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anonymous

  • 5 years ago

Please, please help with substitution rule for indefinite integrals:

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{}^{} (4z + \sin(3z))e ^{6z ^{2}-\cos(3z)} dz\]

  2. anonymous
    • 5 years ago
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    whew, that's a doozy.

  3. anonymous
    • 5 years ago
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    So what do you think you want to use for substitution?

  4. anonymous
    • 5 years ago
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    \[\int\limits_{?}^{?} \sin(x) d x = -\cos(x) + constant\]

  5. anonymous
    • 5 years ago
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    Not sure...possibly something to do with 3z? I have no idea to be honest

  6. anonymous
    • 5 years ago
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    I would let u = that yucky exponent on the e

  7. anonymous
    • 5 years ago
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    Ignore the two ?'s

  8. anonymous
    • 5 years ago
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    Okay but how does that help?

  9. anonymous
    • 5 years ago
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    Well. \[u = 6x^2 - cos(3z)\] \[\implies du = 12z +3sin(3z)dz = 3(4z + sin(3z))dz\]

  10. anonymous
    • 5 years ago
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    Oh so then \[du = 2z ^{3}-\sin(3z)dx\]

  11. anonymous
    • 5 years ago
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    Close, but you integrated. You want to take the derivative

  12. anonymous
    • 5 years ago
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    oh

  13. anonymous
    • 5 years ago
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    Okay. Where do I plug u into?

  14. anonymous
    • 5 years ago
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    Well first rewrite dz in terms of du using that last equation I had.

  15. anonymous
    • 5 years ago
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    ?

  16. anonymous
    • 5 years ago
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    \(du = 3(4z + sin(3z))dz \implies dz =\ ?\)

  17. anonymous
    • 5 years ago
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    I don't understand - what am I solving for?

  18. anonymous
    • 5 years ago
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    Well for this part you're solving for dz.

  19. anonymous
    • 5 years ago
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    Err...I divide 3(4z + sin(3z)) on both sides?

  20. anonymous
    • 5 years ago
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    Yes

  21. anonymous
    • 5 years ago
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    So \[dz = \frac{1}{3(4z+sin(3z))} du\]

  22. anonymous
    • 5 years ago
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    So now. Go back to your integral, and replace the exponent of e with u, and the dz with your expression for du, and you'll be pleasantly surprised.

  23. anonymous
    • 5 years ago
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    Wait a minute...Why do I replace dz?

  24. anonymous
    • 5 years ago
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    Because we are no longer integrating with respect to z, we are integrating with respect to u now. Because it turns out u is nicer.

  25. anonymous
    • 5 years ago
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    We are replacing our nasty expression with z for a nice expression with u.

  26. anonymous
    • 5 years ago
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    \[(4z + \sin(3z)e ^{u}\]du

  27. anonymous
    • 5 years ago
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    But remember that dz does not equal du.

  28. anonymous
    • 5 years ago
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    ButI thought we replaced dz

  29. anonymous
    • 5 years ago
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    \[dz = \frac{1}{3(4z+sin(3z))} du\]

  30. anonymous
    • 5 years ago
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    So you have to put that whole thing in for dz

  31. anonymous
    • 5 years ago
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    I'm so lost

  32. anonymous
    • 5 years ago
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    Lets try an easy one to start with.

  33. anonymous
    • 5 years ago
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    \[\int x*e^{x^2}dx\]

  34. anonymous
    • 5 years ago
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    We let \(u = x^2\)

  35. anonymous
    • 5 years ago
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    So \[du = (2x)dx \implies dx = \frac{1}{2x}du\]

  36. anonymous
    • 5 years ago
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    With me so far?

  37. anonymous
    • 5 years ago
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    sure

  38. anonymous
    • 5 years ago
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    If not, please ask

  39. anonymous
    • 5 years ago
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    yes

  40. anonymous
    • 5 years ago
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    Ok. So then \[\int x*e^{x^2}dx = \int x*e^u(\frac{1}{2x} du)\]

  41. anonymous
    • 5 years ago
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    And we can cancel the x in front with the x in the denominator from the du.

  42. anonymous
    • 5 years ago
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    to get \[\frac{1}{2}\int e^udu\]

  43. anonymous
    • 5 years ago
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    Which is nice and easy.

  44. anonymous
    • 5 years ago
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    \[\frac{1}{2}\int e^udu = \frac{1}{2}e^u + c = \frac{1}{2}e^{x^2} + C\]

  45. anonymous
    • 5 years ago
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    So I've got\[1/3 e ^{u}\] for my problem when do I integrate?

  46. anonymous
    • 5 years ago
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    Now.

  47. anonymous
    • 5 years ago
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    Before or after I plug in u

  48. anonymous
    • 5 years ago
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    Don't switch back your u until after you integrate.

  49. anonymous
    • 5 years ago
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    How do I integrate that?

  50. anonymous
    • 5 years ago
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    \[\int \frac{1}{3}e^u du = \frac{1}{3}\int e^udu\]

  51. anonymous
    • 5 years ago
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    What is the integral of \(e^x\)?

  52. anonymous
    • 5 years ago
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    e^x

  53. anonymous
    • 5 years ago
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    Right.

