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LU3 Integration for Pre- Uniersity Level
- 1. Semester 2 Session2021/2022
- 2. Subtopics Integration of Functions 3.1 Integrationof Trigonometric Functions 3.2 Techniques of Integration 3.3 Definite Integrals 3.4
- 3. 3.1 INTEGRATION OFFUNCTIONS
- 4. At the endof this unit, students will be able to : relate integration and differentiation use basic rules of integration find 𝟏 𝒙 𝒅𝒙 = 𝐥𝐧 𝒙 + 𝒄 and 𝟏 𝒂𝒙+𝒃 𝒅𝒙 = 𝟏 𝒂 𝒍𝒏 𝒂𝒙 + 𝒃 + 𝒄 1 2 3 4 determine the integral of the forms (I) 𝒆𝒙 𝒅𝒙 (II) 𝒆𝒂𝒙+𝒃 𝒅𝒙 (III) 𝒂𝒙 𝒅𝒙 (IV) 𝒂𝒂𝒙+𝒃 𝒅𝒙
- 5. Introduction To INTEGRATION Integrationis a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points and many useful things. But it is easiest to start with finding the area under the curve of a function like this: 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑐𝑢𝑟𝑣𝑒 = න 𝑓 𝑥 𝑑𝑥 𝑦 = 𝑓(𝑥) What is the area under the curve of function, 𝑓(𝑥)? Figure 1
- 6. SLICES We could calculatethe function at a few points and add up slices of width Δx like this (but the answer won't be very accurate): We can make Δx a lot smaller and add up many small slices (answer is getting better): And as the slices approach zero in width, the answer approaches the true answer. We now write dx to mean the Δx slices are approaching zero in width. Figure 2a Figure 2b Figure 2c
- 8. Case 1: Aphysicist who knows the velocity of a particle might wish to know its position at a given time. 3.1 Integration of Functions How is Integration Applied ??? Case 2: A biologist who knows the rate at which a bacteria population is increasing wants to deduce what the size of the population will be at some future time. Figure 3a Figure 3b
- 9. • In eachcase, the problem is to find a function F whose derivative is a known function, f. • If such a function F exists, it is called an antiderivative of f. 3.1 Integration of Functions Antiderivatives Definition A function 𝑭 is called an antiderivative of 𝒇 in the given interval, if 𝐹′ 𝑥 = 𝑓 𝑥 for all 𝑥 in the interval.
- 10. 3.1 Integration ofFunctions Recall Chapter 1 Differentiation For example , Differentiate the following function. 𝐹′ 𝑥 = 𝑥2 = 𝑓 𝑥 𝐹(𝑥) = 𝑥3 3 derivative antiderivative
- 11. 3.1 Integration ofFunctions If given a derivative of a function, we can work backwards to find the function from which it is derived and otherwise. f(x) F(x) To get antiderivative 2𝑥 𝑥2 To get derivative To get derivative To get antiderivative Integration as Differentiation in Reverse Figure 4a Figure 4b
- 12. 3.1 Integration ofFunctions Differentiate each of the following functions. 𝐹′ 𝑥 = 𝑥2 = 𝑓 𝑥 𝐹′(𝑥) = 𝑥2 = 𝑓 𝑥 𝐹 𝑥 = 𝑥3 3 − 3 𝐹(𝑥) = 𝑥3 3 + 100 • In fact, for any constant c, we have 𝐹(𝑥) = 𝑥3 3 + 𝑐 𝐹′ 𝑥 = 𝑥2 = 𝑓 𝑥 • Therefore, all the functions 𝐹 above are antiderivatives of 𝑓 . General Notation of Integral and Form of Antiderivatives derivative derivative derivative antiderivative antiderivative antiderivative
- 13. • The derivativeof 𝑥3 3 + 100 is 𝑥2 , and the derivative of 𝑥3 3 − 3 is also 𝑥2 , and so on! Because the derivative of a constant is zero. • So when we reverse the operation (to find the integral) we only know 𝑥2 , but there could have been a constant of any value. • So we wrap up the idea by just writing + C at the end of function.
