Math 141 Calculus II Section 6382 Spring 2020 Quiz #5 Please answer all questions. The maximum score for each question is posted at the beginning of the question, and the maximum score for the quiz is 100 points. Make sure your answers are as complete as possible and show your work/argument. In particular, when there are calculations involved, you should show how you come up with your answers with necessary tables, if applicable. Answers that come straight from program software packages will not be accepted. The quiz is due by 11:59 pm, Sunday, Feb 23, 2020, Eastern Standard Time. IMPORTANT: Per the direction of the Dean's Office, you are requested to include a brief note at the beginning of your submitted quiz, confirming that your work is your own. By typing my signature below, I pledge that this is my own work done in accordance with the UMUC Policy 150.25 - Academic Dishonesty and Plagiarism (http://www.umuc.edu/policies/academicpolicies/aa15025.cfm) on academic dishonesty and plagiarism. I have not received or given any unauthorized assistance on this assignment/examination. _________________________ Electronic Signature Your submitted quiz will be accepted only if you have included this statement. Note: You can use the integral tables if necessary. http://scidiv.bellevuecollege.edu/dh/Calculus_all/CalculusRefFacts 1. (10 pts) Decompose the following fraction. (8.4 #8) 6𝑥 2 −𝑥−1 𝑥 3 −𝑥 2. (10 pts) Integrate the following. (8.4 #22) ∫ 7𝑥 2 +8𝑥−2 𝑥 2 +2𝑥 dx 3. (10 pts) Let k and M be positive constants, and assume the initial population is 𝑀 x(0) = x0. 0 < x0 < 2 (8.4 #38) Hint: Look at the answer to 8.4 #37. a. Solve the differential equation 𝑥 𝑑𝑥 𝑑𝑡 = k x (M-x) for x(t). You can leave it in the form 𝑀−𝑥 = f(t). b. What is the population after a “long” time? (Find the limit for x as t becomes arbitrarily large.) 𝑑𝑥 c. When is the growth rate the largest? (Maximize 𝑑𝑡 .) d. What is the population when the growth rate is the largest? Evaluate the following integrals. 4. (10 pts) ∫ √1 − 36𝑥 2 dx (8.5 #18) 5. (10 pts) ∫ √25− 𝑥 2 𝑥2 𝑑𝑥 (8.5 #22) 1 6. (10 pts) ∫ 𝑥 2−4𝑥+13 𝑑𝑥 (8.5 #40) 𝜋 7. (10 pts) ∫0 𝑐𝑜𝑠 4 (5𝑥)𝑑𝑥 (8.6 #6) 𝜋 8. (10 pts) ∫02 𝑐𝑜𝑠 3 (5𝑥)𝑑𝑥 (similar to 8.6 #8) 9. (10 pts) ∫ 𝑠𝑒𝑐 3 (5𝑥) 𝑡𝑎𝑛2 (5𝑥)𝑑𝑥 (8.6 #18) 10. (10 pts) Show that if m and n are integers with m ≠ n, then 2𝜋 ∫0 sin(𝑚𝑥) sin(𝑛𝑥) 𝑑𝑥 = 0 (8.6 #19) Section 8.4: Partial Fractions, remixed by Jeff Eldridge from work by Dale Hoffman, is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License. © Mathispower 4u. 8.4 Partial Fraction Decomposition 1 Contemporary Calculus 8.4 PARTIAL FRACTION DECOMPOSITION Rational functions (polynomials divided by polynomials) and their integrals are important in mathematics and applications, but if you look through a table of integral formulas, you will find very few formulas for their integrals. Partly that is because the general formulas are rather complicated and have many special cases, and partly it is because they can all be reduced to just a few cases using the algebraic technique discussed in this section, Partial Fraction Decomposition. In algebra you learned to add rational functions to get a single rational function. Partial Fraction Decomposition is a technique for reversing that procedure to "decompose" a single rational function into a sum of simpler rational functions. Then the integral of the single rational function can be evaluated as the sum of the integrals of the simpler functions. Example 1: Solution: Use the algebraic decomposition 17x – 35 2 2x – 5x 7 3 = x + 2x – 5 to evaluate ⌠ ⌡ 17x2– 35 dx . 2x – x The decomposition allows us to exchange the original integral for two much easier ones: 17x – 35 7 3 ⌠ ⌡ 2 ⌡ x + 2x – 5 dx dx = ⌠ 2x – 5x = 7 3 ⌠ ⌡ x dx + ⌠ ⌡ 2x – 5 dx 3 = 7ln| x | + 2 ln| 2x – 5 | + C . Practice 1: Use the algebraic decomposition 7x – 11 4 1 = 3x + 1 + x – 3 to evaluate 2 3x –8x–3 ⌠ 7x2 – 11 dx . ⌡ 3x –8x–3 The Example illustrates how to use a "decomposed" fraction with integrals, but it does not show how to achieve the decomposition. The algebraic basis for the Partial Fraction Decomposition technique is that every polynomial can be factored into a product of linear factors ax + b and irreducible quadratic factors 2 2 ax + bx + c (with b – 4ac < 0). These factors may not be easy to find , and they will typically be more complicated than the examples in this section, but every polynomial has such factors. Before we apply the Partial Fraction Decomposition technique, the fraction must have the following form: (i) (the degree of the numerator) < (degree of the denominator) (ii) The denominator has been factored into a product of linear factors and irreducible quadratic factors. 8.4 Partial Fraction Decomposition 2 Contemporary Calculus If assumption (i) is not true, we can use polynomial division until we get a remainder which has a smaller degree than the denominator. If assumption (ii) is not true, we simply cannot use the Partial Fraction Decomposition technique. Example 2: Put each fraction into a form for Partial Fraction Decomposition: 2 3 2x + 4x – 6 (a) 2 x – 2x 2 3x – 3x – 9x + 8 (b) 2 x –x–6 2 7x + 12x – 12 (c) 3 x – 4x 2 2x + 4x – 6 8x – 6 8x – 6 Solution: (a) = 2+ 2 = 2 + x(x – 2) 2 x – 2x x – 2x 3 2 (b) 3x – 3x – 9x + 8 9x + 8 9x + 8 = 3x + 2 = 3x + (x + 2)(x – 3) 2 x –x–6 x –x–6 (c) 7x + 12x – 12 7x + 12x – 12 = x(x + 2)(x – 2) 3 x – 4x 2 2 Distinct Linear Factors If the denominator can be factored into a product of distinct linear factors, then the original fraction can be number written as the sum of fractions of the form linear factor . Our job is to find the values of the numbers in the numerators, and that typically requires solving a system of equations. 