  54. anonymous
    • 5 years ago
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    \[1/3 *e ^{6z ^{2}-\cos(3z)} \]

  55. anonymous
    • 5 years ago
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    +C

  56. anonymous
    • 5 years ago
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    oh right

  57. anonymous
    • 5 years ago
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    Nice job!

  58. anonymous
    • 5 years ago
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    Oh my goodness. Would you please walk me through one more?

  59. anonymous
    • 5 years ago
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    Sure, but you'll have to do more of it this time ;)

  60. anonymous
    • 5 years ago
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    How's it start?

  61. anonymous
    • 5 years ago
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    Okay I'll try:

  62. anonymous
    • 5 years ago
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    \[\int\limits_{}^{}(4z ^{3} + 6\csc ^{2}(6z)(z ^{4}-\cot(6z))^{4} dz\]

  63. anonymous
    • 5 years ago
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    Ugh

  64. anonymous
    • 5 years ago
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    Do I need to FOIL first?

  65. anonymous
    • 5 years ago
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    No.

  66. anonymous
    • 5 years ago
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    These are all ugly u substitution problems. Foiling will just make you cry.

  67. anonymous
    • 5 years ago
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    You want to find a nice thing to pick for a u substitution.

  68. anonymous
    • 5 years ago
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    Oh okay

  69. anonymous
    • 5 years ago
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    Do you notice anything interesting about these two things?

  70. anonymous
    • 5 years ago
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    Not really sure what to look for, actually.

  71. anonymous
    • 5 years ago
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    \((4z^3 + 6csc^2(6z)) * (z^4 - cot(6z))^4\) You are looking for something in the equation that looks like the derivative of another part of the equation.

  72. anonymous
    • 5 years ago
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    Err expression, not equation

  73. anonymous
    • 5 years ago
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    Oh! I guess the 4z^3 and z^4

  74. anonymous
    • 5 years ago
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    And what about the \(6csc^2(6z)\) and the -cot(6z) ?

  75. anonymous
    • 5 years ago
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    AND the csc^2

  76. anonymous
    • 5 years ago
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    right.

  77. anonymous
    • 5 years ago
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    So what you want to do is pick the one that is not the derivative, and let that be u.

  78. anonymous
    • 5 years ago
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    So let\[u = z ^{4} - \cot(6z)\]

  79. anonymous
    • 5 years ago
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    Yep, then solve for dz in terms of z and du.

  80. anonymous
    • 5 years ago
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    So \[\int\limits_{}^{}(4z ^{3}+6\csc ^{2}(6z))u ^{4}du\]

  81. anonymous
    • 5 years ago
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    Nope. but close.

  82. anonymous
    • 5 years ago
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    \[u = z^4 - cot(6z)\] So \[du =\ ?\]

  83. anonymous
    • 5 years ago
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    4z^3 + 6csc^2(6z)

  84. anonymous
    • 5 years ago
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    close.

  85. anonymous
    • 5 years ago
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    \[du = (4z^3 + 6csc^2(6z))dz\]

  86. anonymous
    • 5 years ago
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    So then what does dz = ?

  87. anonymous
    • 5 years ago
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    1 over all that mess

  88. anonymous
    • 5 years ago
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    yep.

  89. anonymous
    • 5 years ago
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    So plug in 1 over all that mess times du for dz in the original equation. and plug in u for what we said it equaled.

  90. anonymous
    • 5 years ago
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    \[(4z ^{3} + 6\csc ^{2}(6z))u ^{4} * 1/4z ^{3} + 6\csc ^{2}(6z)\]

  91. anonymous
    • 5 years ago
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    simplify!

  92. anonymous
    • 5 years ago
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    u^4?

  93. anonymous
    • 5 years ago
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    yes

  94. anonymous
    • 5 years ago
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    \[(z ^{4} - \cot(6z))^{4}\]

  95. anonymous
    • 5 years ago
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    No

  96. anonymous
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    \[\int(4z^3 + 6csc^2(6z)) * (z^4 - cot(6z))^4dz\] \[ = \int (4z^3 + 6csc^2(6z)) u^4(\frac{1}{4z^3 + 6csc^2(6z)})du \] \[= \int u^4du =\ ?\]

  97. anonymous
    • 5 years ago
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    ? I don't know

  98. anonymous
    • 5 years ago
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    I can't pull anything out in front of the integral sign

  99. anonymous
    • 5 years ago
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    \[\int x^2dx = \frac{x^3}{3} +C\] Right?

  100. anonymous
    • 5 years ago
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    Oh! Integrate!

  101. anonymous
    • 5 years ago
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    Yes. Once you have something nice, you just integrate.

  102. anonymous
    • 5 years ago
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    Then after you integrate you plug back in what u was.

  103. anonymous
    • 5 years ago
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    \[(z ^{4} - \cot(6z)^{5}/ 5 +c\]

  104. anonymous
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    Yep.

  105. anonymous
    • 5 years ago
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    WOW. Good grief. Thanks so much. I am going to use these as examples for the next ones. Thanks again

  106. anonymous
    • 5 years ago
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    Very welcome =)

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