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.

Trigonometric Integrals

Pythagorean Identities

From Pre-Calculus, we know:

• ${\displaystyle {\sin }^2(x)+{\cos }^2(x)=1}$$\displaystyle\sin^{2}{\!(x)}+\cos^{2}{\!(x)}=1$

• ${\displaystyle {\tan }^2(x)+1={\sec }^2(x)}$$\displaystyle\tan^{2}{\!(x)}+1=\sec^{2}{\!(x)}$

• ${\displaystyle 1+{\cot }^2(x)={\csc }^2(x)}$$\displaystyle 1+\cot^{2}{\!(x)}=\csc^{2}{\!(x)}$

Specifically, we may need one of the following equivalent expressions:

• ${\displaystyle {\cos }^2(x)=1-{\sin }^2(x)}$$\displaystyle \cos^{2}{\!(x)}=1-\sin^{2}{\!(x)}$

• ${\displaystyle {\sin }^2(x)=1-{\cos }^2(x)}$$\displaystyle \sin^{2}{\!(x)}=1-\cos^{2}{\!(x)}$

• ${\displaystyle {\sec }^2(x)={\tan }^2(x)+1}$$\displaystyle\sec^{2}{\!(x)}=\tan^{2}{\!(x)}+1$

• ${\displaystyle {\tan }^2(x)={\sec }^2(x)-1}$$\displaystyle\tan^{2}{\!(x)}=\sec^{2}{\!(x)}-1$

• ${\displaystyle {\cot }^2(x)={\csc }^2(x)-1}$$\displaystyle \cot^{2}{\!(x)}=\csc^{2}{\!(x)}-1$

• ${\displaystyle {\csc }^2(x)=1+{\cot }^2(x)}$$\displaystyle \csc^{2}{\!(x)}=1+\cot^{2}{\!(x)}$

NOTE: Each trigonometric function in the Pythagorean Identities have an exponent of 2.

IMPORTANT NOTE: We now have six ($6$$6$) equivalent Pythagorean expressions will be helpful in the trig integrals that have a trig function to an even power after an initial substitution.

Example 01: Evaluate ${\displaystyle \int {\tan }^4(x){\sec }^4(x)\text{d}x}$$\displaystyle\int{\tan^4{\!(x)}\sec^4{\!(x)}\,\text{d}x}$

Solution:

There are two compositions in the above integral. Thus, there are two (2) potential functions for $u$$u$-Substitution:

Consider Case 02:

${\displaystyle \int {\tan }^4(x){\sec }^4(x)\text{d}x=\int {\tan }^{\cancel{4}3}(x)u^{\bcancel{4}3}\frac{\text{d}u}{\bcancel{\sec (x)}\cancel{\tan (x)}}=\int {\tan }^3(x)u^4\text{d}u}$$\displaystyle \int{\tan^4{\!(x)}\sec^4{\!(x)}}\,\text{d}x=\int{\tan^{\cancel{4}3}{\!(x)}\,u^{\bcancel{4}3}\,\frac{\text{d}u}{\bcancel{\sec{(x)}}\cancel{\tan{(x)}}}}=\int{\tan^3{\!(x)}\,u^4\,\text{d}u}$

This cannot be integrated until all $x$$x$ functions are converted to $u$$u$ functions. Unfortunately, the only way we know how to convert ${\tan }^3(x)$$\tan^3{\!(x)}$ into a function of $u$$u$ is to use a Pythagorean Identity. However, Pythagorean Identities require even powers.