17x – 35 Example 3: Find values for A and B so x(2x – 5) A B = x + 2x – 5 . Solution: We can combine the two terms on the right by putting them over the common denominator x(2x A 2x – 5 B x – 5). Multiplying the x term by 2x – 5 and multiplying the 2x–5 term by x , we have A x Since –5 B .x A2x – 5A + Bx (2A + B)x – 5A . 2x = . 2x – 5 + 2x – 5 x = x(2x – 5) x(2x – 5) (2A + B)x – 5A 17x – 35 = x(2x – 5) , the coefficients of like terms in the numerators must be equal: x(2x – 5) coefficients of x: 2A + B = 17 constant terms: –5A = –35 Solving this system of two equations with two unknowns, we get A = 7 and B = 3 so 17x – 35 x(2x – 5) 7 3 = x + 2x – 5 . 8.4 Partial Fraction Decomposition 7 As a check, add x 3 Contemporary Calculus 3 and 2x – 5 17x – 35 and verify that the sum is x(2x – 5) . Practice 2: Find values of A and B so 6x – 7 (x + 3)(x – 2) A B = x+3 + x–2 . In general, there is one unknown coefficient for each distinct linear factor of the denominator. However, if the number of distinct linear factors is large, we would need to solve a large system of equations for the unknowns. 2 Example 4: Solution: 2x + 7x + 9 A B C Find values for A, B, and C so x(x + 1)(x + 3) = x + x + 1 + x + 3 . A B C x + x+1 + x+3 A (x+1)(x+3) B x(x+3) C x(x+1) = x (x+1)(x+3) + (x + 1) x(x+3) + (x + 3) x(x+1) = A(x+1)(x+3) + Bx(x+3) + Cx(x+1) x(x + 1)(x + 3) 2 = 2 (A + B + C)x + (4A + 3B + C)x + (3A) 2x + 7x + 9 = x(x + 1)(x + 3) . x(x + 1)(x + 3) The coefficients of the like terms in the numerators must be equal: 2 coefficients of x : A +B +C =2 coefficients of x: 4A + 3B + C =7 constant terms: 3A =9 so A = 3, B = –2, and C = 1 . 2 2x + 7x + 9 3 –2 1 Finally, x(x + 1)(x + 3) = x + x + 1 + x + 3 . 2 2x + 7x + 9 ⌠ ⌡ x(x + 1)(x + 3) dx . Practice 3: Use the result of Example 4 to evaluate Practice 4: How large would the system be for a Partial Fraction Decomposition of something ? 5 degree polynomial th The next two subsections describe how to decompose fractions whose denominators contain irreducible quadratic factors and repeated factors. We will not discuss why the suggestions work except to note that they provide enough, but not too many, unknown coefficients for the decomposition. 8.4 Partial Fraction Decomposition 4 Contemporary Calculus Distinct Irreducible Quadratic Factors If the factored denominator includes a distinct irreducible quadratic factor, then the Partial Fraction Decomposition sum contains a fraction of the form of a linear polynomial with unknown coefficients divided by the irreducible quadratic factor: linear polynomial irreducible quadratic factor Ax + B or irreducible quadratic factor . Once again we will solve a system of equations to find the values of the unknown coefficients A and B. 2 Example 5: Solution: Find values for A, B, and C so Ax + B C + x 2 x + 2x + 5 = x + 3x – 15 Ax + B C = 2 + x . 2 (x + 2x + 5)x x + 2x + 5 Ax + B 2 x + 2x + 5 2 2 +5 ( xx ) + Cx ( xx2 ++ 2x ) 2x + 5 2 = Ax + Bx + Cx + 2Cx + 5C 2 x(x + 2x + 5) = (A + C)x + (B + 2C)x + 5C 2 x(x + 2x + 5) 2 2 = x + 3x – 15 2 (x + 2x + 5)x . Then A + C = 1, B + 2C = 3, and 5C = –15 so C = – 3, B = 9, and A = 4. In general, there are 2 unknown coefficients for each distinct irreducible quadratic factor of the denominator. We would start the decomposition of 3 2 6x + 36x + 50x + 53 2 2 (x + 4)(x + 4x + 5) by writing it as the sum Ax + B Cx + D + 2 . 2 x +4 x + 4x + 5 We would finish this decomposition by solving the system of 4 equations with 4 unknowns, A + C = 6, 4A + B + D = 36, 5A + 4B + 4C = 50, and 5B + 4D = 53 to get A = 6, B = 5, C = 0, and D = 7. Repeated Factors If the factored denominator contains a linear factor raised to a power (greater than one), then we need to start the decomposition with several terms. There should be one term with one unknown coefficient for each power of the linear factor. For example, something A B C D 3 = x+1 + x–2 + 2 + 3 . (x + 1)(x – 2) (x – 2) (x – 2) 8.4 Partial Fraction Decomposition 5 Contemporary Calculus Similarly, if the factored denominator contains an irreducible quadratic factor raised to a power greater than one), then we need to start the decomposition with several terms. There should be one term with two unknown coefficients for each power of the irreducible quadratic. For example, something 2 2 3 x (x + 9) A B Cx + D Ex + F Gx + H = x + 2 + 2 + 2 2 + 2 3 . x x +9 (x + 9) (x + 9) This leads to a system of 8 equations with 8 unknowns. 2 Example 6: Decompose –4x + 5x + 3 2 x(x – 1) 2 Solution: –4x + 5x + 3 2 x(x – 1) and evaluate 2 –4x + 5x + 3 ⌠ ⌡ dx . 2 x(x – 1) A B C = x + x–1 + 2 (x – 1) A = x 2 ( (x(x –– 1)1)2 ) B + x–1 – 1) ( x(x x(x – 1) ) + C 2 (x – 1) ( xx ) 2 = A(x – 1) + Bx(x – 1) + Cx 2 x(x – 1) = (A + B)x + (–2A – B + C)x + A –4x + 5x + 3 = . 