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Lesson 9 transcendental functions

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Lesson 9 transcendental functions

  1. 1. TRANSCENDENTAL FUNCTIONS
  2. 2. OBJECTIVES At the end of the lesson, the students are expected to: β€’ use the Log Rule for Integration to integrate a rational functions. β€’ integrate exponential functions. β€’ integrate trigonometric functions. β€’ integrate functions of the nth power of the different trigonometric functions. β€’ use Walli’s Formula to shorten the solution in finding the antiderivative of powers of sine and cosine.
  3. 3. β€’ integrate functions whose antiderivatives involve inverse trigonometric functions. β€’ use the method of completing the square to integrate a function. β€’ review the basic integration rules involving elementary functions. β€’ integrate hyperbolic functions. β€’ integrate functions involving inverse hyperbolic functions.
  4. 4. LOG RULE FOR INTEGRATION Let u be a differentiable function of x. 𝑑𝑒 𝑒 = 𝑙𝑛 𝑒 + 𝐢 or the above formula can also be written as 𝑒′ 𝑒 𝑑π‘₯ = 𝑙𝑛 𝑒 + 𝐢 To apply this rule, look for quotients in which the numerator is the derivative of the denominator.
  5. 5. β€’ EXAMPLE β€’ Find the indefinite integral. 1. π‘₯2 5βˆ’π‘₯3 𝑑π‘₯ 5. 𝑠𝑒𝑐π‘₯π‘‘π‘Žπ‘›π‘₯ 𝑠𝑒𝑐π‘₯βˆ’1 𝑑π‘₯ 2. π‘₯3βˆ’6π‘₯βˆ’20 π‘₯+5 𝑑π‘₯ 6. 𝑒2π‘₯ 𝑒 π‘₯βˆ’1 𝑑π‘₯ 3. 1 π‘₯𝑙𝑛π‘₯3 𝑑π‘₯ 4. 1 π‘₯ 2 3 1+π‘₯ 1 3 𝑑π‘₯
  6. 6. INTEGRATION OF EXPONENTIAL FUNCTIONS Let u be a differentiable function of x. 𝒆 𝒖 𝒅𝒖 = 𝒆 𝒖 + 𝒄 𝒂 𝒖 𝒅𝒖 = 𝒂 𝒖 𝒍𝒏𝒂 + c
  7. 7. EXAMPLE β€’ Find the indefinite integral. 1. 𝑒 1 π‘₯ π‘₯2 𝑑π‘₯ 6. π‘₯473π‘₯3 𝑑π‘₯ 2. 𝑒2π‘₯ + π‘’βˆ’π‘₯ 2 𝑑π‘₯ 3. π‘₯𝑒3π‘₯2+4 𝑑π‘₯ 4. 5𝑒 𝑙𝑛π‘₯2 π‘₯ 𝑑π‘₯ 5. 10 π‘₯3 π‘₯2 𝑑π‘₯
  8. 8. BASIC TRIGONOMETRIC FUNCTIONS INTEGRATION FORMULAS β€’ cos 𝑒𝑑𝑒 = sin 𝑒 +c β€’ sin 𝑒𝑑𝑒 = -cos 𝑒 + c β€’ 𝑠𝑒𝑐2 𝑒𝑑𝑒 = tan 𝑒 + c β€’ 𝑐𝑠𝑐2 𝑒𝑑𝑒 = -cot 𝑒 + 𝑐 β€’ sec 𝑒 tan 𝑒𝑑𝑒 = sec 𝑒 + c β€’ csc 𝑒 cot 𝑒 𝑑𝑒 = -csc 𝑒 + c β€’ tan 𝑒𝑑𝑒 = ln sec 𝑒 + c or - lncos 𝑒 + c β€’ cot 𝑒𝑑𝑒 = lnsin 𝑒 + c β€’ sec 𝑒𝑑𝑒 = ln ( sec 𝑒 +tan 𝑒 ) + c β€’ csc 𝑒𝑑𝑒 = -ln ( csc 𝑒 + cot 𝑒 ) + c
  9. 9. β€’ In all these formulas, u is an angle. In dealing with integrals involving trigonometric functions, transformations using the trigonometric identities are almost always necessary to reduce the integral to one or more of the standard forms.
