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}$

Special Case for Integration by Parts

Example 05: Evaluate ${\displaystyle \int \ln (x)dx}$$\displaystyle\int{\ln{(x)}\,dx}$

Solution 05:

1. There is a product of two functions: $f(x)=1$$f(x)=1$ and $g(x)=\ln (x)$$g(x)=\ln{(x)}$. Thus, there are two possibilities for $U$$U$ and $dV$$dV$:

   Case 01	Case 02
   $U=1$$U=1$	$U=\ln (x)$$U=\ln{(x)}$
   $dV=\ln (x)dx$$dV=\ln{(x)}\,dx$	$dV=dx$$dV=dx$

   Note that Case 01 requires the integral $\int UdV=\int \ln (x)dx$$\int{U\,dV}=\int{\ln{(x)}\,dx}$ which we do not know how to integrate! So, use Case 02:

   • Let $U=\ln (x)\Rightarrow dU=\frac{1}{x}dx$${\color{green}U=\ln{(x)}}\Rightarrow{\color{blue}dU=\frac{1}{x}\,dx}$, and

   • $dV=dx\Rightarrow V=x$${\color{red}dV=dx}\Rightarrow{\color{orange}V=x}$

2. Now Integrate by Parts:

   ${\displaystyle \int \ln (x)dx=x\ln (x)-\int \cancel{x}\left(\frac{1}{\cancel{x}}\right)dx=x\ln (x)-\int dx=x\ln (x)-x+C}$$\displaystyle\int{\ln{(x)}\,dx}={\color{orange}x}\,{\color{green}\ln{(x)}}-\int{{\color{orange}\cancel{x}}\left({\color{blue}\frac{1}{\cancel{x}}}\right)\,\color{blue}dx}=x\,\ln{(x)}-\int{dx}=x\,\ln{(x)}-x+C$ ∎

Example 06: Evaluate ${\displaystyle \int {\tan }^{-1}(x)dx}$$\displaystyle \int{\tan^{-1}{\!(x)}\,dx}$

Solution 06:

1. There is a product of two functions: $f(x)=1$$f(x)=1$ and $g(x)={\tan }^{-1}(x)$$g(x)=\tan^{-1}{\!(x)}$. Thus, there are two possibilities for $U$$U$ and $dV$$dV$:

   Case 01	Case 02
   $U=1$$U=1$	$U={\tan }^{-1}(x)$$U=\tan^{-1}{\!(x)}$
   $dV={\tan }^{-1}(x)dx$$dV=\tan^{-1}{\!(x)}\,dx$	$dV=dx$$dV=dx$

   Note that Case 01 requires the integral $\int UdV=\int {\tan }^{-1}(x)dx$$\int{U\,dV}=\int{\tan^{-1}{\!(x)}\,dx}$ which we do not know how to integrate! So, use Case 02:

   • Let $U={\tan }^{-1}(x)\Rightarrow dU=\frac{1}{x^2+1}dx$${\color{green}U=\tan^{-1}{\!(x)}}\Rightarrow{\color{blue}dU=\frac{1}{x^2+1}\,dx}$, and

   • $dV=dx\Rightarrow V=x$${\color{red}dV=dx}\Rightarrow{\color{orange}V=x}$

2. Now Integrate by Parts:

   ${\displaystyle \int {\tan }^{-1}(x)dx=x{\tan }^{-1}(x)-\int x\left(\frac{1}{x^2+1}\right)dx=x{\tan }^{-1}(x)-\int \frac{x}{x^2+1}dx}$$\displaystyle\int{\tan^{-1}{\!(x)}\,dx}={\color{orange}x}\,{\color{green}\tan^{-1}{\!(x)}}-\int{{\color{orange}{x}}\left({\color{blue}\frac{1}{{x^2+1}}}\right)\,\color{blue}dx}=x\,\tan^{-1}{\!(x)}-\int{\frac{x}{x^2+1}\,dx}$

3. We need to use a Substitution for the remaining integral:

   • Let $u=x^2+1$$\color{purple}u=x^2+1$

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

   ${\displaystyle x{\tan }^{-1}(x)-\int \frac{x}{x^2+1}dx=x{\tan }^{-1}(x)-\int \frac{\cancel{x}}{u}\frac{du}{2\cancel{x}}}$$\displaystyle x\,\tan^{-1}{\!(x)}-\int{\frac{x}{x^2+1}\,dx}=x\,\tan^{-1}{\!(x)}-\int{\frac{\cancel{x}}{\color{purple}u}\,\color{brown}\frac{du}{2\cancel{x}}}$

   ${\displaystyle \Rightarrow x{\tan }^{-1}(x)-\frac{1}{2}\int \frac{1}{u}du=x{\tan }^{-1}(x)-\frac{1}{2}\ln (u)+C}$$\displaystyle \Rightarrow x\,\tan^{-1}{\!(x)}-\frac{1}{2}\int{\frac{1}{u}\,du}=x\,\tan^{-1}{\!(x)}-\frac{1}{2}\ln{(u)}+C$

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

   ${\displaystyle \Rightarrow x{\tan }^{-1}(x)-\frac{1}{2}\ln (u)+C=x{\tan }^{-1}(x)-\frac{1}{2}\ln (x^2+1)+C}$$\displaystyle \Rightarrow x\,\tan^{-1}{\!(x)}-\frac{1}{2}\ln{(u)}+C=x\,\tan^{-1}{\!(x)}-\frac{1}{2}\ln{(x^2+1)}+C$ ∎

Practice 11

Evaluate the following integrals:

1. ${\displaystyle \int x\sin (x)\cos (x)dx}$$\displaystyle \int{x\sin{(x)}\cos{(x)}\,dx}$

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

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

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

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 Example 02: Integrate x^2e^x *)
2
Integrate[x^2*E^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$, ${\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 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:

One Last Case for Integration by Parts

Investigation 12

Evaluate: ${\displaystyle \int e^x\cos (x)dx}$$\displaystyle \int{e^x\cos{(x)}\,dx}$.

Homework

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

• Section 7.1: # 13, 17.

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Last Modified: Thursday, 12 August 2021 - 16:09 (EDT)