2 2 x(x – 1) x(x – 1) 2 2 Then A + B = –4, –2A – B + C = 5, and A = 3 so A = 3, B = –7, and C = 4. Finally, 2 –4x + 5x + 3 3 –7 4 –4 ⌠ ⌡ ⌡ x + x–1 + dx = ⌠ 2 2 dx = 3ln| x | – 7ln| x – 1 | + x – 1 + C. x(x – 1) (x – 1) 2 Practice 5: Decompose 2x + 27x + 85 2 (x + 5) and evaluate 2 2x + 27x + 85 ⌠ ⌡ dx . 2 (x + 5) The primary use of the partial fraction technique in this course is to put rational functions in a form that is easier to integrate, but this algebraic technique can also be used to simplify the differentiation of some rational functions. The next example illustrates the use of partial fractions to make a differentiation problem easier. Example 7: For f(x) = 2x + 13 , calculate f '(x), f "(x), and f '''(x). 2 x +x–2 8.4 Partial Fraction Decomposition Solution: 6 Contemporary Calculus You already know how to calculate these derivatives using the quotient rule, but that process is rather tedious for the second and third derivatives here. Instead, we can use the partial fraction technique to 5 3 –1 –1 rewrite f as f(x) = x – 1 – x + 2 = 5(x – 1) – 3(x + 2) . Then the derivatives are very straightforward: f '(x) = –5(x – 1) –2 + 3(x + 2) –3 – 6(x + 2) f "(x) = 10(x – 1) f '''(x) = –30(x – 1) Practice 6: –4 –2 , –3 , and + 18(x + 2) –4 . Use the partial fraction decomposition of g(x) = g "(x), and g 9x + 1 x – 2x – 3 2 to calculate g '(x), (4) (x). PROBLEMS In problems 1 – 12, decompose the fractions. 1. 7x + 2 x(x + 1) 5. 2x + 15x + 25 2 x + 5x 9. 8x – x + 3 3 x +x 2. 7x + 9 (x + 3)(x – 1) 6. 3x + 3x 2 x +x–2 10. 9x + 13x + 15 3 2 x + 2x – 3x 2 3 2 3. 11x + 25 2 x + 9x + 8 7. 11. 2 4. 3x + 7 2 x –1 6x + 9x – 15 x(x + 5)(x – 1) 8. 6x – x – 1 3 x –x 2 11x + 23x + 6 2 x (x + 2) 12. 6x + 14x – 9 2 x(x + 3) 2 2 2 2 In problems 13 – 30, evaluate the integrals. 13. 3x + 13 ⌠ ⌡ (x + 2)(x – 5) dx 5 2x + 11 ⌡ (x – 7)(x – 2) dx 14. ⌠ 15. ⌠ ⌡ 2 2 x –1 2 dx 3 16. 2 5x + 5x + 3 ⌠ ⌡ dx 3 1 x +x 17. Integrate the functions in problems 1 – 4. 18. Integrate the functions in problems 5 – 8. 19. 2 2x + 5x + 3 ⌠ ⌡ dx 2 20. 2 2x + 19x + 22 ⌠ ⌡ dx 2 21. 2 3x + 19x + 24 ⌠ ⌡ dx 2 22. 2 7x + 8x – 2 ⌠ ⌡ dx 2 23. 2 3x – 1 ⌠ ⌡ 3 dx 24. 4 3 x + 5x + x – 15 ⌠ ⌡ dx 2 x –1 x + 2x x + x – 12 x –x x + 6x + 5 x + 5x 8.4 Partial Fraction Decomposition 7 Contemporary Calculus 25. 3 2 x + 3x – 4x + 30 ⌠ ⌡ dx 2 26. 2x + 5 ⌠ ⌡ 2 dx 27. 2 12x + 19x – 6 ⌠ ⌡ dx 3 2 28. 3 2 7x + x + 7x + 10 ⌠ ⌡ dx 4 3 29. 2 7x + 3x + 7 ⌠ ⌡ dx 3 30. 2 7x – 4x + 4 ⌠ ⌡ dx 3 x + 3x – 10 x + 2x (x + 1) x +x x + 3x x +1 31. Integrals are very sensitive to small changes in the integrand. Evaluate (a) ⌠ ⌡ 1 dx x + 2x + 2 2 32. Evaluate (a) ⌠ ⌡ (b) ⌠ ⌡ 1 dx x + 2x + 1 2 1 dx x – 6x + 8 2 (b) ⌠ ⌡ (c) 1 dx x – 6x + 9 2 ⌠ ⌡ 1 dx . x + 2x + 0 2 (c) ⌠ ⌡ 1 dx . x – 6x + 10 2 33. Use the partial fraction decomposition of the functions in problems 1 and 2 to calculate their first and second derivatives. 34. Use the partial fraction decomposition of the functions in problems 3 and 4 to calculate their first and second derivatives. 35. Use the partial fraction decomposition of the functions in problems 5 and 6 to calculate their first and second derivatives. The following two applications involve a type of differential equation which can be solved by separating the variables and using a partial fraction decomposition to help calculate the antiderivatives. The same type of differential equation is also used to model the spread of rumors and diseases as well as some populations and chemical reactions. Logistic Growth: The growth rate of many different populations depends not only on the number of individuals (leading to exponential growth) but also on a "carrying capacity" of the environment. If x is the population at time t and the growth rate of x is proportional to the product of the population and the carrying capacity M minus the population, then the growth rate is described by the differential equation dx . . dt = k x (M – x) where k and M are constants for a given species in a given environment. 8.4 Partial Fraction Decomposition Contemporary Calculus 8 36. Let k = 1 and M = 100, and assume the initial population is x(0) = 5 . dx (a) Solve the differential equation dt = x(100 – x) for x . (b) Graph the population x(t) for 0 ≤ t ≤ 20. (c) When will the population be 20? 50? 90? 100? (d) What is the population after a "long" time? (Find the limit, as t becomes arbitrarily large, of x .) (e) Explain the shape of the graph in (a) in terms of a population of bacteria. (f) When is the growth rate largest? (Maximize dx/dt .) (g) What is the population when the growth rate is largest? 37. Let k = 1 and M = 100, and assume the initial population is x(0) = 150 . dx (a) Solve the differential equation dt = x(100 – x) for x and graph x(t) for 0 ≤ t ≤ 20. (b) When will the population be 120? 110? 100? (c) What is the population after a "long" time? (Find the limit, as t becomes arbitrarily large, of x .) (d) Explain the shape of the graph in (a) in terms of a population of bacteria. 38. Let k and M be positive constants, and assume the initial population is x(0) = x0 . dx (a) Solve the differential equation dt = k.x.