Substitution Rule

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 5.5 – The Substitution Rule

Expected Educational Results

• Objective 02–01: I can use a substitution to evaluate an indefinite integral using FTC-I.

• Objective 02–02: I can use a substitution to evaluate a definite integral.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Special Case for Substitution Methods

Investigation 09

Consider the integral ${\displaystyle \int x{\left(x+3\right)}^{10}\text{d}x}$$\displaystyle \int{x\left({x+3}\right)^{10}\,\text{d}x}$.

1. Explain how is this integral different from the integrals in Investigation 03 in ACT_02c_Substitution_Rule.html?

2. Explain why $u=x+3$$u=x+3$ is required.

Example 05: Consider the integral ${\displaystyle \int x{\left(x+3\right)}^{10}\text{d}x}$$\displaystyle \int{x\left({x+3}\right)^{10}\,\text{d}x}$.

Answer 05:

Let $u=x+3$${\color{red}u=x+3}$. Then $\text{d}u=\text{d}x$$\color{blue}\,\text{d}u=\,\text{d}x$ and

${\displaystyle \int x{\left(x+3\right)}^{10}\text{d}x=\int x{u}^{10}\text{d}u}$$\displaystyle \int{x\left({\color{red}{x+3}}\right)^{10}\,\text{d}x}=\int{x\,{\color{red}u}^{10}\,{\color{blue}\text{d}u}}$.

Investigation 10

Explain why ${\displaystyle \int x{u}^{10}\text{d}u}$$\displaystyle \int{x\,{\color{red}u}^{10}\,{\color{blue}\text{d}u}}$ is not integrable as written.

Answer 05 (continued):

Rewrite $u=x+3$${\color{red}u=x+3}$ as $x=u-3$$\color{green}x=u-3$ and substitute:

${\displaystyle \int x{u}^{10}\text{d}u=\int (u-3)u^{10}\text{d}u}$$\displaystyle \int{x\,{\color{red}u}^{10}\,{\color{blue}\text{d}u}}=\int{({\color{green}u-3})\,{\color{red}u}^{10}\,{\color{blue}\text{d}u}}$

Investigation 11

Explain why ${\displaystyle \int (u-3)u^{10}\text{d}u}$$\displaystyle \int{({\color{green}u-3})\,{\color{red}u}^{10}\,{\color{blue}\text{d}u}}$ IS integrable as written

Answer 05 (continued):

Simplify the integral by expanding using the Distributive Property:

${\displaystyle \int (u-3)u^{10}\text{d}u=\int (u^{11}-3u^{10})\text{d}u}$$\displaystyle \int{({\color{green}u-3})\,{\color{red}u}^{10}\,{\color{blue}\text{d}u}}=\int{\left(u^{11}-3u^{10}\right)\,\text{d}u}$

Integrate in terms of $u$$u$ using Power Rule for Antiderivatives:

${\displaystyle \int (u^{11}-3u^{10})\text{d}u=\frac{u^{12}}{12}-3\frac{u^{11}}{11}+C}$$\displaystyle \int{\left(u^{11}-3u^{10}\right)\,\text{d}u}=\frac{u^{12}}{12}-3\frac{u^{11}}{11}+C$

NOTE: $+C$$+C$ is needed for all steps that have an antiderivative.

Use $u=x+3$${\color{red}u=x+3}$ to substitute answer back into terms of $x$$x$:

${\displaystyle \frac{u^{12}}{12}-3\frac{u^{11}}{11}+C=\frac{(x+3{)}^{12}}{12}-3\frac{(x+3{)}^{11}}{11}+C}$$\displaystyle \frac{u^{12}}{12}-3\frac{u^{11}}{11}+C=\frac{({\color{red}x+3})^{12}}{12}-3\frac{({\color{red}x+3})^{11}}{11}+C$

NOTE: There is NO need to rewrite the antiderivative into another form.

Verify the answer using:

• FTC-1; or

• technology.

Investigation 12

Consider the integral ${\displaystyle \int \frac{x^5}{\sqrt{1-x^2}}\text{d}x}$$\displaystyle \int{\frac{x^5}{\sqrt{1-x^2}}\,\text{d}x}$.

