Integration by Parts

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

• Section 7.1 – Integration by Parts

Expected Educational Results

• Objective 03–01: I can use integration by parts to evaluate an indefinite integral using FTC-I.

• Objective 03–02: I can use integration by parts to evaluate a definite integral using FTC-II.

• Objective 03–03: I can use substitution and integration by parts to evaluate an indefinite integral using FTC-I.

• Objective 03–04: I can use substitution and integration by parts to evaluate a definite integral using FTC-II.

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.

Integration by Parts

Definition: Integration by Parts

If $U=f(x)$$U=f(x)$ is a differentiable function and $V=g(x)$$V=g(x)$ is a integrable function, then $dU=f^{\prime }(x)$$dU=f^{\,\prime}(x)$ and $dV=g^{\prime }(x)$$dV=g^{\,\prime}(x)$. Thus,

$\int UdV=UV-\int VdU$$\int{U\,dV}=UV-\int{V\,dU}$

Using Integration by Parts with a Substitution

Example 03: Evaluate ${\displaystyle \int \sin \left(\sqrt{x}\right)dx}$$\displaystyle\int{\sin{\left(\sqrt{x}\right)}\,dx}$

Solution 03:

1. Since the function in the integrand is a composition of functions, use $u$$u$-substitution as a first step AND convert all $x$$x$-variables into $u$$u$-variables:

   • Let $u=\sqrt{x}$$\color{green}u=\sqrt{x}$, and

   • $du=\frac{dx}{2\sqrt{x}}\Rightarrow dx=2\sqrt{x}du$$du=\frac{dx}{2\sqrt{x}}\Rightarrow dx=2\sqrt{x}\,du$

   ${\displaystyle \int \sin \left(\sqrt{x}\right)dx=\int \sin \left(u\right)2\sqrt{x}du}$$\displaystyle\int{\sin{\left(\sqrt{x}\right)}\,dx}=\int{\sin{\left({\color{green}u}\right)}\,2{\color{green}\sqrt{x}}du}$

2. Since the integral still has two different variables after the substitution, we need to make another substitution:

   ${\displaystyle \int \sin \left(u\right)2\sqrt{x}du=\int \sin \left(u\right)2udu}$$\displaystyle\int{\sin{\left({\color{green}u}\right)}\,2{\color{green}\sqrt{x}}du}=\int{\sin{\left({\color{green}u}\right)}\,2{\color{green}u}du}$

3. Now we can integrate by parts:

   • Let $U=2u\Rightarrow dU=2du$$U=2u\Rightarrow dU=2du$

     NOTE: $U$$U$ is used for Integration by Parts and $u$$u$ is used for Substitution to avoid confusion.

   • $dV=\sin (u)du\Rightarrow V=-\cos (u)$$dV=\sin{(u)}du\Rightarrow V=-\cos{(u)}$

4. Now Integrate by Parts:

   ${\displaystyle \int \sin \left(u\right)2udu=2u(-\cos (u))-\int -\cos (u)2du=-2u\cos (u)+2\sin (u)+C}$$\displaystyle\int{\sin{\left({\color{green}u}\right)}\,2{\color{green}u}du}=2u(-\cos{(u)})-\int{-\cos{(u)}\,2du}=-2u\cos{(u)}+2\sin{(u)}+C$

5. Convert the antiderivative back into the original variable $x$$x$:

   ${\displaystyle \Rightarrow -2u\cos (u)+2\sin (u)+C=-2\sqrt{x}\cos (\sqrt{x})+2\sin (\sqrt{x})+C}$$\displaystyle\Rightarrow -2u\cos{(u)}+2\sin{(u)}+C = -2\sqrt{x}\cos{(\sqrt{x})}+2\sin{(\sqrt{x})}+C$ ∎

Investigation 07

Example 04: Evaluate ${\displaystyle \int x^5{e}^{x^2}dx}$$\displaystyle\int{x^5\,e^{x^2}\,dx}$

Solution 04:

1. Since the function in the integrand contains a composition of functions, use $u$$u$-substitution as a first step AND convert all $x$$x$-variables into $u$$u$-variables:

   • Let $u=x^2$$\color{green}u=x^2$, and

   • $du=2xdx\Rightarrow dx=\frac{du}{2x}$$du=2x\,dx\Rightarrow \color{red}dx=\frac{du}{2x}$

   ${\displaystyle \int x^5{e}^{x^2}dx=\int x^{\cancel{5}4}{e}^u\frac{du}{2\cancel{x}}=\frac{1}{2}\int x^4{e}^{u}du}$$\displaystyle\int{x^5\,e^{\color{green}x^2}\,dx}=\int{x^{\cancel{5}4}\,e^{\color{green}u}\,\color{red}\frac{du}{2\cancel{x}}}=\frac{1}{2}\int{x^4\,e^u\,du}$