(M – x) for x. (b) What is the population after a "long" time? (Find the limit, as t becomes arbitrarily large, of x .) (c) When is the growth rate largest? (Maximize dx/dt .) (d) What is the population when the growth rate is largest? Chemical Reaction: In some chemical reactions, a new material X is formed from materials A and B, and the rate at which X forms is proportional to the product of the amount of A and the amount of B remaining in the solution. Let x represent the amount of material X present at time t, and assume that the reaction begins with a grams of A, b grams of B, and no material X ( x(0) = 0 ). Then the rate of formation of material X can be described by the differential equation dx dt = k(a – x)(b – x). dx 39. Solve the differential equation dt = k(a – x)(b – x) for x if k = 1 and the reaction begins with (i) 7 grams of A and 5 grams of B, and (ii) 6 grams of A and 6 grams of B. dx 40. Solve the differential equation dt = k(a – x)(b – x) for x if k = 1 and the reaction begins with (i) a grams of A and b grams of B with a ≠ b, and (ii) c grams of A and c grams of B (c ≠ 0). 8.4 Partial Fraction Decomposition Section 8.4 Practice 1: 9 Contemporary Calculus PRACTICE Answers ⌠ ⌡ 7x – 11 dx = 2 3x – 8x – 3 4 1 ⌠ ⌡ 3x + 1 dx + ⌠ ⌡ x – 3 dx 4 = 3 .ln|3x + 1| + ln|x – 3| + C . Practice 2: 6x – 7 A B A(x–2) + B(x+3) (A + B)x + (–2A + 3B) = . (x + 3)(x – 2) = x + 3 + x – 2 = (x + 3)(x – 2) (x + 3)(x – 2) This gives us the system: A + B = 6 and –2A + 3B = –7 so (solving) A = 5 and B = 1. 6x – 7 5 1 (x + 3)(x – 2) = x + 3 + x – 2 . Practice 3: From Example 4, 2 2x + 7x + 7 3 –2 1 ⌠ ⌡ x(x + 1)(x + 3) dx = ⌠ ⌡ x + x + 1 + x + 3 dx = 3.ln| x | – 2.ln| x + 1 | + ln| x + 3 | + C . Practice 4: th If the 5 degree polynomial can be factored into a product of 5 distinct linear terms, then we would have something A B E = st + nd + . . . + th 5 degree polynomial 1 term 2 term 5 term th 2 Practice 5: 2 2x + 27x + 85 2x + 27x + 85 7x + 35 7 = 2 = 2+ 2 2 = 2+x+5 . (x + 5) x + 10x + 25 (x + 5) Then Practice 6: . 2 ⌠ 2x + 27x 2+ 85 dx = ⌡ ⌠ 2 + x +7 5 dx = 2x + 7.ln| x + 5 | + C . ⌡ (x + 5) 9x + 1 A B A(x+1) + B(x–3) (A + B)x + (A – 3B) g(x) = (x – 3)(x + 1) = x – 3 + x + 1 = (x – 3)(x + 1) = (x – 3)(x + 1) This gives us the system A + B = 9 and A – 3B = 1 so (solving) A = 7 and B = 2. 7 2 –1 –1 g(x) = x – 3 + x + 1 = 7(x – 3) + 2(x + 1) . Then g '(x) = –7(x – 3) –2 g "(x) = 14(x – 3) – 2(x + 1) –3 g '''(x) = –42(x – 3) + 4(x + 1) –4 g ''''(x) = 168(x – 3) –2 –5 , –3 – 12(x + 1) , –4 , and –5 . + 48(x + 1) . Section 8.5: Trig Substitution, remixed by Jeff Eldridge from work by Dale Hoffman, is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License. © Mathispower 4u. 8.5 Trigonometric Substitution 1 Contemporary Calculus 8.5 Trigonometric Substitution –– Another Change of Variable Changing the variable is a very powerful technique for finding antiderivatives, and by now you have probably found a lot of integrals by setting u = something. This section also involves a change of variable, but for more specialized patterns, and the change is more complicated. Another difference from previous work is that instead of setting u equal to a function of x we will be replacing x with a function of θ. The next three examples illustrate the typical steps involved making trigonometric substitutions. After these examples, we examine each step in more detail and consider how to make the appropriate decisions. Example 1: Solution: In the expression 9–x 2 replace x with 3 sin(θ) and simplify the result. Replacing x with 3sin(θ), 9 – ( 3sin(θ) ) Example 2: Evaluate 2 9–x 2 2 = 9 – 9sin (θ) = becomes 9.(1 – sin (θ) ) = 3 cos(θ) . 2 ⌠ ⌡ 9 – x2 dx using the change of variable x = 3 sin(θ) and then use the 3 antiderivative to evaluate ⌠ 9 – x2 dx. ⌡ 0 Solution: 2 If x = 3 sin(θ), then dx = 3 cos(θ) dθ and 9 – x = 3 cos(θ). With this change of variable, the integral becomes ⌠ ⌡ 9 – x2 dx = θ sin(2θ) ⌠ } +C ⌡ 3 cos(θ) 3 cos(θ) dθ = 9 ⌠ ⌡ cos2(θ) dθ = 9 { 2 – 4 θ 2sin(θ)cos(θ) 9 = 9{2 – } + C = 2 { θ – sin(θ)cos(θ) } + C 4 This antiderivative, a function of the variable θ, can be converted back to a function of the variable x. Since x = 3 sin(θ) we can solve for θ to get θ = arcsin( x/3 ). Replacing θ with arcsin( x/3 ) in the antiderivative, we get 9 2 { θ – sin(θ)cos(θ) } + C 9 = 2 9 = 2 x { arcsin( 3 { arcsin(x/3) – sin( arcsin(x/3) )cos( arcsin(x/3) ) x ) + 3 2 9–x 3 }+C= 9 x 1 2 arcsin( 3 ) + 2 x Using this antiderivative, we can evaluate the definite integral: }+C 2 9 – x + C. 8.5 Trigonometric Substitution 2 Contemporary Calculus 3 9 x x ⌠ ⌡ 9 – x2 dx = 2 .arcsin( 3 ) + 2 9 – x2 0 9 3 3 = { 2 arcsin( 3 ) + 2 9–3 3 |0 2 9 }–{2 0 0 arcsin( 3 ) + 2 9–0 2 } 9π = 4 . 3 Example 3: The definite integral ⌠ ⌡ 9 – x2 dx represents the area of what region? 0 Solution: The area of one fourth of the circle of radius 3 which lies in the first quadrant (Fig. 1). The area of this quarter circle is area of whole circle 1 1 9 2 2 = 4 πr = 4 π3 = 4 π 4 which agrees with the value found in the previous example. Each Trigonometric Substitution involves four major steps: 1. Choose which substitution to make, x = a trigonomteric function of θ . 2. Rewrite the original integral in terms of θ and dθ . 3. Find an antiderivative of the new integral. 