- 14. 3.1 Integration ofFunctions • The general antiderivatives of 𝑓 are 𝒙𝟑 𝟑 + 𝒄. • By assigning specific values to the constant 𝑐, we obtain a family of antiderivatives as in Figure 5. Figure 5
- 15. 3.1 Integration ofFunctions • When we want to differentiate a function 𝑓(𝑥) we use the notation 𝑑 𝑑𝑥 as an instruction to differentiate. • In a similar way, when we want to integrate a function we use a special notation : 𝒇 𝒙 𝒅𝒙 • The symbol for integration, ∫ is known as an integral sign
- 16. 3.1 Integration ofFunctions • Hence, the notation 𝑓 𝑥 𝑑𝑥 which means “integrate the integrand 𝑓 𝑥 with respect to 𝑥” is traditionally used for an antiderivative of f and is called an indefinite integral. • An indefinite integral 𝑓(𝑥) 𝑑𝑥 is a function (or family of functions), whereas a definite integral 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 is a number.
- 17. An Indefinite Integralhas no actual values or interval to calculate. The result is a function. A Definite Integral has actual values to calculate between (they are put at the bottom and top of the " "). The result is a value. Figure 6
- 18. Simple Example: ConstantFlow Rate Integration: With a flow rate of 1, the tank volume increases by 𝑥. Derivative: If the tank volume increases by 𝑥, then the flow rate is 1. Figure 7a
- 19. Now For AnIncreasing Flow Rate As the flow rate increases, the tank fills up faster and faster. Integration: With a flow rate of 2𝑥, the tank volume increases by 𝑥2. Derivative: If the tank volume increases by 𝑥2 , then the flow rate must be 2𝑥. Imagine the flow starts at 0 and gradually increases (maybe a motor is slowly opening the tap). Figure 7b
- 20. Based on theprevious example, if the flow rate of water is 2𝑥 and the tank volume increases by 𝑥2 + 𝑐, explain what it means by + 𝑐 ??