1. Explain how is this integral different from the integrals in Investigation 03 in ACT_02c_Substitution_Rule.html?

2. Explain why $u=1-x^2$$\color{red}{u=1-x^2}$ is required.

Example 06: Consider the integral ${\displaystyle \int \frac{x^5}{\sqrt{1-x^2}}\text{d}x}$$\displaystyle \int{\frac{x^5}{\sqrt{1-x^2}}\,\text{d}x}$.

Answer 06:

Let $u=1-x^2$${\color{red}u=1-x^2}$. Then ${\displaystyle \text{d}u=-2x\text{d}x\Rightarrow \text{d}x=\frac{\text{d}u}{-2x}}$$\displaystyle\color{blue}\,\text{d}u=-2x\,\text{d}x\Rightarrow \text{d}x=\frac{\text{d}u}{-2x}$ and

${\displaystyle \int \frac{x^5}{\sqrt{1-x^2}}\text{d}x=\int \frac{x^5}{\sqrt{u}}\frac{\text{d}u}{-2x}}$$\displaystyle \int{\frac{x^5}{\sqrt{\color{red}1-x^2}}\,{\color{blue}\text{d}x}}=\int{\frac{x^5}{\sqrt{\color{red}u}}\,{\color{blue}\frac{\text{d}u}{-2x}}}$

Use algebra to simplify:

${\displaystyle \int \frac{\cancel{x^5}{x}^4}{\sqrt{u}}\frac{\text{d}u}{-2\cancel{x}}=\frac{-1}{2}\int \frac{x^4}{\sqrt{u}}\text{d}u}$$\displaystyle \int{\frac{\cancel{x^5}x^4}{\sqrt{\color{red}u}}\,{\color{blue}\frac{\text{d}u}{-2\cancel{x}}}}=\frac{-1}{2}\int{\frac{x^4}{\sqrt{\color{red}u}}\,{\color{blue}\text{d}u}}$

Investigation 13

Explain why ${\displaystyle \frac{-1}{2}\int \frac{x^4}{\sqrt{u}}\text{d}u}$$\displaystyle\frac{-1}{2}\int{\frac{x^4}{\sqrt{\color{red}u}}\,{\color{blue}\text{d}u}}$ is not integrable as written.

Answer 06 (continued):

Rewrite $u=1-x^2$${\color{red}u=1-x^2}$ as $x^2=1-u$$\color{green}x^2=1-u$ and substitute:

${\displaystyle \frac{-1}{2}\int \frac{x^4}{\sqrt{u}}\text{d}u=\frac{-1}{2}\int \frac{(x^2{)}^2}{\sqrt{u}}\text{d}u=\frac{-1}{2}\int \frac{(1-u{)}^2}{\sqrt{u}}\text{d}u}$$\displaystyle\frac{-1}{2}\int{\frac{x^4}{\sqrt{\color{red}u}}\,{\color{blue}\text{d}u}}=\frac{-1}{2}\int{\frac{({\color{green}{x^2}})^2}{\sqrt{\color{red}u}}\,{\color{blue}\text{d}u}}=\frac{-1}{2}\int{\frac{({\color{green}{1-u}})^2}{\sqrt{\color{red}u}}\,{\color{blue}\text{d}u}}$.

Investigation 14

Explain why ${\displaystyle \frac{-1}{2}\int \frac{(1-u{)}^2}{\sqrt{u}}\text{d}u}$$\displaystyle \frac{-1}{2}\int{\frac{({\color{green}{1-u}})^2}{\sqrt{\color{red}u}}\,{\color{blue}\text{d}u}}$ IS integrable as written.

Answer 06 (continued):

Simplify the integral by expanding the numerator using the Distributive Property:

${\displaystyle \frac{-1}{2}\int \frac{(1-u{)}^2}{\sqrt{u}}\text{d}u=\frac{-1}{2}\int \frac{1-2u+u^2}{\sqrt{u}}\text{d}u}$$\displaystyle \frac{-1}{2}\int{\frac{({\color{green}{1-u}})^2}{\sqrt{\color{red}u}}\,{\color{blue}\text{d}u}}= \frac{-1}{2}\int{\frac{{\color{green}{1-2u+u^2}}}{\sqrt{\color{red}u}}\,{\color{blue}\text{d}u}}$

Simplify the integral using the Distributive Property:

${\displaystyle \frac{-1}{2}\int (\frac{1}{\sqrt{u}}-2\frac{u}{\sqrt{u}}+\frac{u^2}{\sqrt{u}})\text{d}u}$$\displaystyle \frac{-1}{2}\int{\left(\frac{1}{\sqrt{u}}-2\frac{u}{\sqrt{u}}+\frac{u^2}{\sqrt{u}}\right)\,\text{d}u}$

Simplify the rational expressions using a Property of Exponential Expressions:

${\displaystyle \frac{-1}{2}\int (\frac{1}{\sqrt{u}}-2\frac{u}{\sqrt{u}}+\frac{u^2}{\sqrt{u}})\text{d}u=\frac{-1}{2}\int (u^{-1/2}-2u^{1/2}+u^{3/2})\text{d}u}$$\displaystyle \frac{-1}{2}\int{\left(\frac{1}{\sqrt{u}}-2\frac{u}{\sqrt{u}}+\frac{u^2}{\sqrt{u}}\right)\,\text{d}u}=\frac{-1}{2}\int{\left(u^{-1/2}-2u^{1/2}+u^{3/2}\right)\,\text{d}u}$

Integrate in terms of $u$$u$ using Power Rule for Antiderivatives:

${\displaystyle \frac{-1}{2}\int (u^{-1/2}-2u^{1/2}+u^{3/2})\text{d}u=\frac{-1}{2}(\frac{u^{1/2}}{1/2}-2\frac{u^{3/2}}{3/2}+\frac{u^{5/2}}{5/2})+C}$$\displaystyle \frac{-1}{2}\int{\left(u^{-1/2}-2u^{1/2}+u^{3/2}\right)\,\text{d}u}=\frac{-1}{2}\left(\frac{u^{1/2}}{1/2}-2\frac{u^{3/2}}{3/2}+\frac{u^{5/2}}{5/2}\right)+C$

NOTE: $+C$$+C$ is needed for all steps that have an antiderivative.

NOTE: $+C$$+C$ can be included inside the parentheses or not – both ways to include $+C$$+C$ are equivalent.

Use $u=1-x^2$${\color{red}u=1-x^2}$ to substitute answer back into terms of $x$$x$:

${\displaystyle -\frac{1}{2}(\frac{u^{1/2}}{1/2}-2\frac{u^{3/2}}{3/2}+\frac{u^{5/2}}{5/2})+C=-\frac{1}{2}(\frac{(1-x^2{)}^{1/2}}{1/2}-2\frac{(1-x^2{)}^{3/2}}{3/2}+\frac{(1-x^2{)}^{5/2}}{5/2})+C}$$\displaystyle -\frac{1}{2}\left(\frac{u^{1/2}}{1/2}-2\frac{u^{3/2}}{3/2}+\frac{u^{5/2}}{5/2}\right)+C=-\frac{1}{2}\left(\frac{({\color{red}1-x^2})^{1/2}}{1/2}-2\frac{({\color{red}1-x^2})^{3/2}}{3/2}+\frac{({\color{red}1-x^2})^{5/2}}{5/2}\right)+C$

NOTE: There is NO need to rewrite the antiderivative into another form.

Verify the answer using:

• FTC-1; or

• technology.

Use Technology to Verify Indefinite Integrals

Mathematica

NOTE: Mathematica is used here to verify indefinite integrals. On all Assessments, you must show that you understand the Calculus I concept of using $u$$u$​-Substitution to evaluate an indefinite integral.

​x
1
(* Example 03 *)
2
(* Evaluate the indefinite integral csc^(-1)(x)/(x sqrt(x^2-1)) *)
3

4
Integrate[ArcCsc[x]/(x Sqrt[x^2-1]), x]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: https://www.jjw3.com/Common_Mathematica_Code.html.

2. Integrate[ ] for indefinite integrals has two arguments:

   1. The integrand, i.e., function of the definite integral;

   2. $x$$x$​, where $x$$x$​ is the variable used in the integrand.

3. The Mathematica syntax for $e^x$$e^x$​​ is $\text{E}\wedge x$$\text{E}\!\wedge\!x$​​; $\cos (x)$$\cos{(x)}$​​ is $\text{Cos[x]}$$\text{Cos[x]}$​​, $\ln (x)$$\ln{(x)}$​​ is $\text{Log[x]}$$\text{Log[x]}$​​, ${\tan }^{-1}(x)$$\tan^{-1}{(x)}$​​ is $\text{ArcTan[x]}$$\text{ArcTan[x]}$​​, etc.

