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.

Substitution Rule

Procedure for Substitution Rule for Indefinite Integrals

NOTE: When using the Substitution Rule, you must completely convert all $x$$x$​-variables to $u$$u$​-variable(s). Show the following work:

• Let $u=\text{expression}$$u=\text{expression}$

• ${\displaystyle du=\text{derivative expression}dx}$$\displaystyle du=\text{derivative expression}\,dx$; rewrite as: $dx=\frac{du}{\text{derivative expression}}$$dx=\frac{du}{\text{derivative expression}}$

1. Substitute $u$$u$ and $dx$$dx$​ into the indefinite integral.

2. Appropriately cancel factors.

3. For each converted integral, integrate with respect to $u$$u$​​​​.

4. Substitute result back into terms of $x$$x$.

5. Verify answer:

   1. Use FTC-I; or

   2. Use technology.

Example 04: Evaluate: ${\displaystyle \int \frac{{\csc }^{-1}(x)}{x\sqrt{x^2-1}}dx}$$\displaystyle \int{\frac{\csc^{-1}{(x)}}{x\sqrt{x^2-1}}\,dx}$​​​​​​​

Solution 04:

NOTE: ${\csc }^{-1}(x)$$\csc^{-1}{(x)}$ is not a composition of two functions. ${\csc }^{-1}(x)$$\csc^{-1}{(x)}$ is the inverse cosecant function.

Let $u={\csc }^{-1}(x)$$\color{green}u=\csc^{-1}{(x)}$​​​​​ not because ${\csc }^{-1}(x)$$\csc^{-1}{(x)}$​​​​​ is an “inside” function​​.

Then the differential for the above let statement is: $du=-\frac{1}{x\sqrt{x^2-1}}dx$$du=-\frac{1}{x\sqrt{x^2-1}}dx$​​​​.

Rewrite the differential as $dx=-x\sqrt{x^2-1}du$$\color{red}dx=-x\sqrt{x^2-1}du$​​​.

1. Substitute $u$$\color{green}u$ and $dx$$\color{red}dx$ into the indefinite integral: ${\displaystyle \int \frac{{\csc }^{-1}(x)}{x\sqrt{x^2-1}}dx\Rightarrow \int \frac{u}{x\sqrt{x^2-1}}\left(-x\sqrt{x^2-1}du\right)}$$\displaystyle \int{\frac{\csc^{-1}{(x)}}{x\sqrt{x^2-1}}\,dx}\Rightarrow \int{\frac{{\color{green}u}}{x\sqrt{x^2-1}}\,\left(\color{red}-x\sqrt{x^2-1}du\right)}$

2. Appropriately cancel factors: ${\displaystyle \int \frac{u}{x\sqrt{x^2-1}}\left(-x\sqrt{x^2-1}du\right)\Rightarrow \int \frac{u}{\cancel{x}\cancel{\sqrt{x^2-1}}}\left(-\cancel{x}\cancel{\sqrt{x^2-1}}du\right)\Rightarrow -\int udu}$$\displaystyle \int{\frac{{\color{green}u}}{x\sqrt{x^2-1}}\,\left(\color{red}-x\sqrt{x^2-1}du\right)}\Rightarrow \int{\frac{{\color{green}u}}{\cancel{x}\cancel{\sqrt{x^2-1}}}\,\left(\color{red}-\cancel{x}\cancel{\sqrt{x^2-1}}du\right)}\Rightarrow -\int{{\color{green}u}\,{\color{red}du}}$​​​​​​ NOTE: There is only a single variable in the simplified indefinite integral above.

3. Integrate with respect to $u$$u$: ${\displaystyle -\int udu=-\frac{u^2}{2}+C}$$\displaystyle -\int{{\color{green}u}\,{\color{red}du}}=-\frac{{\color{green}u}^2}{2}+C$​​

4. Substitute result back into terms of $x$$x$​. ${\displaystyle -\frac{u^2}{2}+C=-\frac{{{\csc }^{-1}(x)}^2}{2}+C}$$\displaystyle -\frac{{\color{green}u}^2}{2}+C=-\frac{{\color{green}\csc^{-1}{(x)}}^2}{2}+C$​

5. Verify answer: ${\displaystyle \frac{d}{dx}[-\frac{1}{2}{\csc }^{-1}(x{)}^2+C]=-\frac{1}{2}\left(2\right){\csc }^{-1}(x)\frac{d}{dx}({\csc }^{-1}(x))}$$\displaystyle \frac{d}{dx}\left[-\frac{1}{2}\csc^{-1}{\!(x)^2}+C\right]=-\frac{1}{2}\left(2\right)\csc^{-1}{\!(x)}\frac{d}{dx}\left(\csc^{-1}{\!(x)}\right)$ ${\displaystyle \Rightarrow \cancel{-}\cancel{\frac{1}{2}\left(2\right)}{\csc }^{-1}(x)\left(\cancel{-}\frac{1}{x\sqrt{x^2-1}}\right)=\frac{{\csc }^{-1}(x)}{x\sqrt{x^2-1}}}$$\displaystyle \Rightarrow \cancel{-}{\cancel{\frac{1}{2}\left(2\right)}}\csc^{-1}{\!(x)}\left(\cancel{-}\frac{1}{x\,\sqrt{x^2-1}}\right)=\frac{\csc^{-1}{\!(x)}}{x\,\sqrt{x^2-1}}$ The derivative function is equivalent to the function in the integrand of the indefinite integral. NOTE: $\frac{d}{dx}C=0$$\frac{d}{dx}C=0$ and Chain Rule for Derivatives is required to differentiate $-{\csc }^{-1}{\left(x\right)}^2=-{({\csc }^{-1}\left(x\right))}^2$$-\csc^{-1}{\left(x\right)}^2=-\left(\csc^{-1}{\left(x\right)}\right)^2$.

Use Technology to Verify Definite Integrals

Mathematica

NOTE: Mathematica is used here to verify definite 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 *)
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(* Evaluate the indefinite integral csc^(-1)(x)/(x sqrt(x^2-1)) *)
3

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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 definite 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 08

Evaluate the following indefinite integrals:

1. ${\displaystyle \int \frac{{\tan }^{-1}(x)}{1+x^2}dx}$$\displaystyle \int{\frac{{\color{c8}\tan^{-1}{(x)}}}{{\color{c1}1+x^2}}\,{\color{c1}dx}}$

2. ${\displaystyle \int \frac{{\sin }^{-1}(x)}{\sqrt{1-x^2}}dx}$$\displaystyle \int{\frac{{\color{c8}\sin^{-1}{(x)}}}{{\sqrt{\color{c1}1-x^2}}}\,{\color{c1}dx}}$​

3. ${\displaystyle \int \frac{\ln (x)}{x}dx}$$\displaystyle \int{\frac{\ln{(x)}}{x}\,dx}$

4. ${\displaystyle \int \tan (x){\sec }^2(x)dx}$$\displaystyle \int{\tan{(x)}\sec^2{(x)}\,{\color{c1}dx}}$​

5. ${\displaystyle \int \sin (x)\cos (x)dx}$$\displaystyle \int{{\color{c8}\sin}{\color{c8}(x)}\,{\color{c1}\cos{(x)}}\,{\color{c1}dx}}$

Homework

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

• Section 5.5: # 51.

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Created: Monday, 9 August 2021 18:16 EDT Last Modified: Thursday, 8 June 2023 – 11:34 (EDT)