- 21. …maybe the tankalready has water in it! •The flow still increases the volume by the same amount •And the increase in volume can give us back the flow rate. Which teaches us to always add "+ C". Answer
- 22. Example 3.1 Given that𝑦 = ln 𝑥 2𝑥 , 𝑑𝑦 𝑑𝑥 = − ln 𝑥 −1 2𝑥2 and 𝑑2𝑦 𝑑𝑥2 = 2 ln 𝑥 −3 2𝑥3 . Hence, determine a) − ln 𝑥 −1 2𝑥2 𝑑𝑥 b) 2 ln 𝑥 −3 2𝑥3 𝑑𝑥 3.1 Integration of Functions
- 23. Solution 3.1 Integration ofFunctions 𝑦 = ln 𝑥 2𝑥 𝑑𝑦 𝑑𝑥 = − ln 𝑥 − 1 2𝑥2 𝑑2𝑦 𝑑𝑥2 = 2 ln 𝑥 − 3 2𝑥3 differentiate differentiate integrate integrate a) − ln 𝑥 −1 2𝑥2 𝑑𝑥 = ln 𝑥 2𝑥 + 𝑐 b) 2 ln 𝑥 −3 2𝑥3 𝑑𝑥 = − ln 𝑥 −1 2𝑥2 + 𝑐
- 24. Example 3.2 a) (i)Given 𝑦 = 1 2 𝑥2 , find 𝑑𝑦 𝑑𝑥 . (ii) Hence, find 𝑥 𝑑𝑥. b) (i) Differentiate 𝑦 = 2 ln 𝑥 3𝑥 with respect to 𝑥. (ii) Hence, find 4 1−ln 𝑥 7𝑥2 𝑑𝑥 . 3.1 Integration of Functions
- 25. Example 3.2 Answer: a) (i) 𝑑𝑦 𝑑𝑥 =𝑥. (ii) 𝑥 𝑑𝑥 = 1 2 𝑥2 + 𝑐. b) (i) 𝑑𝑦 𝑑𝑥 = 2 3 1−ln 𝑥 𝑥2 . (ii) 4 1−ln 𝑥 7𝑥2 𝑑𝑥 = 4 ln 𝑥 7𝑥 + 𝑐. 3.1 Integration of Functions
- 26. Exercise 1 a) If𝑦 = 4𝑥 + 3 7 , then find 𝑑𝑦 𝑑𝑥 . Hence, find 28 4𝑥 + 3 6 𝑑𝑥. Exercise 2 b) If 𝑦 = 5𝑥 − 10 9 , then find 𝑑𝑦 𝑑𝑥 . Hence, find 180 5𝑥 − 10 8 𝑑𝑥. 3.1 Integration of Functions
- 27. Exercise 1 Answer: a) 𝑑𝑦 𝑑𝑥 = 284𝑥 + 3 6 . 28 4𝑥 + 3 6 𝑑𝑥 = 4𝑥 + 3 7 + 𝑐. Exercise 2 b) 𝑑𝑦 𝑑𝑥 = 45 5𝑥 − 10 8 . 180 5𝑥 − 10 8 𝑑𝑥 = 4 5𝑥 − 10 9 + 𝑐. 3.1 Integration of Functions
- 28. 𝑑𝑥= 𝑥 + 𝑐, where 𝑐 is a constant. 𝑘 𝑑𝑥 = 𝑘𝑥 + 𝑐, where 𝑘 and 𝑐 are constants. 𝑥𝑛 𝑑𝑥 = 𝑥𝑛+1 𝑛+1 + 𝑐, where 𝑛 ≠ −1. 𝑘 𝑓 𝑥 𝑑𝑥 = 𝑘 𝑓 𝑥 𝑑𝑥, where 𝑘 is a constant. 𝑓(𝑥) ± 𝑔(𝑥) 𝑑𝑥 = 𝑓 𝑥 𝑑𝑥 ± 𝑔 𝑥 𝑑𝑥 Basic Rules of Integration 3.1 Integration of Functions
