example 1

Example 1 x 2 dx x 2 16 Example 1 x 2 dx x 2 16 We want - PowerPoint PPT Presentation

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Example 1 x 2 dx x 2 16 Example 1 x 2 dx x 2 16 We want to use the identity: Example 1 x 2 dx x 2 16 We want to use the identity: 1 + tan 2 t = sec 2 t Example 1 x 2 dx x 2 16 We want to

  3. Example 1 � x 2 dx √ x 2 − 16

  4. Example 1 � x 2 dx √ x 2 − 16 We want to use the identity:

  5. Example 1 � x 2 dx √ x 2 − 16 We want to use the identity: 1 + tan 2 t = sec 2 t

  6. Example 1 � x 2 dx √ x 2 − 16 We want to use the identity: 1 + tan 2 t = sec 2 t We can write the integral as:

  7. Example 1 � x 2 dx √ x 2 − 16 We want to use the identity: 1 + tan 2 t = sec 2 t We can write the integral as: � x 2 dx �� x � 2 − 1 4 4

  8. Example 1 � x 2 dx √ x 2 − 16 We want to use the identity: 1 + tan 2 t = sec 2 t We can write the integral as: � x 2 dx �� x � 2 − 1 4 4 So, we make the substitution:

  9. Example 1 � x 2 dx √ x 2 − 16 We want to use the identity: 1 + tan 2 t = sec 2 t We can write the integral as: � x 2 dx �� x � 2 − 1 4 4 So, we make the substitution: x 4 = sec t

  10. Example 1 � x 2 dx √ x 2 − 16 We want to use the identity: 1 + tan 2 t = sec 2 t We can write the integral as: � x 2 dx �� x � 2 − 1 4 4 So, we make the substitution: x 4 = sec t ⇒ x = 4 sec t

  11. Example 1 � x 2 dx √ x 2 − 16 We want to use the identity: 1 + tan 2 t = sec 2 t We can write the integral as: � x 2 dx �� x � 2 − 1 4 4 So, we make the substitution: x 4 = sec t ⇒ x = 4 sec t dx dt = 4 tan t sec t

  12. Example 1 � x 2 dx √ x 2 − 16 We want to use the identity: 1 + tan 2 t = sec 2 t We can write the integral as: � x 2 dx �� x � 2 − 1 4 4 So, we make the substitution: x 4 = sec t ⇒ x = 4 sec t dx dt = 4 tan t sec t ⇒ dx = 4 tan t sec tdt

  13. Example 1 So, we can make the substitution:

  14. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt

  15. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � x 2 dx √ x 2 − 16

  16. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16

  17. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt x 2 � �� � 16 sec 2 t . 4 tan t sec t � � x 2 dx √ √ = x 2 − 16 16 sec 2 t − 16

  18. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt dx � 16 sec 2 t . � �� � � x 2 dx 4 tan t sec t √ = √ x 2 − 16 16 sec 2 t − 16

  19. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = � x 2 − 16 16 sec 2 t − 16 � �� � x 2

  20. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16

  21. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt = � 16 (sec 2 t − 1)

  22. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = = √ � 16 (sec 2 t − 1) sec 2 t − 1 4

  23. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = = √ � 16 (sec 2 t − 1) sec 2 t − 1 4 But:

  24. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = = √ � 16 (sec 2 t − 1) sec 2 t − 1 4 But: 1 + tan 2 t = sec 2 t

  25. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = = √ � 16 (sec 2 t − 1) sec 2 t − 1 4 But: 1 + tan 2 t = sec 2 t ⇒ sec 2 t − 1 = tan 2 t

  26. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = = √ � 16 (sec 2 t − 1) sec 2 t − 1 4 But: 1 + tan 2 t = sec 2 t ⇒ sec 2 t − 1 = tan 2 t � 64 tan t sec 3 tdt √ sec 2 t − 1 4

  27. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = = √ � 16 (sec 2 t − 1) sec 2 t − 1 4 But: 1 + tan 2 t = sec 2 t ⇒ sec 2 t − 1 = tan 2 t � 64 tan t sec 3 tdt ✘✘✘✘✘✘ ✿ tan t √ sec 2 t − 1 4