4. Remember to use parens $($$\left(\right.$​ and $)$$\left.\right)$​​ to group multiple terms or factors in the numerator or denominator of a rational expression.

5. Remember, correct Mathematica code will be all black except for variables.

6. Mathematica does not return the antiderivative with $+C$$+C$​. You need to use $+C$$+C$ for all steps that are antiderivatives.

7. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

Python

NOTE: Python is used here to verify indefinite integrals. On all Assessments, you must show that you understand the Calculus I concept of using $u$$u$-Substitution to evaluate an indefinite integral.

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Python syntax is located at: https://www.jjw3.com/Common_Python_Code.html.

2. All code is included above with comment codes that explain the executable code.

3. integrate( ) for indefinite integrals has two arguments:

   1. The integrand, i.e., function of the definite integral;

   2. $x$$x$​, where $x$$x$​​ is the variable used in the integrand.

4. You only need to change the function in the integrate( ) command.

5. The Python syntax for $e^x$$e^x$​​ is $\text{exp(x)}$$\text{exp(x)}$​​; ${\cos }^2(x)$$\cos^2{(x)}$​​ is $\text{cos(x)**2}$$\text{cos(x)**2}$​​; $x^4$$x^4$​​ is $\text{x**4}$$\text{x**4}$​​, $\ln (x)$$\ln{(x)}$​​ is $\text{log(x)}$$\text{log(x)}$​​, ${\tan }^{-1}(x)$$\tan^{-1}{(x)}$​​ is $\text{atan(x)}$$\text{atan(x)}$​​, etc.

6. Python requires $\ast$$*$ for explicit multiplication; $\ast \ast$$**$​ for exponents.

7. Remember to use parens $($$\left(\right.$ and $)$$\left.\right)$​ to group multiple terms in the numerator or denominator of a rational expression.

8. Python does not return the antiderivative with $+C$$+C$. You need to use $+C$$+C$ for all steps that are antiderivatives.

9. To execute code, press the “View the result” button:

Investigation 15

Integrate the following:

1. ${\displaystyle \int \frac{x}{\sqrt{1-x}}\text{d}x}$$\displaystyle \int{\frac{x}{\sqrt{1-x}}\,\text{d}x}$

2. ${\displaystyle \int \frac{x}{1+x}\text{d}x}$$\displaystyle \int{\frac{x}{1+x}\,\text{d}x}$

3. ${\displaystyle \int x^2(1-x{)}^{12}\text{d}x}$$\displaystyle \int{x^2(1-x)^{12}\,\text{d}x}$

4. ${\displaystyle \int x^3\sqrt{x^2-1}\text{d}x}$$\displaystyle \int{x^3\sqrt{x^2-1}\,\text{d}x}$

5. ${\displaystyle \int x^8\sqrt{1+x^3}\text{d}x}$$\displaystyle \int{x^8\sqrt{1+x^3}\,\text{d}x}$

Investigation 16

Explain why the above method is not valid for the following integrals:

1. ${\displaystyle \int \frac{x^2}{\sqrt{1-x^2}}\text{d}x}$$\displaystyle \int{\frac{x^2}{\sqrt{1-x^2}}\,\text{d}x}$

2. ${\displaystyle \int \frac{x^4}{1+x^3}\text{d}x}$$\displaystyle \int{\frac{x^4}{1+x^3}\,\text{d}x}$

3. ${\displaystyle \int x^2(1-x^2{)}^{12}\text{d}x}$$\displaystyle \int{x^2(1-x^2)^{12}\,\text{d}x}$

4. ${\displaystyle \int x^2\sqrt{x^2-1}\text{d}x}$$\displaystyle \int{x^2\sqrt{x^2-1}\,\text{d}x}$

5. ${\displaystyle \int x^6\sqrt{1+x^3}\text{d}x}$$\displaystyle \int{x^6\sqrt{1+x^3}\,\text{d}x}$

Homework

At this time, you should be able to complete the following assignments:

• Section 5.5: # 53, [and 73, 79].

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Created: Tuesday, 25 August 2020 07:18 EDT Last Modified: Thursday, 8 June 2023 – 12:22 (EDT)