- 29. Example 3.3 Integrate thefollowing with respect to the variable used : a) 𝑡4 𝑑𝑡 b) 1 𝑠 + 5 𝑑𝑠 c) 𝑥2 (𝑥 + 3) 𝑑𝑥 d) 𝑟5+2𝑟2−1 𝑟4 𝑑𝑟 e) Discuss, can we apply product rule and quotient rule in integration? Give your reason. 3.1 Integration of Functions
- 30. Example 3.3 Answer: a) 𝑡4 𝑑𝑡 = 𝑡5 5 + 𝑐 b) 1 𝑠 + 5 𝑑𝑠 = 2 𝑠 + 5𝑠 + 𝑐 c) 𝑥2(𝑥 + 3) 𝑑𝑥 = 𝑥4 4 + 𝑥3 + 𝑐 d) 𝑟5+2𝑟2−1 𝑟4 𝑑𝑟 = 𝑟2 2 − 2 𝑟 + 1 3𝑟3 + 𝑐 3.1 Integration of Functions
- 31. Exercise 3 Integrate thefollowing with respect to the variable used: a) 𝑡4 − 2𝑡2 + 3 𝑡 𝑑𝑡 b) 𝑎𝑥2 + 𝑏 2 𝑑𝑥 c) 𝑠3−5𝑠−1 𝑠3 𝑑𝑠 d) 1+ 𝑥 𝑥2 𝑑𝑥 3.1 Integration of Functions
- 32. Answer: a) 𝑡4 −2𝑡2 + 3 𝑡 𝑑𝑡 = 𝑡5 5 − 2𝑡3 3 − 6 𝑡 + 𝑐 b) 𝑎𝑥2 + 𝑏 2 𝑑𝑥 = 𝑎2𝑥5 5 + 2𝑎𝑏𝑥3 3 + 𝑏2 𝑥 + 𝑐 c) 𝑠3−5𝑠−1 𝑠3 𝑑𝑠 = 𝑠 + 5 𝑠 + 1 2𝑠2 + 𝑐 d) 1+ 𝑥 𝑥2 𝑑𝑥 = − 1 𝑥 − 2 𝑥 + 𝑐 3.1 Integration of Functions
- 33. 3.1 Integration ofFunctions 1 𝑥 𝑑𝑥 = 𝑙𝑛 𝑥 + 𝑐, where 𝑥 ≠ 0 In general, න 1 𝑎𝑥 + 𝑏 𝑑𝑥 = 1 𝑎 𝑙𝑛 𝑎𝑥 + 𝑏 + 𝑐 Integration of Rational Function
- 34. 3.1 Integration ofFunctions Example 3.4: Integrate the following: a) 2 𝑥 𝑑𝑥 b) 5 3−2𝑥 𝑑𝑥 Answer: a) 2 𝑥 𝑑𝑥 = 2𝑙𝑛 𝑥 + 𝑐 b) 5 3−2𝑥 𝑑𝑥 = − 5𝑙𝑛 3−2𝑥 2 + 𝑐
- 35. 3.1 Integration ofFunctions Exercise 4: Integrate the following: a) 5 𝑥 + 3 2𝑥+1 𝑑𝑥 b) 5 2−6𝑥 + 5 5𝑥+3 𝑑𝑥 Answer: a) 5 𝑥 + 3 2𝑥+1 𝑑𝑥 = 5𝑙𝑛 𝑥 + 3𝑙𝑛 2𝑥+1 2 + 𝑐 b) 5 2−6𝑥 + 5 5𝑥+3 𝑑𝑥 = − 5𝑙𝑛 2−6𝑥 6 + 𝑙𝑛 5𝑥 + 3 + c
- 36. 3.1 Integration ofFunctions Integration of the Improper Form of Rational Function Given 𝑃(𝑥) 𝑄(𝑥) is an improper form, where the degree of 𝑃(𝑥) is greater than or equal to the degree of 𝑄(𝑥). We need to perform long division before integrating the rational function.
- 37. 3.1 Integration ofFunctions Example 3.5: Integrate the following: a) 𝑥−3 𝑥+1 𝑑𝑥 b) 4𝑥2+6𝑥+1 2𝑥+3 𝑑𝑥 Answer: a) 𝑥−3 𝑥+1 𝑑𝑥 = 𝑥 − 4 ln 𝑥 + 1 + 𝑐 b) 4𝑥2+6𝑥+1 2𝑥+3 𝑑𝑥 = 𝑥2 + ln 2𝑥+3 2 + 𝑐