4. Write the antiderivative in step 3 in terms of the original variable x. The rest of this section discusses each of these steps. The first step requires you to make a decision. Then the other three steps follow from that decision. For most students, the key to success with the Trigonometric Substitution technique is to THINK TRIANGLES. Step 1: Choosing the substitution The first step requires that you make a decision, and the pattern of the familiar Pythagorean Theorem can help you make the correct choice. Pythagorean Theorem: 2 2 2 2 2 The pattern 3 + x matches the Pythagorean pattern if 3 and x are sides of a right triangle. For a right triangle with sides 3 and x (Fig. 2), we know tan( θ ) = opposite/adjacent = x/3 so x = 3 tan( θ ). 2 2 2 (side) + (side) = (hypotenuse) or (side) = (hypotenuse) – (side) . 8.5 Trigonometric Substitution 2 3 Contemporary Calculus 2 The pattern 3 – x matches the Pythagorean pattern if 3 is the hypotenuse and x is a side of a right triangle (Fig. 3). Then sin( θ ) = opposite/hypotenuse = x/3 so x = 3 sin( θ ). 2 2 The pattern x – 3 matches the Pythagorean pattern if x is the hypotenuse and 3 is a side of a right triangle (Fig. 4). Then sec( θ ) = hypotenuse/adjacent = x/3 so x = 3 sec( θ ). Once the choice has been made for the substitution, then several things follow automatically: dx can be calculated by differentiating x with respect to θ , θ can be found by solving the substitution equation for θ , ( if x = 3 tan( θ ) then tan( θ ) = x/3 so θ = arctan( x/3 ) ) , and 2 2 2 2 2 2 the patterns 3 + x , 3 – x , and x – 3 can be simplified using algebra and the 2 2 2 2 2 2 trigonometric identities 1 + tan (θ) = sec (θ), 1 – sin (θ) = cos (θ), and sec (θ) – 1 = tan (θ). These results are collected in the table below. 2 3 +x 2 2 (Fig. 2) 3 –x 2 2 (Fig. 3) x –3 2 (Fig. 4) Put x = 3 tan( θ ) . Put x = 3 sin( θ ) . Put x = 3 sec( θ ) . Then Then Then 2 dx = 3 sec ( θ ) dθ dx = 3 cos( θ ) dθ dx = 3 sec( θ ) tan( θ ) dθ x θ = arctan( 3 ) x θ = arcsin( 3 ) x θ = arcsec( 3 ) 2 3 +x 2 2 3 –x 2 2 = 3 – 3 sin ( θ ) 2 2 = 3 ( 1 – sin ( θ ) ) 2 2 = 3 sec ( θ ) Example 4: 2 2 = 3 ( 1 + tan ( θ ) ) 2 2 x –3 2 = 3 + 3 tan ( θ ) 2 2 2 2 = 3 sec ( θ ) – 3 2 = 3 ( sec ( θ ) – 1 ) = 3 cos ( θ ) 2 2 2 2 2 2 2 2 2 = 3 tan ( θ ) 2 For the patterns 16 – x and 5 + x , (a) decide on the appropriate substitution for x, (b) calculate dx and θ , and (c) use the substitution to simplify the pattern. 8.5 Trigonometric Substitution 4 Contemporary Calculus 2 Solution: 16 – x : This matches the Pythagorean pattern if 4 is a hypotenuse and x is the side of a right triangle. Then sin( θ ) = opposite/hypotenuse = x/4 so x = 4 sin( θ ) . For x = 4 sin(θ), dx = 4 cos(θ) dθ and θ = arcsin( x/4 ). Finally, 2 2 2 2 2 16 – x = 16 – ( 4 sin(θ) ) = 16 – 16sin (θ) = 16( 1 – sin (θ) ) = 16 cos (θ) . 2 5 + x : This matches the Pythagorean pattern if x and 5 are the sides of a right triangle. Then tan( θ ) = opposite/adjacent = x/ 5 so x = 5 tan( θ ) . For x = 5 tan(θ), dx = 5 2 sec (θ) dθ and θ = arctan( x/ 5 ). Finally, 2 2 2 2 2 5 + x = 5 + ( 5 tan(θ) ) = 5 + 5 tan (θ) = 5( 1 + tan (θ) ) = 5 sec (θ) . 2 Practice 1: 2 For the patterns 25 + x and x – 13, (a) decide on the appropriate substitution for x, (b) calculate dx and θ , and (c) use the substitution to simplify the pattern. Step 2: Rewriting the integral in terms of θ and dθ Once we decide on the appropriate substitution, calculate dx , and simplify the the pattern, then the second step is very straightforward. Example 5: Solution: Use the substitution x = 5 tan(θ) to rewrite the integral 1 ⌠ ⌡ 25 + x 2 dx in terms of θ and dθ. 2 Since x = 5 tan(θ) , then dx = 5 sec (θ) dθ and 2 2 2 2 2 25 + x = 25 + ( 5 tan(θ) ) = 25 + 25 tan (θ) = 25{ 1 + tan (θ) ) = 25 sec (θ) . Finally, ⌠ ⌡ Practice 2: Steps 3 & 4: 1 25 + x 2 dx = ⌠ ⌡ 1 2 5 sec (θ) 5 sec (θ) dθ = ⌠ ⌡ 5 sec(θ) dθ = ⌠ ⌡ sec(θ) dθ . 2 2 25 sec (θ) Use the substitution x = 5 sin(θ) to rewrite the integral ⌠ ⌡ 1 25 – x 2 dx in terms of θ and dθ. Finding an antiderivative of the new integral & writing the answer in terms of x After changing the variable, the new integral typically involves trigonometric functions and we can use any of our previous methods (a change of variable, integration by parts, a trigonometric identity, or the integral tables) to find an antiderivative. 8.5 Trigonometric Substitution 5 Contemporary Calculus Once we have an antiderivative, usually a trigonometric function of θ , we can replace θ with the appropriate inverse trigonometric function of x and simplify. Since the antiderivatives commonly contain trigonometric functions, we frequently need to simplify a trigonometric function of an inverse trigonometric function, and it is very helpful to refer back to the right triangle we used at the beginning of the substitution process. Example 6: By replacing x with 5 tan(θ) , becomes 1 ⌠ ⌡ 25 + x 2 dx ⌠ ⌡ sec(θ) dθ . Evaluate ⌠ ⌡ sec(θ) dθ and write the resulting antiderivative in terms of the variable x. Solution: x = 5 tan(θ) so θ = arctan( x/5 ) (Fig. 5). Then ⌠ ⌡ sec(θ) dθ = ln| sec(θ) + tan(θ) | + C = ln| sec( arctan(x/5) ) + tan( arctan(x/5) ) | + C By referring to the right triangle in Fig. 5, we see that sec( arctan(x/5) ) = 25 + x 5 2 x and tan( arctan(x/5) ) = 5 25 + x 5 | ln| sec( arctan(x/5) ) + tan( arctan(x/5) ) | + C= ln Putting these pieces together, we have Practice 3: ⌠ ⌡ 1 x 25 + x Show that by replacing x with 3 sin(θ) , 2 2 x + 5 dx = ln ⌠ ⌡ so | +C. 25 + x 5 | 1 2 2 2 x + 5 | +C. 1⌠ 2 dx becomes 9 ⌡ csc (θ) dθ . x 9–x 1⌠ 2 Evaluate 9 ⌡ csc (θ) dθ and write the resulting antiderivative in terms of the variable x. Sometimes it is useful to "complete the square" in an irreducible quadratic to make the pattern more obvious. Example 7: ⌠ Rewrite x + 2x + 26 by completing the square and evaluate ⌡ 2 1 2 dx . x + 2x + 26 Solution: 2 2 x + 2x + 26 = (x + 1) + 25 so ⌠ ⌡ 1 2 x + 2x + 26 Put u = x + 1. Then du = dx , and dx = ⌠ ⌡ 1 2 (x + 1) + 25 dx . 