  28. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = = √ � 16 (sec 2 t − 1) sec 2 t − 1 4 But: 1 + tan 2 t = sec 2 t ⇒ sec 2 t − 1 = tan 2 t � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = ✘✘✘✘✘✘ ✿ tan t √ 4 tan t sec 2 t − 1 4

  29. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = = √ � 16 (sec 2 t − 1) sec 2 t − 1 4 But: 1 + tan 2 t = sec 2 t ⇒ sec 2 t − 1 = tan 2 t � 64 tan t sec 3 tdt � 64 ✘✘ ✘ tan t sec 3 tdt = ✘✘✘✘✘✘ ✿ tan t 4 ✘✘ ✘ √ tan t sec 2 t − 1 4

  30. Example 1 So, we can make the substitution: x = 4 sec t dx = 4 tan t sec tdt � 16 sec 2 t . 4 tan t sec t � x 2 dx √ = √ x 2 − 16 16 sec 2 t − 16 � 64 tan t sec 3 tdt � 64 tan t sec 3 tdt = = √ � 16 (sec 2 t − 1) sec 2 t − 1 4 But: 1 + tan 2 t = sec 2 t ⇒ sec 2 t − 1 = tan 2 t � 64 tan t sec 3 tdt � 64 ✘✘ ✘ tan t sec 3 tdt � sec 3 tdt = = 16 ✘✘✘✘✘✘ ✿ tan t 4 ✘✘ ✘ √ tan t sec 2 t − 1 4

  31. Example 1 So, we ”only” need to solve this trigonometric integral:

  32. Example 1 So, we ”only” need to solve this trigonometric integral: � sec 3 tdt 16

  33. Example 1 So, we ”only” need to solve this trigonometric integral: � sec 3 tdt 16 We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts):

  34. Example 1 So, we ”only” need to solve this trigonometric integral: � sec 3 tdt 16 We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): � sec 3 tdt = 8 sec t tan t + 8 ln | sec t + tan t | + C 16

  35. Example 1 So, we ”only” need to solve this trigonometric integral: � sec 3 tdt 16 We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): � sec 3 tdt = 8 sec t tan t + 8 ln | sec t + tan t | + C 16 Now we only need to substitute back. Remember that our substitution was:

  36. Example 1 So, we ”only” need to solve this trigonometric integral: � sec 3 tdt 16 We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): � sec 3 tdt = 8 sec t tan t + 8 ln | sec t + tan t | + C 16 Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4

  37. Example 1 So, we ”only” need to solve this trigonometric integral: � sec 3 tdt 16 We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): � sec 3 tdt = 8 sec t tan t + 8 ln | sec t + tan t | + C 16 Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4 We need to have tan t also as a function of x :

  38. Example 1 So, we ”only” need to solve this trigonometric integral: � sec 3 tdt 16 We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): � sec 3 tdt = 8 sec t tan t + 8 ln | sec t + tan t | + C 16 Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4 We need to have tan t also as a function of x : � sec 2 t − 1 tan t =

  39. Example 1 So, we ”only” need to solve this trigonometric integral: � sec 3 tdt 16 We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): � sec 3 tdt = 8 sec t tan t + 8 ln | sec t + tan t | + C 16 Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4 We need to have tan t also as a function of x : � x 2 � sec 2 t − 1 = tan t = 16 − 1

  40. Example 1 So, we ”only” need to solve this trigonometric integral: � sec 3 tdt 16 We won’t solve this one, I’ll only give you the answer (it is solved using integration by parts): � sec 3 tdt = 8 sec t tan t + 8 ln | sec t + tan t | + C 16 Now we only need to substitute back. Remember that our substitution was: x = 4 sec t ⇒ sec t = x 4 We need to have tan t also as a function of x : � x 2 16 − 1 = 1 � � sec 2 t − 1 = x 2 − 16 tan t = 4

  41. Example 1 So, we need to substitute:

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