- 38. 3.1 Integration ofFunctions Exercise 5: Integrate the following: a) 15𝑥2−9𝑥−1 5𝑥−3 𝑑𝑥 b) 𝑥3+4𝑥−2 𝑥−1 𝑑𝑥 Answer: a) 15𝑥2−9𝑥−1 5𝑥−3 𝑑𝑥 = 3𝑥2 2 − 𝑙𝑛 5𝑥−3 5 + 𝑐 b) 𝑥3+4𝑥−2 𝑥−1 𝑑𝑥 = 𝑥3 3 + 𝑥2 2 + 5𝑥 + 3𝑙𝑛 𝑥 − 1 + 𝑐
- 39. 3.1 Integration ofFunctions Integration of Exponential Function 𝑒𝑥 𝑑𝑥 = 𝑒𝑥 + 𝑐 𝑎𝑥 𝑑𝑥 = 𝑎𝑥 ln 𝑎 + 𝑐 𝑒𝑎𝑥+𝑏 𝑑𝑥 = 1 𝑎 𝑒𝑎𝑥+𝑏 + 𝑐 𝑎𝑚𝑥+𝑛 𝑑𝑥 = 𝑎𝑚𝑥+𝑛 𝑚 ln 𝑎 + 𝑐
- 40. 3.1 Integration ofFunctions Example 3.6: Integrate the following: a) 𝑒3𝑥 + 1 𝑒𝑥 𝑑𝑥 b) 1+𝑒2𝑥+1 𝑒𝑥+1 𝑑𝑥 c) 2𝑥+1 𝑑𝑥
- 41. 3.1 Integration ofFunctions Example 3.6: Answer: a) 𝑒3𝑥 + 1 𝑒𝑥 𝑑𝑥 = 𝑒3𝑥 3 − 1 𝑒𝑥 + 𝑐 b) 1+𝑒2𝑥+1 𝑒𝑥+1 𝑑𝑥 = − 1 𝑒𝑥+1 + 𝑒𝑥 + 𝑐 c) 2𝑥+1 𝑑𝑥 = 2𝑥+1 ln 2 + 𝑐
- 42. 3.1 Integration ofFunctions Exercise 6: Integrate the following: a) 𝑒2𝑥 + 3 𝑒𝑥 𝑑𝑥 b) 22−3𝑥 − 3𝑒4𝑥−3 𝑑𝑥 c) 4 + 𝑒𝑥 3 + 1 𝑒𝑥 𝑑𝑥
- 43. 3.1 Integration ofFunctions Exercise 6: Answer: a) 𝑒2𝑥 + 3 𝑒𝑥 𝑑𝑥 = 𝑒2𝑥 2 − 3 𝑒𝑥 + 𝑐 b) 22−3𝑥 − 3𝑒4𝑥−3 𝑑𝑥 = 22−3𝑥 −3 ln 2 − 3𝑒4𝑥−3 4 + 𝑐 c) 4 + 𝑒𝑥 3 + 1 𝑒𝑥 𝑑𝑥 = 13𝑥 − 4 𝑒𝑥 + 3𝑒𝑥 + 𝑐
- 44. PSS 3.2 Discuss howcan we prevent the integration mistake as below? (a) 1 𝑥 𝑑𝑥 = 𝑥−2 −2 + 𝑐 𝑜𝑟 𝑥0 0 + 𝑐 (b) 32𝑥 𝑑𝑥 = 32𝑥 2 𝑙𝑛 3 + 𝑐 3.1 Integration of Functions
- 45. Self – Evaluationof 3.1 Items Very Poor Very Good 1 2 3 4 5 Introduction to integration How integration is applied Antiderivative Integration as differentiation in reverse General notation of integral and form of antiderivatives Simple Example of flow rate (constant rate and increase rate) Basic rules of integration Integration of rational Function Integration of the improper form of rational function Integration of exponential function Please rate your achievement for each topic you have learned from your lecture. My performance of Unit 3.1: 𝟓𝟎 × 𝟓 = Please check your strength and weakness for each topic you have learned from your previous lecture.