  10. 10. EXAMPLE Find the indefinite integral. 1. cos π‘₯ sec π‘₯+tan π‘₯ 𝑑π‘₯ 2. cot 3π‘₯ sin 3π‘₯𝑑π‘₯ 7. 1βˆ’cos π‘₯ 𝑠𝑖𝑛2 π‘₯ 𝑑π‘₯ 3. π‘₯ csc π‘₯2 𝑑π‘₯ 8. 2 π‘π‘œπ‘ 22π‘₯ 𝑑π‘₯ 4. sin 2π‘₯ π‘π‘œπ‘ 2 π‘₯ sin π‘₯ 𝑑π‘₯ 5. cos π‘₯ 1βˆ’ cos π‘₯ 𝑑π‘₯ 6. (csc π‘₯ sin 2π‘₯ + 1 sin π‘₯ sec π‘₯ ) 𝑑π‘₯
  11. 11. TRANSFORMATION OF TRIGONOMETRIC FUNCTIONS If we are given the product of an integral power of sin π‘₯ and an integral power of cos π‘₯, where in the powers may be equal or unequal, both even, both odd, or one is even the other odd, we use the trigonometric identities and express the given integrand as a power of a trigonometric function times the derivative of that function or as the sum of powers of a function times the derivative of the function β€’ We shall now see how to perform the details under specified conditions.
  12. 12. POWERS OF SINE AND COSINE β€’ CASE 1. π’”π’Šπ’ 𝒏 𝒖𝒄𝒐𝒔 π’Ž 𝒖 𝒅𝒖 Transformations: a) If n is odd and m is even, π’”π’Šπ’ 𝒏 𝒖𝒄𝒐𝒔 π’Ž 𝒖 = π’”π’Šπ’ π’βˆ’πŸ 𝒖𝒄𝒐𝒔 π’Ž (π’”π’Šπ’ 𝒖) b) If m isoddand n is even, π’”π’Šπ’ 𝒏 𝒖𝒄𝒐𝒔 π’Ž 𝒖 = π’”π’Šπ’ 𝒏 𝒖𝒄𝒐𝒔 π’Žβˆ’πŸ 𝒖(𝒄𝒐𝒔𝒖) c) If n and m are both odd, transform the lesser power. If n and m are same degree either can be transformed
  13. 13. CASE II. π’”π’Šπ’ 𝒏 𝒙𝒄𝒐𝒔 π’Ž 𝒙 𝒅𝒙 where m and n are positive even integers. When both m and n are even, the method of type 1 fails. In this case, the identities, π’”π’Šπ’ 𝟐 𝒙 = 𝟏 βˆ’ π’„π’π’”πŸπ’™ 𝟐 , 𝒄𝒐𝒔 𝟐 𝒙 = 𝟏+π’„π’π’”πŸπ’™ 𝟐 , π’”π’Šπ’ 𝒙 𝒄𝒐𝒔 𝒙 = π’”π’Šπ’ πŸπ’™ 𝟐 will be used.