8.5 Trigonometric Substitution ⌠ ⌡ 1 dx = 2 6 Contemporary Calculus ⌠ ⌡ 1 dx 2 x + 2x + 26 (x + 1) + 25 = ⌠ ⌡ 1 du 2 u + 25 2 = ln | 25 + u 5 u + 5 = ln | 25 + (x+1) 5 |+C 2 + x+1 5 (using the result of Example 6) |+C = ln 2 x + 2x + 26 x+1 + 5 5 | |+C THINK TRIANGLES. The first and last steps of the method (choosing the substitution and writing the answer interms of x) are easier if you understand the triangles (Figures 2, 3, and 4) and have drawn the appropriate triangle for the problem. Of course, you also need to practice the method. PROBLEMS In problems 1–6, (a) make the given substitution and simplify the result, and (b) calculate dx. 1. x = 3.sin(θ) in 1 9–x 3. x = 3.sec(θ) in 2. 2 1 2 1 x = 2 .tan(θ) in 2+x In problems 7–12, 2 x +9 4. x = 6.sin(θ) in 6. x = sec(θ) in x –9 5. 1 x = 3.tan(θ) in 2 1 2 36 – x 1 x –1 2 (a) solve for θ as a function of x, (b) replace θ in f(θ) with you result in part (a), and (c) simplify. 7. x = 3.sin(θ), f(θ) = cos(θ).tan(θ) 8. 9. x = 3.sec(θ), f(θ) = cos(θ) 10. x = 5.sin(θ), f(θ) = 1 + sec(θ) 2 1 + sin (θ) 2 cos (θ) 11. x = 5.tan(θ), f(θ) = 1 + cot(θ) x = 3.tan(θ), f(θ) = sin(θ).tan(θ) 12. x = 5.sec(θ), f(θ) = cos(θ) + 7.tan (θ) 2 8.5 Trigonometric Substitution 7 Contemporary Calculus In problems 13–36, evaluate the integrals. (More than one method works for some of the integrals.) 13. ⌠ ⌡ 1 x 9–x 2 14. ⌠ ⌡ x 2 9–x dx 2 15. ⌠ ⌡ 1 dx 2 x + 49 dx 17. ⌠ ⌡ 36 – x2 dx 18. ⌠ ⌡ 1 – 36x2 dx 2 dx 20. ⌠ ⌡ 21. ⌠ ⌡ ⌠ 25 – x2 ⌡ dx 2 23. 25. ⌠ ⌡ x dx 2 x + 49 26. ⌠ ⌡ 28. ⌠ ⌡ 1 2 3/2 dx (4x – 1) 29. ⌠ ⌡ 1 2 dx 25 – x 32. 16. ⌠ ⌡ 2 dx 1 2 x +1 19. 22. 31. 34. ⌠ ⌡ 1 36 + x x 25 – x x ⌠ ⌡ ⌠ ⌡ 1 2 2 x. a + x 1 ⌠ ⌡ 2 35. 2 dx dx 24. ⌠ ⌡ 1 dx 2 x + 49 1 dx 2 49x + 25 27. ⌠ ⌡ 1 2 3/2 dx (x – 9) 30. ⌠ ⌡ 2 5 dx 2 2x x – 25 ⌠ ⌡ 1 49 + x x 25 – x dx dx x 3–x 1 2 dx a +x 33. 2 ⌠ ⌡ 1 x 2 2 a +x 2 1 ⌠ ⌡ 1 2 a +x dx 36. ⌠ ⌡ 2 2 dx dx 1 2 2 3/2 dx (a + x ) In problems 37–42, first complete the square, make the appropriate substitutions, and evaluate the integral. 37. ⌠ ⌡ 1 2 dx 38. ⌠ ⌡ (x+1) + 9 40. ⌠ ⌡ 1 dx 2 x – 4x + 13 1 2 dx 39. ⌠ ⌡ 42. ⌠ ⌡ (x+3) + 1 41. ⌠ ⌡ 1 2 x + 4x + 3 dx 1 dx 2 x + 10x + 29 1 2 x – 6x – 16 dx 8.5 Trigonometric Substitution Section 8.5 8 Contemporary Calculus Practice Answers 2 Practice 1: 25 + x : (a) Put x = 5.tan(θ) (b) Then dx = 5.sec (θ) dθ and θ = arctan( x/5 ) 2 (c) 25 + x = 25 + 25.tan (θ) = 25( 1 + tan (θ) ) = 25.sec (θ) 2 2 2 2 (a) Put x = 13 .sec(θ) 2 x – 13 : (b) Then dx = 13 .sec(θ).tan(θ) dθ and θ = arcsec( x/ 13 ) (c) x – 13 = 13.sec (θ) – 13 = 13( sec (θ) – 1 ) = 13.tan (θ) 2 = 1 25 – x 2 25.cos (θ) ⌠ ⌡ dx = 2 1 ⌠ ⌡ 1 2 2 2 5.cos(θ) dθ 25 – sin (θ) 5.cos(θ) dθ = ⌠ ⌡ 1 dθ = θ + C = arcsin( x/5 ) + C x = 3.sin(θ) so dx = 3.cos(θ) dθ and 9 – x = 9( 1 – sin (θ) ) = 9.cos (θ) . 2 Practice 3: Then 2 2 ⌠ ⌡ ⌠ ⌡ 2 x = 5.sin(θ) so dx = 5.cos(θ) dθ and 25 – x = 25( 1 – sin (θ) ) = 25.cos (θ) . Practice 2: Then 2 1 x 2 9–x 1 = 9 2 dx = 2 2 1 1 ⌠ 3.cos(θ) dθ = ⌠ 3.cos(θ) dθ ⌡ . 2 ⌡ 2 2 2 . . 9 sin (θ) 9 – sin (θ) 9 sin (θ) 9 cos (θ) 2 1 1 1 9–x ⌠ + C. ⌡ csc2(θ) dθ = – 9 cot() + C = – 9 cot( arcsin(x/3) ) + C = – 9 x Section 8.6: Trigonometric Integration, remixed by Jeff Eldridge from work by Dale Hoffman, is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License. © Mathispower 4u. 8.6 Integrals of Trigonometric Functions 1 Contemporary Calculus 8.6 Integrals of Trigonometric Functions There are an overwhelming number of combinations of trigonometric functions which appear in integrals, but fortunately they fall into a few patterns and most of their integrals can be found using reduction formulas and tables of integrals. This section examines some of the patterns of these combinations and illustrates how some of their integrals can be derived. Products of Sine and Cosine: ⌠ ⌡ sin(ax).sin(bx) dx , ⌠ ⌡ cos(ax). cos(bx) dx , ⌠ ⌡ sin(ax).cos(bx) dx All of these integrals are handled by referring to the trigonometric identities for sine and cosine of sums and differences: sin(A + B) = sin(A)cos(B) + cos(A)sin(B) sin(A – B) = sin(A)cos(B) – cos(A)sin(B) cos(A + B) = cos(A)cos(B) – sin(A)sin(B) cos(A – B) = cos(A)cos(B) + sin(A)sin(B) By adding or subtracting the appropriate pairs of identities, we can write the various products such as sin(ax)cos(bx) as a sum or difference of single sines or cosines. For example, by adding the first two 1 identities we get 2sin(A)cos(B) = sin(A + B) + sin(A – B) so sin(A)cos(B) = 2 { sin(A+B) + sin(A–B) }. Using this last identity, the integral of sin(ax)cos(bx) for a ≠ b is relatively easy: –cos( (a–b)x ) –cos( (a+b)x ) ⌠ ⌡ 12 { sin( (a+b)x ) + sin( (a–b)x ) } dx = 12 { + } + C. ⌡ sin(ax)cos(bx) dx = ⌠ a–b a+b The other integrals of products of sine and cosine follow in a similar manner. If a ≠ b, then ⌠ ⌡ sin(ax).sin(bx) dx ⌠ ⌡ cos(ax).cos(bx) dx = " ! 