- 46. 3.2 INTEGRATION OF TRIGONOMETRICFUNCTIONS
- 47. න cos 𝑎𝑥𝑑𝑥 න 𝑠𝑖𝑛 𝑎𝑥 𝑑𝑥 𝑠𝑒𝑐2 𝑎𝑥 𝑑𝑥 At the end of this unit, students will be able to find the following integrals ;
- 48. න cos 𝑥𝑑𝑥 = sin 𝑥 + 𝑐 න sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐 න 𝑠𝑒𝑐2𝑥 𝑑𝑥 = tan 𝑥 + 𝑐 3.2 Integration of Trigonometric Functions න cos 𝑎𝑥 𝑑𝑥 = 1 𝑎 sin 𝑎𝑥 + 𝑐 න sin 𝑎𝑥 𝑑𝑥 = − 1 𝑎 cos 𝑎𝑥 + 𝑐 න 𝑠𝑒𝑐2𝑎𝑥 𝑑𝑥 = 1 𝑎 tan 𝑎𝑥 + 𝑐 The integral of trigonometric functions given above can be generalised as follows,
- 49. Similarly, 3.2 Integration ofTrigonometric Functions න cos 𝑎𝑥 + 𝑏 𝑑𝑥 = 1 𝑎 sin 𝑎𝑥 + 𝑏 + 𝑐 න sin 𝑎𝑥 + 𝑏 𝑑𝑥 = − 1 𝑎 cos 𝑎𝑥 + 𝑏 + 𝑐 න 𝑠𝑒𝑐2 𝑎𝑥 + 𝑏 𝑑𝑥 = 1 𝑎 tan 𝑎𝑥 + 𝑏 + 𝑐
- 50. Example 3.7: Find thefollowing integrals. (a) (cos 𝑥 − 2 sin 3𝑥 ) 𝑑𝑥 (b) cos 3𝑥 + 𝜋 4 𝑑𝑥 (c) 𝑒2𝑥 − 𝑠𝑒𝑐2 2𝑥 + 𝜋 𝑑𝑥 Solution (a) (cos 𝑥 − 2 sin 3𝑥 ) 𝑑𝑥 = cos 𝑥 𝑑𝑥 − 2 sin 3𝑥 𝑑𝑥 = sin 𝑥 − 2(− 1 3 cos 3𝑥) + 𝑐 3.2 Integration of Trigonometric Functions = sin 𝑥 + 2 3 cos 3𝑥 + 𝑐
- 51. (b) cos3𝑥 + 𝜋 4 𝑑𝑥 = sin 3𝑥+ 𝜋 4 3 + 𝑐 3.2 Integration of Trigonometric Functions (c) 𝑒2𝑥 − 𝑠𝑒𝑐2 2𝑥 + 𝜋 𝑑𝑥 = 𝑒2𝑥𝑑𝑥 − 𝑠𝑒𝑐2 2𝑥 + 𝜋 𝑑𝑥 = 𝑒2𝑥 2 − 1 2 𝑡𝑎𝑛 2𝑥 + 𝜋 + 𝑐
- 52. Exercise 7: Find thefollowing integrals. (a) (s𝑖𝑛 3𝑡 − cos 4𝑡 ) 𝑑𝑡 (b) 𝑠𝑒𝑐2 3 𝑥 − cos 𝑥 4 𝑑𝑥 (c) 1−𝑐𝑜𝑠2𝑥−𝑐𝑜𝑠3𝑥 𝑐𝑜𝑠2𝑥 𝑑𝑥 3.2 Integration of Trigonometric Functions
- 53. Exercise 7: Answer: (a) (s𝑖𝑛3𝑡 − cos 4𝑡 ) 𝑑𝑡 = cos 3𝑡 3 − 1 4 sin 4𝑡 + 𝑐 (b) 𝑠𝑒𝑐2 3 𝑥 − cos 𝑥 4 𝑑𝑥 = 1 3 tan 3𝑥 − 4 sin 𝑥 4 + 𝑐 (c) 1−𝑐𝑜𝑠2𝑥−𝑐𝑜𝑠3𝑥 𝑐𝑜𝑠2𝑥 𝑑𝑥 = tan 𝑥 − 𝑥 − sin 𝑥 + 𝑐 3.2 Integration of Trigonometric Functions
- 54. Integration of 𝒔𝒊𝒏𝒏 𝒙 𝒅𝒙 and 𝒄𝒐𝒔𝒏 𝒙 𝒅𝒙 where n is an even integer 3.2 Integration of Trigonometric Functions Double angle formula can be used for trigonometric integrals of the form, න 𝑠𝑖𝑛2 𝑎𝑥 𝑑𝑥 and න 𝑐𝑜𝑠2 𝑎𝑥 𝑑𝑥 If n is even integer, double angle formulas are used. Recall Semester 1 Chapter 10 Trigonometric Functions Double-angle Formulae
