0^2 \frac{x^3}{x^4+4}\,dx \text{?}\)

  1. \(\displaystyle \frac{8}{20}\)

  2. \(\displaystyle -\frac{8}{20}\)

  3. \(\displaystyle \frac{1}{4}\ln(20)-\frac{1}{4}\ln(4)\)

  4. \(\displaystyle \frac{1}{4}\ln(4)-\frac{1}{4}\ln(20)\)

Activity 5.1.14.

What's wrong with the following computation?

\begin{align*} \int_0^2 \frac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4\\ &&& du = 4x^3\,dx\\ &&& \frac{1}{4} du = x^3\,dx\\ \int_0^2 \frac{x^3}{x^4+4}\,dx &= \int_0^2 \frac{1/4}{u}\, du\\ &= \left[\frac{1}{4}\ln|u|\right]_0^2\\ &= \frac{1}{4}\ln 2-\frac{1}{4}\ln 0 \end{align*}

  1. The wrong \(u\) substitution was made.

  2. The antiderivative of \(\frac{1/4}{u}\) was wrong.

  3. The \(x\) values \(0,2\) were plugged in for the variable \(u\text{.}\)

Activity 5.1.15.

When \(x=2\text{,}\) we know that \(u=2^4+4=20\text{.}\) Likewise, when \(x=0\text{,}\) we know that \(u=0^4+4=4\text{.}\) Use these facts to complete the below computation.

\begin{align*} \int_0^2 \frac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4\\ &&& du = 4x^3\,dx\\ &&& \frac{1}{4} du = x^3\,dx\\ \int_{x=0}^{x=2} \frac{x^3}{x^4+4}\,dx &= \int_{u=\unknown}^{u=\unknown} \frac{1/4}{u}\, du\\ &= \dots\\ &= \frac{1}{4}\ln(20)-\frac{1}{4}\ln(4) \end{align*}

Example 5.1.16.

Here is how one might write out the explanation of how to find \(\int_1^3 x^2e^{x^3}\,dx\) from start to finish by leaving bounds in terms of \(x\text{:}\)

\begin{align*} \int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=x^3\\ &&& du = 3x^2\,dx\\ &&& \frac{1}{3}du = x^2\,dx\\ \int_1^3 x^2e^{x^3}\,dx &= \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx)\\ &= \int_{x=1}^{x=3} e^{u} \frac{1}{3}du\\ &= \left[\frac{1}{3}e^{u}\right]_{x=1}^{x=3}\\ &= \left[\frac{1}{3}e^{x^3}\right]_{x=1}^{x=3}\\ &= \frac{1}{3}e^{3^3} - \frac{1}{3}e^{1^3}\\ &= \frac{1}{3}e^{27} - \frac{1}{3}e \end{align*}

Example 5.1.17.

Here is how one might write out the explanation of how to find \(\int_1^3 x^2e^{x^3}\,dx\) from start to finish by transforming bounds into terms of \(u\text{:}\)

\begin{align*} \int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=x^3\\ &&&du = 3x^2\,dx\\ &&&\frac{1}{3}du = x^2\,dx\\ \int_1^3 x^2e^{x^3}\,dx &= \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx)\\ &= \int_{u=1^3}^{u=3^3} e^{u} \frac{1}{3}du\\ &= \left[\frac{1}{3}e^{u}\right]_{1}^{27}\\ &= \frac{1}{3}e^{27} - \frac{1}{3}e \end{align*}

Activity 5.1.18.

Use substitution to show that

\begin{equation*} \int_1^4 \frac{e^{\sqrt x}}{\sqrt x}\,dx=2e^2-2e\text{.} \end{equation*}

Activity 5.1.19.

Use substitution to show that

\begin{equation*} \int_0^{\pi/4} \sin(2\theta)\,d\theta=\frac{1}{2}\text{.} \end{equation*}

Activity 5.1.20.

Use substitution to show that

\begin{equation*} \int u^5(u^3+1)^{1/3}\,du= \frac{1}{7}(u^3+1)^{7/3}- \frac{1}{4}(u^3+1)^{4/3}+C\text{.} \end{equation*}

Activity 5.1.21.

Consider \(\int (3x-5)^2\,dx\text{.}\)

(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 \(\int \tan(x)\,dx\text{.}\)

(a)

Replace \(\tan(x)\) in the integral with a fraction involving sine and cosine.

(b)

Use substitution to solve the integral.

Subsection 5.1.1 Videos

Figure 74. Video: Evaluate various integrals via the substitution method