1 { sin( (a–b)x ) sin( (a+b)x ) – a–b a+b }+C 1 2 { sin( (a–b)x ) sin( (a+b)x ) + a–b a+b }+C = 2 sin(ax).cos(bx) dx = 2 –1 { cos( (a–b)x ) cos( (a+b)x ) + a–b a+b }+C 8.6 Integrals of Trigonometric Functions 2 Contemporary Calculus If a = b, we have patterns we have already used. ⌠ sin2(ax) dx ⌡ x sin(2ax) x sin(ax).cos(ax) = 2 – + C = 2 – +C 4a 2a ⌠ ⌡ cos2(ax) dx x sin(2ax) x sin(ax).cos(ax) = 2 + + C = + +C 4a 2 2a 2 ⌠ sin(ax).cos(ax) dx = sin2a(ax) + C = 1 – cos(2ax) +C ⌡ 4a 2 The first and second of these integral formulas follow from the identities sin (ax) = 2 cos (ax) = 1 – cos(2ax) and 2 1 + cos(2ax) , and the third can be derived by changing the variable to u = sin(ax). 2 Powers of Sine and Cosine Alone: ⌠ ⌡ sinn(x) dx , ⌠ ⌡ cosn(x) dx All of these antiderivatives can be found using integration by parts or the reduction formulas (formulas 19 and 20 in the integral tables) which were derived using integration by parts. For small values of m and n it is just as easy to find the antiderivatives directly. Even Powers of Sine or Cosine Alone For even powers of sine or cosine, we can successfully reduce the size of the exponent by repeatedly 2 applying the identities sin (x) = Example 1: Solution: Evaluate 1 – cos(2x) 2 2 and cos (x) = 1 + cos(2x) . 2 ⌠ ⌡ sin4(x) dx . 1 1 4 2 2 2 2 sin (x) = { sin (x) } = { 2 [ 1 – cos(2x) ] } = 4 { 1 – 2cos(2x) + cos (2x) } so 1 ⌠ sin4(x) dx = ⌠ ⌡ 4 { 1 – 2cos(2x) + cos2(2x) } dx ⌡ 1 = 4 Practice 1: Evaluate {x + sin(2x) + ⌠ ⌡ cos4(x) dx . x sin(2x)cos(2x) 2 + 2 } +C. 8.6 Integrals of Trigonometric Functions 3 Contemporary Calculus Odd Powers of Sine or Cosine Alone For odd powers of sine or cosine we can split off one factor of sine or cosine, reduce the remaining even 2 2 2 2 exponent using the identities sin (x) = 1 – cos (x) or cos (x) = 1 – sin (x) , and finally integrate by changing the variable. Example 2: Solution: Evaluate ⌠ ⌡ sin5(x) dx . 5 4 2 2 sin (x) = sin (x) sin(x) = { sin (x) } sin(x) 2 2 = { 1 – cos (x) } sin(x) 2 4 = { 1 – 2 cos (x) + cos (x) } sin(x) . Then ⌠ sin5(x) dx = ⌡ ⌠ sin(x) dx – 2 ⌡ ⌠ cos2(x)sin(x) dx + ⌡ ⌠ cos4(x)sin(x) dx . ⌡ The first integral is easy, and the last two can be evaluated by changing the variable to u = cos(x) : 3 5 cos (x) cos (x) ⌠ } + {– 5 }+C. ⌡ sin5(x) dx = – cos(x) – 2{ – 3 Practice 2: Patterns for Evaluate ⌠ ⌡ cos5(x) dx . ⌠ ⌡ sinm(x) cosn(x) dx If the exponent of sine is odd, we can split off one factor sin(x) and use the identity 2 2 sin (x) = 1 – cos (x) to rewrite the remaining even power of sine in terms of cosine. Then the change of variable u = cos(x) makes all of the integrals straightforward. Example 3: Evaluate 3 6 ⌠ sin3(x) cos6(x) dx . ⌡ 2 6 2 6 Solution: sin (x) cos (x) = sin(x) sin (x) cos (x) = sin(x) { 1 – cos (x) } cos (x) 6 8 = sin(x)cos (x) – sin(x)cos (x) . Then ⌠ ⌡ sin3(x) cos6(x) dx = ⌠ ⌡ sin(x)cos6(x) – sin(x)cos8(x) dx ( put u = cos(x) ) 7 = – Practice 3: Evaluate ⌠ sin3(x) cos4(x) dx . ⌡ 9 cos (x) cos (x) + +C. 7 9 8.6 Integrals of Trigonometric Functions 4 Contemporary Calculus If the exponent of cosine is odd, we can split off one factor cos(x) and use the identity 2 2 cos (x) = 1 – sin (x) to rewrite the remaining even power of cosine in terms of sine. Then the change of variable u = sin(x) makes all of the integrals straightforward. 1 2 If both exponents are even, we can use the identities sin (x) = 2 (1 – cos(2x) ) and 1 2 cos (x) = 2 (1 + cos(2x) ) to rewrite the integral in terms of powers of cos(2x) and then proceed with integrating even powers of cosine. Powers of Secant and Tangent Alone: ⌠ ⌡ secn(x) dx , ⌠ ⌡ tann(x) dx All of the integrals of powers of secant and tangent can be evaluated by knowing ⌠ ⌡ sec(x) dx = ln| sec(x) + tan(x) | + C and ⌠ ⌡ tan(x) dx = – ln| cos(x) | + C = ln| sec(x) | + C and then using the reduction formulas n–2 . tan(x) n–2 ⌠ n–2 ⌠ secn(x) dx = sec n(x) + n–1 ⌡ sec (x) dx and ⌡ –1 n–1 tan (x) ⌠ ⌡ tann(x) dx = n – 1 – ⌠ ⌡ tann–2(x) dx . Example 4: Solution: Evaluate ⌠ ⌡ sec3(x) dx . Using the reduction formula with n = 3, . . ⌠ sec3(x) dx = sec(x)2tan(x) + 12 ⌡ ⌠ sec(x) dx = sec(x)2tan(x) + ⌡ 1 2 ln| sec(x) + tan(x) | + C. ⌠ ⌡ tan3(x) dx and ⌠ ⌡ sec5(x) dx . Practice 4: Evaluate Patterns for ⌠ ⌡ secm(x).tann(x) dx The patterns for evaluating ⌠ secm(x).tann(x) dx are similar to those for ⌡ ⌠ sinm(x).cosn(x) dx ⌡ 2 2 because we treat the even and odd powers differently and we use the identities tan (x) = sec (x) – 1 and 2 2 sec (x) = tan (x) + 1. 