- 55. 3.2 Integration ofTrigonometric Functions sin 2𝑥 = cos 2𝑥 = = = tan 2𝑥 = DOUBLE-ANGLE FORMULAE (RECALL LU 10 Trigonometric)
- 57. 3.2 Integration ofTrigonometric Functions a) Since 𝑐𝑜𝑠 2𝑥 = 1 − 2 𝑠𝑖𝑛2𝑥 (recall double-angle formula) 𝑠𝑖𝑛2 𝑥 = 1 2 1 − cos 2𝑥 b) Since 𝑐𝑜𝑠 2𝑥 = 2 𝑐𝑜𝑠2𝑥 − 1 (recall double-angle formula) 𝑐𝑜𝑠2 𝑥 = 1 2 1 + cos 2𝑥 න 𝑠𝑖𝑛2 𝑎𝑥 𝑑𝑥 = න 1 2 1 − cos 2𝑎𝑥 𝑑𝑥 න 𝑐𝑜𝑠2 𝑎𝑥 𝑑𝑥 = න 1 2 1 + cos 2𝑎𝑥 𝑑𝑥
- 58. 3.2 Integration ofTrigonometric Functions Note: Double angle formula for sine, is also useful in solving trigonometric integrals. sin 2𝑥 = 2 sin 𝑥 cos 𝑥
- 59. Example 3.8: Find thefollowing integrals. (a) 𝑠𝑖𝑛2 2𝑥 𝑑𝑥 (b) 𝑐𝑜𝑠4 3𝑥 𝑑𝑥 (c) sin 2𝑥 cos 𝑥 𝑑𝑥 Answer: a. 𝑠𝑖𝑛2 2𝑥 𝑑𝑥 = 1 8 4𝑥 − sin 4𝑥 + 𝑐 b. 𝑐𝑜𝑠4 3𝑥 𝑑𝑥 = 1 96 36𝑥 + 8 sin 6𝑥 + sin 12𝑥 + 𝑐 c. sin 2𝑥 cos 𝑥 𝑑𝑥 = −2 cos 𝑥 + 𝑐 3.2 Integration of Trigonometric Functions
- 60. Exercise 8: Find thefollowing integrals. (a) 2 𝑐𝑜𝑠2 3𝑥 𝑑𝑥 (b) sin 4𝑥 cos 2𝑥 𝑑𝑥 (c) 𝑠𝑖𝑛4 2𝑥 𝑑𝑥 3.2 Integration of Trigonometric Functions
- 61. Exercise 8: Answer: (a) 2 𝑐𝑜𝑠2 3𝑥 𝑑𝑥 = 𝑥 + sin 6𝑥 6 + 𝑐 (b) sin 4𝑥 cos 2𝑥 𝑑𝑥 = − cos 2𝑥 + 𝑐 (c) 𝑠𝑖𝑛4 2𝑥 𝑑𝑥 = 1 64 24𝑥 − 8 sin 4𝑥 + sin 8𝑥 + 𝑐 3.2 Integration of Trigonometric Functions
- 62. Self – Evaluationof 3.2 Items Very Poor Very Good 1 2 3 4 5 Memorize basic integral of trigonometric Apply basic integral of trigonometric Retrieve prior knowledge of double angle formula of trigonometric Integration of 𝑠𝑖𝑛𝑛 𝑥 𝑑𝑥 where n is an even integer Integration of 𝑐𝑜𝑠𝑛 𝑥 𝑑𝑥 where n is an even integer Please rate your achievement for each topic you have learned from your previous lecture. My performance of Unit 3.2: 𝟐𝟓 × 𝟓 = Please check your strength and weakness for each topic you have learned from your previous lecture.
- 63.