Trigonometric Integrals

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

• Section 7.2 – Trigonometric Integrals

Expected Educational Results

• Objective 04–01: I can evaluate an indefinite trigonometric integral.

• Objective 04–02: I can evaluate a definite trigonometric 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.

Definite Trigonometric Integrals

Example 03: Evaluate ${\displaystyle {\int }_{\pi /6}^{\pi /4}{\sin }^2(x){\cos }^5(x)\text{d}x}$$\displaystyle\int_{\pi/6}^{\pi/4}{\sin^2{\!(x)}\,\cos^5{\!(x)}\,\text{d}x}$

Solution 03:

Try $u$$u$-Substitution:

• Let $u=\sin (x)$$\color{red}u=\sin{(x)}$

• $du=\cos (x)dx\Rightarrow dx=\frac{du}{\cos (x)}$$du=\cos{(x)}\,dx\Rightarrow dx=\frac{du}{\cos{(x)}}$

The limits of integration must be converted to $u$$u$-values:

• When $x=\frac{\pi }{6}$$x=\frac{\pi}{6}$, $u=\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$$u=\sin{\left(\frac{\pi}{6}\right)}=\frac{1}{2}$

• When $x=\frac{\pi }{4}$$x=\frac{\pi}{4}$, $u=\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$$u=\sin{\left(\frac{\pi}{4}\right)}=\frac{\sqrt{2}}{2}$

So,

${\displaystyle {\int }_{1/2}^{\sqrt{2}/2}{u}^2{\cos }^{\cancel{5}4}(x)\frac{\text{d}u}{\cancel{\cos (x)}}={\int }_{1/2}^{\sqrt{2}/2}{u}^2{\cos }^4(x)\text{d}u={\int }_{1/2}^{\sqrt{2}/2}{u}^2{({\cos }^2(x))}^2\text{d}u}$$\displaystyle\int_{1/2}^{\sqrt{2}/2}{u^2\,\cos^{\cancel{5}4}{\!(x)}\,\frac{\text{d}u}{\cancel{\cos{(x)}}}}=\int_{1/2}^{\sqrt{2}/2}{u^2\,\cos^{4}{\!(x)}\,\text{d}u}=\int_{1/2}^{\sqrt{2}/2}{u^2\,\left(\cos^{2}{\!(x)}\right)^2\,\text{d}u}$

${\displaystyle \Rightarrow {\int }_{1/2}^{\sqrt{2}/2}{u}^2{({\cos }^2(x))}^2\text{d}u={\int }_{1/2}^{\sqrt{2}/2}{u}^2{(1-{\sin }^2(x))}^2\text{d}u={\int }_{1/2}^{\sqrt{2}/2}{u}^2{(1-u^2)}^2\text{d}u}$$\displaystyle\Rightarrow\int_{1/2}^{\sqrt{2}/2}{u^2\,\left(\cos^{2}{\!(x)}\right)^2\,\text{d}u}=\int_{1/2}^{\sqrt{2}/2}{u^2\,\left(1-{\color{red}\sin^{2}{\!(x)}}\right)^2\,\text{d}u}=\int_{1/2}^{\sqrt{2}/2}{u^2\,\left(1-{\color{red}u^{2}}\right)^2\,\text{d}u}$

Since we converted all $x$$x$-values and functions to $u$$u$-values and functions, $u$$u$-Substitution is a valid method for this integral.

Since there is no product rule or chain rule for antiderivatives, use algebra to rewrite:

${\displaystyle \Rightarrow {\int }_{1/2}^{\sqrt{2}/2}{u}^2{(1-u^2)}^2\text{d}u={\int }_{1/2}^{\sqrt{2}/2}{u}^2(1-2u^2+u^4)\text{d}u={\int }_{1/2}^{\sqrt{2}/2}(u^2-2u^4+u^6)\text{d}u}$$\displaystyle\Rightarrow\int_{1/2}^{\sqrt{2}/2}{u^2\,\left(1-{u^{2}}\right)^2\,\text{d}u}=\int_{1/2}^{\sqrt{2}/2}{u^2\,\left(1-2{u^{2}}+u^4\right)\,\text{d}u}=\int_{1/2}^{\sqrt{2}/2}{\left(u^2-2{u^{4}}+u^6\right)\,\text{d}u}$