  14. 14. EXAMPLE β€’ Evaluate the following integrals: 1. π‘π‘œπ‘ 3 π‘₯𝑠𝑖𝑛7 π‘₯ 𝑑π‘₯ 2. 𝑠𝑖𝑛5 2π‘₯π‘π‘œπ‘ 5 2π‘₯ 𝑑π‘₯ 3 π‘ π‘–π‘›βˆ’3 π‘₯π‘π‘œπ‘ 5 π‘₯ 𝑑π‘₯ 4. 𝑠𝑖𝑛2 π‘₯π‘π‘œπ‘ 4 π‘₯ 𝑑π‘₯ 5. 𝑠𝑖𝑛4 2π‘₯ 𝑑π‘₯ 6. π‘π‘œπ‘ 2π‘₯ + 2𝑠𝑖𝑛π‘₯ 2 𝑑π‘₯ 7. 𝑠𝑖𝑛6 π‘₯π‘π‘œπ‘ 4 π‘₯ 𝑑π‘₯ 8. 𝑠𝑖𝑛4 π‘₯π‘π‘œπ‘ 5 π‘₯ 𝑑π‘₯ 9. 0 πœ‹ 2 𝑠𝑖𝑛2 π‘₯π‘π‘œπ‘ 5 π‘₯ 𝑑π‘₯ 10. 0 πœ‹ 2 𝑠𝑖𝑛2 π‘₯π‘π‘œπ‘ 2 π‘₯ 𝑑π‘₯
  15. 15. PRODUCT OF SINE AND COSINE β€’ Integration of the products sin π‘Žπ‘₯ sin 𝑏π‘₯ , cos π‘Žπ‘₯ cos 𝑏π‘₯ , sin π‘Žπ‘₯ cos 𝑏π‘₯ , where a and b are constants is carried out by using the formulas: sin 𝐴 sin 𝐡 = 1 2 cos 𝐴 βˆ’ 𝐡 - 1 2 cos 𝐴 + 𝐡 sin 𝐴 cos 𝐡 = 1 2 sin 𝐴 βˆ’ 𝐡 + 1 2 sin 𝐴 + 𝐡 cos 𝐴 cos 𝐡 = 1 2 cos 𝐴 βˆ’ 𝐡 + 1 2 cos 𝐴 + 𝐡
  16. 16. EXAMPLE β€’ Perform the indicated integrations: 1. cos 8π‘₯ cos 5π‘₯ 𝑑π‘₯ 2. sin 6π‘₯ cos 8π‘₯ 𝑑π‘₯ 3. 2 cos 6π‘₯ cos βˆ’4π‘₯ 𝑑π‘₯ 4. 2 sin(2π‘₯ βˆ’ πœ‹) sin 3πœ‹ βˆ’ 2π‘₯ 𝑑π‘₯ 5. cos 5π‘₯ cos 7π‘₯ sin 3π‘₯ 𝑑π‘₯ 6. sin 4π‘₯ sin 10π‘₯ 𝑑π‘₯ 7. 2 cos 2π‘₯ cos π‘₯ 𝑑π‘₯ 8. 3 sin π‘₯ cos 3π‘₯ 𝑑π‘₯
  17. 17. WALLIS’ FORMULA 𝟎 𝝅 𝟐 π’”π’Šπ’ π’Ž 𝒙𝒄𝒐𝒔 𝒏 𝒙 𝒅𝒙 = π‘šβˆ’1 π‘šβˆ’3 ... 2 π‘œπ‘Ÿ 1 π‘›βˆ’1 π‘›βˆ’3 … 2 π‘œπ‘Ÿ 1 π‘š+𝑛 π‘š+π‘›βˆ’2 … 2 π‘œπ‘Ÿ 1 βˆ™ πœƒ where in m and n are integers β‰₯ 0, πœƒ = πœ‹ 2 , if m and n are both even, πœƒ = 1 , if either one or both are odd, and that the lower and upper limits are 0 and πœ‹ 2
  18. 18. EXAMPLE β€’ Evaluate by Wallis’ Formula. 1. 0 πœ‹ 2 𝑠𝑖𝑛4 π‘₯𝑑π‘₯ 2. 0 πœ‹ 2 𝑠𝑖𝑛5 π‘₯π‘π‘œπ‘ 6 π‘₯𝑑π‘₯ 3. 0 πœ‹ 2 𝑠𝑖𝑛4 π‘₯π‘π‘œπ‘ 8 π‘₯𝑑π‘₯ 4. 0 πœ‹ 6 𝑠𝑖𝑛6 3π‘¦π‘π‘œπ‘ 3 3𝑦𝑑𝑦 5. 0 πœ‹ 3 𝑠𝑖𝑛2 3π‘₯ 2 π‘π‘œπ‘ 2 3π‘₯ 2 𝑑π‘₯
  19. 19. POWERS OF TANGENT AND SECANT (COTANGENT AND COSECANT) I. 𝒕𝒂𝒏 𝒏 𝜽 π’…πœ½ or 𝒄𝒐𝒕 𝒏 𝜽 π’…πœ½ where n is a positive integer. When n=1 𝒕𝒂𝒏 𝒏 𝜽 π’…πœ½= - ln𝒄𝒐𝒔 𝜽 + c 𝒄𝒐𝒕 𝒏 𝜽 π’…πœ½ =ln sin 𝜽 + c When nβ‰₯ 1, we set π‘‘π‘Žπ‘› 𝑛 πœƒ equal to π‘‘π‘Žπ‘› π‘›βˆ’2 πœƒ π‘‘π‘Žπ‘›2 πœƒ π‘œπ‘Ÿ π‘π‘œπ‘‘2 πœƒ 𝑏𝑦 π‘π‘œπ‘‘ π‘›βˆ’2 πœƒπ‘π‘œπ‘‘2 πœƒ , replace π‘‘π‘Žπ‘›2 πœƒ 𝑏𝑦 𝑠𝑒𝑐2 πœƒ βˆ’ 1 π‘œπ‘Ÿ π‘π‘œπ‘‘2 πœƒ by (𝑐𝑠𝑐2 πœƒ βˆ’ 1). Thus we get powers of tanπœƒ and by power formula, we can evaluate the integral.