Consider Case 01:

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

• ${\displaystyle \text{d}u={\sec }^2(x)\text{d}x\Rightarrow \text{d}x=\frac{\text{d}u}{{\sec }^2(x)}}$$\displaystyle\text{d}u=\sec^2{\!(x)}\,\text{d}x\Rightarrow \color{blue}\text{d}x=\frac{\text{d}u}{\sec^2{\!(x)}}$

So, ${\displaystyle \int {\tan }^4(x){\sec }^4(x)\text{d}x=\int u^4{\sec }^{\cancel{4}2}(x)\frac{\text{d}u}{\cancel{{\sec }^2(x)}}=\int u^4{\sec }^2(x)\text{d}u}$$\displaystyle\int{\tan^4{\!(x)}\sec^4{\!(x)}}\,\text{d}x=\int{{\color{red}u}^4\sec^{{\cancel 4}2}{\!(x)}}\,{\color{blue}\frac{\text{d}u}{\cancel{\sec^2{\!(x)}}}}=\int{u^4\sec^2{\!(x)}\,\text{d}u}$

The simplified integral is not integrable since there are two variables, $x$$x$ and $u$$u$. Thus, the integral needs to be rewritten. Since the $\sec (x)$$\sec{(x)}$ is raised to an even power, a Pythagorean Identity, ${\displaystyle {\sec }^2(x)={\tan }^2(x)+1}$$\displaystyle \color{purple}\sec^{\,2}{(x)}=\tan^{\,2}{(x)}+1$, may be used:

${\displaystyle \int u^4{\sec }^2(x)\text{d}u=\int u^4\left({\tan }^2(x)+1\right)\text{d}u}$$\displaystyle\int{u^4\,{\color{purple}\sec^2{\!(x)}}\,\text{d}u}=\int{u^4\left({\color{purple}\tan^2{\!(x)}+1}\right)\,\text{d}u}$

Since $u=\tan (x)$$\color{orange}u=\tan{(x)}$:

${\displaystyle \int u^4({\tan }^2(x)+1)\text{d}u=\int u^4(u^2+1)\text{d}u}$$\displaystyle\int{u^4\left({\color{orange}\tan^2{\!(x)}}+1\right)\,\text{d}u}=\int{u^4\left({\color{orange}u^2}+1\right)\,\text{d}u}$

The last integral above contains only the variable $u$$u$, so $u$$u$-Substitution was the correct method to use.

Since there is no product rule for Antiderivatives, we need to expand the integrand:

${\displaystyle \int u^4(u^2+1)\text{d}u=\int (u^6+u^4)\text{d}u}$$\displaystyle\int{u^4\left(u^2+1\right)\,\text{d}u}=\int{\left(u^6+u^4\right)\,\text{d}u}$

Power Rule for Antiderivatives is now valid:

${\displaystyle \int (u^6+u^4)\text{d}u=\frac{u^7}{7}+\frac{u^5}{5}+C}$$\displaystyle\int{\left(u^6+u^4\right)\,\text{d}u}=\frac{u^7}{7}+\frac{u^5}{5}+C$

Use $u=\tan (x)$$u=\tan{(x)}$ to convert the antiderivative back into terms of $x$$x$:

${\displaystyle \frac{u^7}{7}+\frac{u^5}{5}+C=\frac{{\tan }^7(x)}{7}+\frac{{\tan }^5(x)}{5}+C}$$\displaystyle\frac{u^7}{7}+\frac{u^5}{5}+C=\frac{\tan^7{\!(x)}}{7}+\frac{\tan^5{\!(x)}}{5}+C$

Thus, ${\displaystyle \int {\tan }^4(x){\sec }^4(x)\text{d}x=\frac{{\tan }^7(x)}{7}+\frac{{\tan }^5(x)}{5}+C}$$\displaystyle\int{\tan^4{\!(x)}\,\sec^4{\!(x)}}\,\text{d}x=\frac{\tan^7{\!(x)}}{7}+\frac{\tan^5{\!(x)}}{5}+C$ ∎

Investigation 06

Evaluate the following, using $u$$u$-substitution, then rewriting the integrand using a Pythagorean Identity, then rewriting using algebra:

1. ${\displaystyle \int {\sin }^2(\theta ){\cos }^3(\theta )d\theta }$$\displaystyle\int{\sin^2{\!(\theta)}\cos^{3}{\!(\theta)}\,d\theta}$