Integrate using Power Rule for Antiderivatives:

${\displaystyle \Rightarrow {\int }_{1/2}^{\sqrt{2}/2}(u^2-2u^4+u^6)\text{d}u={(\frac{u^3}{3}-2\frac{u^5}{5}+\frac{u^7}{7})}_{1/2}^{\sqrt{2}/2}}$$\displaystyle\Rightarrow\int_{1/2}^{\sqrt{2}/2}{\left(u^2-2{u^{4}}+u^6\right)\,\text{d}u}=\left(\frac{u^3}{3}-2\frac{u^5}{5}+\frac{u^7}{7}\right)_{1/2}^{\sqrt{2}/2}$

${\displaystyle \Rightarrow \frac{{\left(\frac{\sqrt{2}}{2}\right)}^3}{3}-2\frac{{\left(\frac{\sqrt{2}}{2}\right)}^5}{5}+\frac{{\left(\frac{\sqrt{2}}{2}\right)}^7}{7}-(\frac{{\left(\frac{1}{2}\right)}^3}{3}-2\frac{{\left(\frac{1}{2}\right)}^5}{5}+\frac{{\left(\frac{1}{2}\right)}^7}{7})}$$\displaystyle\Rightarrow \frac{\left(\frac{\sqrt{2}}{2}\right)^3}{3}-2\frac{\left(\frac{\sqrt{2}}{2}\right)^5}{5}+\frac{\left(\frac{\sqrt{2}}{2}\right)^7}{7}-\left(\frac{\left(\frac{1}{2}\right)^3}{3}-2\frac{\left(\frac{1}{2}\right)^5}{5}+\frac{\left(\frac{1}{2}\right)^7}{7}\right)$ ∎

Investigation 11

Evaluate the following definite integrals.

1. ${\displaystyle {\int }_0^{\frac{\pi }{4}}{\sec }^2(x){\tan }^4(x)dx}$$\displaystyle \int_0^{\frac{\pi}{4}}{\sec^2{(x)}\tan^4{(x)}\,dx}$

2. ${\displaystyle {\int }_0^{\frac{\pi }{6}}{\sin }^2(x){\cos }^2(x)dx}$$\displaystyle \int_0^{\frac{\pi}{6}}{\sin^2{(x)}\cos^2{(x)}\,dx}$

3. ${\displaystyle {\int }_0^{\pi }{\sin }^3(x){\cos }^2(x)dx}$$\displaystyle \int_0^{\pi}{\sin^3{(x)}\cos^2{(x)}\,dx}$

4. ${\displaystyle {\int }_0^{\frac{\pi }{3}}\sec (x)dx}$$\displaystyle \int_0^{\frac{\pi}{3}}{\sec{(x)}\,dx}$

5. ${\displaystyle {\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}{\sec }^3(x)\tan (x)dx}$$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{\sec^3{(x)}\tan{(x)}\,dx}$

Use Technology to Verify Definite Integrals

Mathematica

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

1
(* Example from Investigation 10-1: Integrate sec(x)^2 tan(x)^4 from x=0 to x=pi/4 *)
2
Integrate[Sec[x]^2 Tan[x]^4, {x, 0, Pi/4}]

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 ${\cos }^2(x)$$\cos^2{(x)}$ is $\text{Cos[x]}\wedge 2$$\text{Cos[x]}\!\wedge\!2$​, etc.

3. Integrate[ ] for definite integrals has two arguments:

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

   2. $\{x,0,\text{Pi}/4\}$$\{x,0,\text{Pi}/4\}$, where $x$$x$​ is the variable used in the integrand, $a=0$$a=0$ is the lower limit of integration, $b=\pi /4$$b=\pi/4$ is the upper limit of integration.

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

5. 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 concept of the definite 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. The Python syntax for ${\cos }^2(x)$$\cos^2{(x)}$​​​ is $\text{cos(x)**2}$$\text{cos(x)**2}$​​​; etc.

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

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

Homework

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

• Section 7.2: # 3, 7, 9, 11.

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Created: Tuesday, 1 September 2020 04:06 EDT Last Modified: Tuesday, 16 August 2022 - 21:49 (EDT)