2. Since the integral still has two different variables after the substitution, we need to make another substitution:

   ${\displaystyle \frac{1}{2}\int x^4{e}^{u}du=\frac{1}{2}\int {\left(x^2\right)}^2{e}^{u}du=\frac{1}{2}\int u^2{e}^{u}du}$$\displaystyle\frac{1}{2}\int{x^4\,e^u\,du}=\frac{1}{2}\int{\left({\color{green}x^2}\right)^2\,e^u\,du}=\frac{1}{2}\int{{\color{green}u^2}\,e^u\,du}$

3. Now Integrate by Parts (using the Tanzalin Method as shown in CPT_03h_IBP_Procedure.html):

   ${\displaystyle \Rightarrow \frac{1}{2}\int u^2{e}^{u}du=\frac{1}{2}(u^2{e}^u-2u{e}^u+2e^u)+C}$$\displaystyle \Rightarrow\frac{1}{2}\int{{u^2}\,e^u\,du}=\frac{1}{2}\left(u^2\,e^u-2u\,e^u+2e^u\right)+C$

   NOTE: The $+C$$+C$ may be placed inside the parens or outside the parens – it does not make any difference.

4. Convert the antiderivative back into the original variable $x$$x$:

   ${\displaystyle \Rightarrow \frac{1}{2}(u^2{e}^u-2u{e}^u+2e^u)+C=\frac{1}{2}({\left(x^2\right)}^2{e}^{x^2}-2x^2{e}^{x^2}+2e^{x^2})+C}$$\displaystyle\Rightarrow \frac{1}{2}\left(u^2\,e^u-2u\,e^u+2e^u\right)+C=\frac{1}{2}\left(\left({\color{green}x^2}\right)^2\,e^{\color{green}x^2}-2{\color{green}x^2}\,e^{\color{green}x^2}+2e^{\color{green}x^2}\right)+C$ ∎

Use Technology to Verify Antiderivatives

Mathematica

NOTE: Mathematica is used here to verify antiderivatives. You must show that you understand the Calculus concept of the antiderivative.

1
(* Example from Investigation 06: Integrate sin(sqrt(x)) *)
2
Integrate[Sin[Sqrt[x]], 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. The Mathematica syntax for ${\sin }^{-1}(x)$$\sin^{-1}{(x)}$​ is $\text{ArcSin[x]}$$\text{ArcSin[x]}$​, $e^x$$e^x$​ is $\text{E}\wedge x$$\text{E}\!\wedge\!x$​, $\sqrt{x}$$\sqrt{x}$ is $\text{Sqrt[x]}$$\text{Sqrt[x]}$, ${\sec }^2(x)$$\sec^2{(x)}$ is $\text{Sec[x]}\wedge 2$$\text{Sec[x]}\!\wedge\!2$​, etc.

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

4. 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.

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

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

7. Mathematica does not show $+C$$+C$ for antiderivatives. You will need to remember to include $+C$$+C$ in your answers.

8. Mathematica may return a result in a different, but equivalent, form than your answer. In these cases, use FTC-I to check your work.

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)}$​​​, $\sqrt{x}$$\sqrt{x}$ is $\text{sqrt(x)}$$\text{sqrt(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: ​​

Practice 08

Integrate the following:

1. ${\displaystyle \int \cos \left(\sqrt{x}\right)dx}$$\displaystyle\int{\cos{\left(\sqrt{x}\right)}\,dx}$

2. ${\displaystyle \int x^3{e}^{x^2}dx}$$\displaystyle\int{x^3e^{x^2}\,dx}$

3. ${\displaystyle \int e^{\sqrt{x}}dx}$$\displaystyle \int{e^{\sqrt{x}}\,dx}$

4. ${\displaystyle \int \sin (e^x)dx}$$\displaystyle \int{\sin{(e^x)}\,dx}$

5. ${\displaystyle \int e^{4x}\cos (e^{2x})dx}$$\displaystyle \int{e^{4x}\cos{(e^{2x})}\,dx}$

6. ${\displaystyle \int \frac{x^3}{(x^2+7{)}^2}dx}$$\displaystyle \int{\frac{x^3}{(x^2+7)^2}\,dx}$

7. ${\displaystyle \int \frac{\ln (x)}{x^2}dx}$$\displaystyle \int{\frac{\ln{(x)}}{x^2}\,dx}$

Homework

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

• Section 7.1: # 19, 43.

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Created: Thursday, 12 August 2021 - 15:40 (EDT) Last Modified: Tuesday, 16 August 2022 - 16:48 (EDT)