8.6 Integrals of Trigonometric Functions 5 Contemporary Calculus 2 If the exponent of secant is even, factor off sec (x), replace the other even powers (if any) of secant using 2 2 2 sec (x) = tan (x) + 1, and make the change of variable u = tan(x) (then du = sec (x) dx ). If the exponent of tangent is odd, factor off sec(x)tan(x), replace the remaining even powers (if any) of 2 2 tangent using tan (x) = sec (x) – 1, and make the change of variable u = sec(x) (then du = sec(x)tan(x) dx ). If the exponent of secant is odd and the exponent of tangent is even, replace the even powers of tangent 2 2 using tan (x) = sec (x) – 1. Then the integral contains only powers of secant, and we can use the patterns for integrating powers of secant alone. Example 5: Solution: ⌠ ⌡ sec(x).tan2(x) dx . Evaluate Since the exponent of secant is odd and and the exponent of tangent is even, we can use the 2 2 last method mentions: replace the even powers of tangent using tan (x) = sec (x) – 1. Then ⌠ ⌡ sec(x).tan2(x) dx = ⌠ ⌡ sec(x).{ sec2(x) – 1 } dx = ⌠ sec3(x) – sec(x) dx = ⌡ ⌠ sec3(x) dx – ⌡ ⌠ sec(x) dx ⌡ sec(x).tan(x) 2 ={ = Practice 5: sec(x).tan(x) 2 Evaluate 1 + 2 ln| sec(x) + tan(x) | } – ln| sec(x) + tan(x) | + C 1 – 2 ln| sec(x) + tan(x) | + C. ⌠ sec4(x).tan2(x) dx . ⌡ Wrap Up Even if you use tables of integrals (or computers) for most of your future work, it is important to realize that most of the integral formulas can be derived from some basic facts using the techniques we have discussed in this and earlier sections. PROBLEMS Evaluate the integrals. (More than one method works for some of the integrals.) 1. ⌠ 2 ⌡ sin (3x) dx ⌠1 2 4. ⌡ x .sin ( ln(x) ) dx 2. ⌠ 2 ⌡ cos (5x) dx π ⌠ 4 5. ⌡ sin (3x) dx 0 3. ⌠ x. x x ⌡ e sin(e ).cos(e ) dx π ⌠ 4 6. ⌡ cos (5x) dx 0 8.6 Integrals of Trigonometric Functions π ⌠ 3 7. ⌡ sin (7x) dx 6 Contemporary Calculus π ⌠ 3 8. ⌡ cos (5x) dx 0 9. ⌠ ⌡ sin(7x).cos(7x) dx 0 10. ⌠ ⌡ sin(7x).cos2(7x) dx 11. ⌠ ⌡ sin(7x).cos3(7x) dx 12. ⌠ ⌡ sin2(3x).cos(3x) dx 13. ⌠ ⌡ sin2(3x).cos2(3x) dx 14. ⌠ ⌡ sin2(3x).cos3(3x) dx 15. ⌠ ⌡ sec2(5x).tan(5x) dx 16. ⌠ ⌡ sec2(3x).tan2(3x) dx 17. ⌠ ⌡ sec3(3x).tan(3x) dx 18. ⌠ ⌡ sec3(5x).tan2(5x) dx The definite integrals of various combinations of sine and cosine on the interval [0, 2π] exhibit a number of interesting patterrns. For now these patterns are simply curiousities and a source of additional problems for practice, but the patterns are very important as the foundation for an applied topic, Fourier Series, that you may encounter in more advanced courses. The next three problems ask you to show that the definite integral on [0, 2π] of sin(mx) multiplied by almost any other combination of sin(nx) or cos(nx) is 0. The only nonzero value comes when sin(mx) is multiplied by itself. 2π 19. Show that if m and n are integers with m ≠ n, then ⌠ ⌡ sin(mx).sin(nx) dx = 0. 0 2π 20. Show that if m and n are integers, then ⌠ ⌡ sin(mx).cos(nx) dx = 0. (Consider m = n and m ≠ n.) 0 2π 21. Show that if m ≠ 0 is an integer, then ⌠ ⌡ sin(mx).sin(mx) dx = π. 0 22. Suppose P(x) = 5.sin(x) + 7.cos(x) – 4.sin(2x) + 8.cos(2x) – 2.sin(3x). (This is called a trigonometric polynomial.) Use the results of problems 19–21 to quickly evaluate 2π (a) (c) 1 a1 = π 1 a3 = π ⌠ ⌡ sin(1x).P(x) dx 0 2π ⌠ ⌡ sin(3x).P(x) dx 0 2π 1 (b) a2 = π 1 (d) a4 = π ⌠ ⌡ sin(2x).P(x) dx 0 2π ⌠ ⌡ sin(4x).P(x) dx 0 (e) Describe how the values of ai are related to the coeffiecients of P(x). (f) Make up your own trigonometric polynomial P(x) and see if your description in part (e) holds for the ai values calculated from the new P(x). (g) Just by knowing the ai values we can "rebuild" part of P(x). Find a similar method for getting the coefficients of the cosine terms of P(x): bi = ?? 8.6 Integrals of Trigonometric Functions 7 Contemporary Calculus 2π 23. Show that if n is a positive, odd integer, then ⌠ ⌡ sinn(x) dx = 0. 0 2π 2π 0 0 ⌠ 2 ⌠ 24. It is straightforward (using formula 19 in the integral table) to show that ⌡ sin (x) dx = π, ⌡ 2π ⌠ ⌡ sin6(x) dx = 5 3 π. 64 3 sin (x) dx = 4 π, and 4 2π (a) Evaluate 0 ⌠ ⌡ sin8(x) dx . 0 2π (b) Predict the value of ⌠ 10 ⌡ sin (x) dx and then evalaute the integral. 0 Section 8.6 Practice 1: Practice Answers ⌠ ⌡ cos4(x) dx 1 2 { Use cos (x) = 2 ( 1 + cos(2x) ) } =⌠ ⌡ cos (x).cos (x) dx 2 Practice 2: 2 1 1 =⌠ ⌡ 2 ( 1 + cos(2x) ) 2 ( 1 + cos(2x) ) dx 1 = 4 1 1 ⌠ ⌡ 1 + 2cos(2x) + cos2(2x) dx = 4 ⌠ ⌡ 1 + 2cos(2x) + 2 { 1 + cos(4x) } dx 1 = 4 3 1 3 1 1 ⌠ ⌡ 2 + 2cos(2x) + 2 cos(4x) dx = 8 x + 4 sin(2x) + 32 sin(4x) + C . ⌠ ⌡ cos5(x) dx = ⌠ ⌡ cos2(x).cos2(x).cos(x) dx = ⌠ ⌡ ( 1 – sin2(x) )( 1 – sin2(x) ) cos(x) dx = ⌠ { 1 – 2sin2(x) + sin4(x) }cos(x) dx ⌡ = ⌠ cos(x) dx – 2 ⌡ ⌠ sin2(x).cos(x) dx + ⌡ ⌠ sin4(x).cos(x) dx (Use u = sin(x), du = cos(x) dx ) ⌡ 2 1 3 5 = sin(x) – 3 sin (x) + 5 sin (x) + C . Practice 3: ⌠ ⌡ sin3(x).cos4(x) dx = ⌠ ⌡ sin(x).sin2(x).cos4(x) dx = ⌠ ⌡ sin(x).(1 – cos2(x) ).cos4(x) dx = ⌠ ⌡ sin(x).cos4(x) dx – ⌠ ⌡ sin(x).cos6(x) dx 1 1 5 7 = – 5 cos (x) + 7 cos (x) + C (Use u = cos(x), du = – sin(x) dx ) 8.6 Integrals of Trigonometric Functions Practice 4: 8 Contemporary Calculus ⌠ tan3(x) dx = 12 tan2(x) – ⌡ ⌠ tan(x) dx = 12 tan2(x) – ln| sec(x) | + C . ⌡ 1 3 ⌠ ⌡ sec5(x) dx = 2 sec3(x).tan(x) + 4 ⌠ ⌡ sec3(x) dx 1 3 3 = 2 sec (x).tan(x) + 4 1 {2 1 sec(x).tan(x) + 2 ⌠ ⌡ sec(x) dx } 1 3 3 3 = 2 sec (x).tan(x) + 8 sec(x).tan(x) + 8 ln| sec(x) + tan(x) | + C. Practice 5: ⌠ ⌡ sec4(x).tan2(x) dx = ⌠ ⌡ sec2(x).sec2(x).tan2(x) dx = ⌠ ⌡ sec2(x).(tan2(x) + 1).tan2(x) dx = ⌠ ⌡ sec2(x).tan4(x) dx + ⌠ ⌡ sec2(x).tan2(x) dx 1 1 5 3 = 5 tan (x) + 3 tan (x) + C . 2 (Use u = tan(x), du = sec (x) dx )
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