2. ${\displaystyle \int {\sin }^4(\theta ){\cos }^3(\theta )d\theta }$$\displaystyle\int{\sin^{4}{\!(\theta)}\cos^{3}{\!(\theta)}\,d\theta}$

3. ${\displaystyle \int {\sin }^3(\theta ){\cos }^3(\theta )d\theta }$$\displaystyle\int{\sin^{3}{\!(\theta)}\cos^{3}{\!(\theta)}\,d\theta}$

4. ${\displaystyle \int {\sin }^3(\theta ){\cos }^4(\theta )d\theta }$$\displaystyle\int{\sin^{3}{\!(\theta)}\cos^{4}{\!(\theta)}\,d\theta}$

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

6. ${\displaystyle \int {\sin }^3(x){\cos }^2(x)dx}$$\displaystyle\int{\sin^{3}{\!(x)}\cos^{2}{\!(x)}\,dx}$

7. ${\displaystyle \int {\sec }^4(t){\tan }^3(t)dt}$$\displaystyle\int{\sec^{4}{\!(t)}\tan^3{\!(t)}\,dt}$

8. ${\displaystyle \int {\sec }^4(t){\tan }^2(t)dt}$$\displaystyle\int{\sec^{4}{\!(t)}\tan^{2}{\!(t)}\,dt}$

9. ${\displaystyle \int {\sec }^6(x){\tan }^3(x)dx}$$\displaystyle\int{\sec^{6}{\!(x)}\tan^{3}{\!(x)}\,dx}$

10. ${\displaystyle \int {\sec }^6(x){\tan }^3(x)dx}$$\displaystyle\int{\sec^{6}{\!(x)}\tan^{3}{\!(x)}\,dx}$

11. ${\displaystyle \int {\sec }^2(x){\tan }^5(x)dx}$$\displaystyle\int{\sec^{2}{\!(x)}\tan^{5}{\!(x)}\,dx}$

12. ${\displaystyle \int {\sec }^3(t){\tan }^3(t)dt}$$\displaystyle\int{\sec^{3}{\!(t)}\tan^{3}{\!(t)}\,dt}$

13. ${\displaystyle \int {\csc }^4(x){\cot }^5(x)dx}$$\displaystyle\int{\csc^{4}{\!(x)}\cot^{5}{\!(x)}\,dx}$

14. ${\displaystyle \int {\csc }^4(x){\cot }^5(x)dx}$$\displaystyle\int{\csc^{4}{\!(x)}\cot^{5}{\!(x)}\,dx}$

15. ${\displaystyle \int {\csc }^2(x){\cot }^3(x)dx}$$\displaystyle\int{\csc^{2}{\!(x)}\cot^{3}{\!(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 Investigation 05: Integrate sin(x)^3 cos(x)^3 *)
2
Integrate[Sin[x]^3 Cos[x]^3, 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 ${\sec }^2(x)$$\sec^2{(x)}$ is $\text{Sec[x]}\wedge 2$$\text{Sec[x]}\!\wedge\!2$​, etc.

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. 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.

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

7. 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 Mathematica syntax is located at: https://www.jjw3.com/Common_Mathematica_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 ${\cos }^2(x)$$\cos^2{(x)}$​​​ is $\text{cos(x)**2}$$\text{cos(x)**2}$​​​; 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 07

Consider ${\displaystyle \int {\sec }^4(t){\tan }^2(t)dt}$$\displaystyle\int{\sec^{\,4}{\!(t)}\,\tan^{\,2}{\!(t)}\,dt}$.

1. What do you notice after the initial substitution of $u=\tan (x)$$u=\tan{(x)}$?

2. What do you notice after the initial substitution of $u=\sec (x)$$u=\sec{(x)}$?

3. Which substitution is required for this integral? Explain.

Homework

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

• Section 7.2: # 1, 5, 13, 15, 21, 23, 25, 27, 29, 31.

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Created: Tuesday, 1 September 2020 03:32 EDT Last Modified: Tuesday, 16 August 2022 - 20:55 (EDT)