  20. 20. II. 𝒔𝒆𝒄 π’Ž πœ½π’•π’‚π’ 𝒏 𝜽 π’…πœ½ 𝒐𝒓 𝒄𝒔𝒄 π’Ž πœ½π’„π’π’• 𝒏 πœ½π’…πœ½ where m and n are positive integers. β€’ When m is even, we let 𝒔𝒆𝒄 π’Ž 𝜽 = 𝒔𝒆𝒄 π’Žβˆ’πŸ 𝜽 𝒔𝒆𝒄 𝟐 𝜽, and express 𝒔𝒆𝒄 π’Žβˆ’πŸ 𝜽 = (𝒕𝒂𝒏 𝟐 𝜽 + 𝟏) π’Žβˆ’πŸ .We will then obtain products of powers of tan πœƒ 𝑏𝑦 𝑠𝑒𝑐2 πœƒ. The integral could be integrated by means of power formula.
  21. 21. β€’ If n is odd, we express 𝒔𝒆𝒄 π’Ž 𝒕𝒂𝒏 𝒏 𝜽 = 𝒔𝒆𝒄 π’Žβˆ’πŸ πœ½π’•π’‚π’ π’βˆ’πŸ 𝜽(𝐬𝐞𝐜 𝜽 𝐭𝐚𝐧 𝜽).Then we transform π‘‘π‘Žπ‘› π‘›βˆ’1 into power of secπœƒ using the identity 𝒕𝒂𝒏 𝟐 𝜽 = 𝒔𝒆𝒄 𝟐 𝜽 βˆ’ 𝟏. β€’ If m is odd and n is even this can be evaluated using integration by parts
  22. 22. EXAMPLE β€’ Find the indefinite integral. 1. π‘‘π‘Žπ‘›5 π‘₯𝑑π‘₯ 2. π‘‘π‘Žπ‘›3 π‘₯𝑠𝑒𝑐4 π‘₯𝑑π‘₯ 3. 𝑐𝑠𝑐4 π‘₯𝑑π‘₯ 4. π‘π‘œπ‘‘2 π‘₯𝑐𝑠𝑐4 π‘₯𝑑π‘₯ 5. π‘‘π‘Žπ‘›3 π‘₯𝑠𝑒𝑐5 π‘₯𝑑π‘₯
  23. 23. INTEGRALS INVOLVING INVERSE TRIGONOMETRIC FUNCTIONS β€’ Let u be a differentiable function of x, and let a> 0. 1. 𝑑𝑒 π‘Ž2βˆ’π‘’2 = π‘Žπ‘Ÿπ‘π‘ π‘–π‘› 𝑒 π‘Ž + 𝐢 2. 𝑑𝑒 π‘Ž2+𝑒2 = 1 π‘Ž π‘Žπ‘Ÿπ‘π‘‘π‘Žπ‘› 𝑒 π‘Ž + 𝐢 3. 𝑑𝑒 𝑒 𝑒2βˆ’π‘Ž2 = 1 π‘Ž π‘Žπ‘Ÿπ‘π‘ π‘’π‘ 𝑒 π‘Ž + 𝐢
  24. 24. EXAMPLE β€’ Find or evaluate the integral. 1. π‘₯βˆ’3 π‘₯2+1 𝑑π‘₯ 6. 𝑙𝑛2 𝑙𝑛4 π‘’βˆ’π‘₯ 1βˆ’π‘’βˆ’2π‘₯ 𝑑π‘₯ 2. 𝑠𝑒𝑐2 π‘₯ 25βˆ’π‘‘π‘Žπ‘›2 π‘₯ 𝑑π‘₯ 7. π‘₯ 9+8π‘₯2βˆ’π‘₯4 𝑑π‘₯ 3. 3 2 π‘₯(1+π‘₯) 𝑑π‘₯ 8. 3 6 1 25+(π‘₯βˆ’3)2 𝑑π‘₯ 4. 2 3 2π‘₯βˆ’3 4π‘₯βˆ’π‘₯2 𝑑π‘₯ 9. π‘₯+2 βˆ’π‘₯2βˆ’4π‘₯ 𝑑π‘₯ 5. βˆ’2 2 𝑑π‘₯ π‘₯2+4π‘₯+13 10. 2π‘₯βˆ’5 π‘₯2+2π‘₯+2 𝑑π‘₯
  25. 25. HYPERBOLIC FUNCTIONS β€’ Definitions of the Hyperbolic Function π‘ π‘–π‘›β„Žπ‘₯ = 𝑒π‘₯ βˆ’ π‘’βˆ’π‘₯ 2 π‘π‘ π‘β„Žπ‘₯ = 1 π‘ π‘–π‘›β„Žπ‘₯ , π‘₯ β‰  0 cπ‘œπ‘ β„Žπ‘₯ = 𝑒 π‘₯+π‘’βˆ’π‘₯ 2 π‘ π‘’π‘β„Žπ‘₯ = 1 π‘π‘œπ‘ β„Žπ‘₯ π‘‘π‘Žπ‘›β„Žπ‘₯ = π‘ π‘–π‘›β„Žπ‘₯ π‘π‘œπ‘ β„Žπ‘₯ π‘π‘œπ‘‘β„Žπ‘₯ = 1 π‘‘π‘Žπ‘›β„Žπ‘₯ , π‘₯ β‰  0
  26. 26. β€’ HYPERBOLIC IDENTITIES π‘π‘œπ‘ β„Ž2 𝑒 βˆ’ π‘ π‘–π‘›β„Ž2 𝑒 = 1 π‘‘π‘Žπ‘›β„Ž2 𝑒 + π‘ π‘’π‘β„Ž2 𝑒 = 1 π‘π‘œπ‘‘β„Ž2 𝑒 βˆ’ π‘π‘ π‘β„Ž2 𝑒 = 1 cosh2u = π‘π‘œπ‘ β„Ž2 𝑒 + π‘ π‘–π‘›β„Ž2 𝑒 π‘ π‘–π‘›β„Ž2π‘₯ = 2π‘ π‘–π‘›β„Žπ‘₯π‘π‘œπ‘ β„Žπ‘₯ π‘ π‘–π‘›β„Ž2 1 2 𝑒 = 1 2 (π‘π‘œπ‘ β„Ž 𝑒 βˆ’ 1) π‘π‘œπ‘ β„Ž2 π‘₯ = 1 + π‘π‘œπ‘ β„Ž2π‘₯ 2 tanh(x + y) = (π‘‘π‘Žπ‘›β„Ž π‘₯+π‘‘π‘Žπ‘›β„Ž 𝑦) 1+π‘‘π‘Žπ‘›β„Ž π‘₯ π‘‘π‘Žπ‘›β„Žπ‘¦
  27. 27. INTEGRALS OF HYPERBOLIC FUNCTIONS Let u be a differentiable function of x. 1. π‘ π‘–π‘›β„Žπ‘’ 𝑑𝑒 = π‘π‘œπ‘ β„Žπ‘’ + 𝐢 2. π‘π‘œπ‘ β„Žπ‘’ 𝑑𝑒 = π‘ π‘–π‘›β„Žπ‘’ + 𝐢 3. π‘‘π‘Žπ‘›β„Žπ‘’ 𝑑𝑒 = 𝑙𝑛 π‘π‘œπ‘ β„Žπ‘’ + 𝐢 4. π‘π‘œπ‘‘β„Žπ‘’ 𝑑𝑒 = 𝑙𝑛 π‘ π‘–π‘›β„Ž 𝑒 + 𝐢 5. π‘ π‘’π‘β„Žπ‘’ 𝑑𝑒 = π‘‘π‘Žπ‘›βˆ’1 𝑠𝑖𝑛 6. π‘π‘ π‘β„Žπ‘’ 𝑑𝑒 = 𝑙𝑛(π‘π‘œπ‘‘β„Ž 𝑒 βˆ’ π‘π‘ π‘β„Žπ‘’) + 𝐢 7. π‘ π‘’π‘β„Ž2 𝑒 𝑑𝑒 = π‘‘π‘Žπ‘›β„Žπ‘’ + 𝐢 8. π‘π‘ π‘β„Ž2 𝑒 𝑑𝑒 = βˆ’ π‘π‘œπ‘‘β„Ž 𝑒 + 𝐢 9. π‘ π‘’π‘β„Žπ‘’ π‘‘π‘Žπ‘›β„Žπ‘’ 𝑑𝑒 = βˆ’π‘ π‘’π‘β„Žπ‘’ + 𝐢 10. π‘π‘ π‘β„Ž 𝑒 π‘π‘œπ‘‘β„Žπ‘’ 𝑑𝑒 = βˆ’π‘π‘ π‘β„Žπ‘’ + 𝐢
  28. 28. INVERSE HYPERBOLIC FUNCTIONS β€’ Function Domain β€’ π‘ π‘–π‘›β„Žβˆ’1 π‘₯ = 𝑙𝑛(π‘₯ + π‘₯2 + 1)(βˆ’βˆž, +∞) β€’ π‘π‘œπ‘ β„Žβˆ’1 π‘₯ = 𝑙𝑛(π‘₯ + π‘₯2 βˆ’ 1) 1, ∞ β€’ π‘‘π‘Žπ‘›β„Žβˆ’1 π‘₯ = 1 2 𝑙𝑛 1+π‘₯ 1βˆ’π‘₯ βˆ’1,1 β€’ π‘π‘œπ‘‘β„Žβˆ’1 π‘₯ = 1 2 𝑙𝑛 π‘₯+1 π‘₯βˆ’1 βˆ’βˆž, βˆ’1 U(1, ∞) β€’ π‘ π‘’π‘β„Žβˆ’1 π‘₯ = 𝑙𝑛 1+ 1βˆ’π‘₯2 π‘₯ (0,1 β€’ π‘π‘ π‘β„Žβˆ’1 π‘₯ = 𝑙𝑛 1 π‘₯ + 1+π‘₯2 π‘₯ βˆ’βˆž, 0 U(0, ∞)
  29. 29. INTEGRATION INVOLVING INVERSE HYPERBOLIC FUNCTION β€’ Let u be a differentiable function of x. β€’ 𝑑𝑒 𝑒2Β±π‘Ž2 = 𝑙𝑛 𝑒 + 𝑒2 Β± π‘Ž2 + 𝐢 β€’ 𝑑𝑒 π‘Ž2βˆ’π‘’2 = 1 2π‘Ž 𝑙𝑛 π‘Ž+𝑒 π‘Žβˆ’π‘’ + 𝐢 β€’ 𝑑𝑒 𝑒 π‘Ž2±𝑒2 = βˆ’ 1 π‘Ž 𝑙𝑛 π‘Ž+ π‘Ž2±𝑒2 𝑒 + 𝐢
  30. 30. EXAMPLE β€’ Find the indefinite integral. 1. 1 3βˆ’9π‘₯2 𝑑π‘₯ 2. 1 π‘₯ 1+π‘₯ 𝑑π‘₯ 3. 1 1βˆ’4π‘₯βˆ’2π‘₯2 𝑑π‘₯ 4. 𝑑π‘₯ (π‘₯+2) π‘₯2+4π‘₯+8 5. π‘₯ 1+π‘₯3 𝑑π‘₯

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