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anonymous

  • 5 years ago

How would you integrate the following function? Particularly, what would you do after the last following step? integral( (9-x^2)^(1/2) dx) x = 3 sin u u = invsin(x/3) integral( 3 (cos u) d?

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  1. anonymous
    • 5 years ago
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    fuk

  2. anonymous
    • 5 years ago
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    I have the answer sheet, and wolfram gives a decent explanation, but I can't understand what the next step is.

  3. anonymous
    • 5 years ago
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    pull out 3 get 3*integral(cos u du) = 3sin u + C

  4. anonymous
    • 5 years ago
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    You just have to know the integral of cos(u)

  5. anonymous
    • 5 years ago
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    Nah, it's still in the form dx, not du. My worksheet says the answer is. (1/2)x (9-x^2)^(1/2) + 9 invsin(x/3))

  6. anonymous
    • 5 years ago
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    Once you integrate it you can plug back in what you substituted for u

  7. anonymous
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    You need to solve for dx in terms of du.

  8. anonymous
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    plug u back in

  9. anonymous
    • 5 years ago
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    Yeah, that's what I'm having issues with. I've never been strong with trig functions.

  10. anonymous
    • 5 years ago
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    When you plug u in, there's a sin and cosine function, which indicates integration by parts, that seem right?

  11. anonymous
    • 5 years ago
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    \[x = 3 sin(u) \implies dx = 3cos(u)du\] \[(9-x^2)^{1/2} dx= [9-9cos^2(u)]^{1/2} \cdot (3cos(u)\ du)\]

  12. anonymous
    • 5 years ago
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    Then you can factor out a 3 from the sqrted factor and a 3 from the du part and have a 9 out front and rewrite it as \((1-cos^2(u))^{1/2}cos(u) du\). This solves nicely with another substitution.

  13. anonymous
    • 5 years ago
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    oh wait, no.

  14. anonymous
    • 5 years ago
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    It's a tough one XD

  15. anonymous
    • 5 years ago
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    u use sin(x)^2 + cos(x)^2 = 1

  16. anonymous
    • 5 years ago
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    yeah, 1-cos^2(u) = sin^2(u)

  17. anonymous
    • 5 years ago
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    which you can take the square root of easily and also u-sub easily.

  18. anonymous
    • 5 years ago
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    so z = sin u dz = cos x dx

  19. anonymous
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    i mean dz = cos u du

  20. anonymous
    • 5 years ago
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    integral(z) = z^2/2 then replace with u and then replace with x

  21. anonymous
    • 5 years ago
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    I have ... \[3\cos(\sin^-(x/3))\] Where would I go from here?

  22. anonymous
    • 5 years ago
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    that's not right. How did you get there?

  23. anonymous
    • 5 years ago
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    \[\cos x =(1- \sin^2 x)^{?}\]

  24. anonymous
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    other way round bob.

  25. anonymous
    • 5 years ago
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    no

  26. anonymous
    • 5 years ago
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    x = 3sin(u) x/3 = sin(u) invsin(x/3) = u integral ( 3cos(u)) integral (3cos(invsin(x/3)))

  27. anonymous
    • 5 years ago
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    you don't want to sub back in your u until after you integrate.

  28. anonymous
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    so that'll be negative sin(u)

  29. anonymous
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    because the integral of cosine is sin, right? See, I'm confused because it's still technically in the form dx, not du.

  30. anonymous
    • 5 years ago
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    negative sin*

  31. anonymous
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    It shouldn't be. When you do the u-sub you need to replace your dx with some function of u du

  32. anonymous
    • 5 years ago
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    This is what you're starting with right? \[\int (9-x^2)^{1/2} dx\]

  33. anonymous
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    Yessir.

  34. anonymous
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    Ok, and we let \(x = 3sin(u)\).

  35. anonymous
    • 5 years ago
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    So you know, I'm studying for a test tomorrow, this isn't homework.

  36. anonymous
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    yeah, np

  37. anonymous
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    Ok, so now what is dx in terms of du?

  38. anonymous
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    dx = 3cosu du

  39. anonymous
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    Right.

  40. anonymous
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    So now lets swap out \(x^2\) and \(dx\) with the appropriate expressions of u and du.

  41. anonymous
    • 5 years ago
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    What do you have?

  42. anonymous
    • 5 years ago
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    (9-9sin^2u)^(1/2) dx (9(1-sin^2u))^(1/2) dx 3 cos u dx

  43. anonymous
    • 5 years ago
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    you didn't swap out dx.

  44. anonymous
    • 5 years ago
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    I don't know how the hell to do that my friend :P

  45. anonymous
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    dx = 3cosu du You said it yourself.

  46. anonymous
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    Agreed, but if you .... OH!

  47. anonymous
    • 5 years ago
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    3cosu * 3 cosu ?

  48. anonymous
    • 5 years ago
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    Indeed (along with a du)

  49. anonymous
    • 5 years ago
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    okay, so we have ... \[\int\limits_{}^{} 3 \cos (u)cos(u) du\] which is ... \[\int\limits_{}^{} 9 \cos ^2 (u) ^ {1/2} du\]

  50. anonymous
    • 5 years ago
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    forget that square root

  51. anonymous
    • 5 years ago
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    no its Integral (3 cos^2 x)

  52. anonymous
    • 5 years ago
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    Right. So then you can use the double/half angle equation to swap the \(cos^2\) for an expression involving the cos(2u)

  53. anonymous
    • 5 years ago
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    She said she would be providing those equations on the test.

  54. anonymous
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    Should be 9 cause there's a 3 from the 9 under the sqrt and a 3 from the du replacement of dx.

  55. anonymous
    • 5 years ago
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    \[\cos^2 x = (1 + \cos 2x )/2\]

  56. anonymous
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    that's the one.

  57. anonymous
    • 5 years ago
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    I think polpak's solution is about the same as the attachment.

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  58. anonymous
    • 5 years ago
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    Yeah but polpak is a better teacher than any pdf :p

  59. anonymous
    • 5 years ago
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    I'm working out the cos squared substitution on paper as we speak.

  60. anonymous
    • 5 years ago
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    particularly a pdf that is a screenshot of wolframalpha.

  61. anonymous
    • 5 years ago
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    ;)

  62. anonymous
    • 5 years ago
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    Though alpha is good for checking your answers and finding where you made a mistake, but sometimes it's process is a little cumbersome.

  63. anonymous
    • 5 years ago
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    its rather. I hate that stupid part of english.

  64. anonymous
    • 5 years ago
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    Tell me about it, I checked wolfram and couldn't find how they went from 3cos(u) to 9cos^2(u). This is getting uglier and uglier as we go along. I have... \[\int\limits_{}^{} 9 {( 1+\cos2u)}/2 du\]

  65. anonymous
    • 5 years ago
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    That's not ugly at all!

  66. anonymous
    • 5 years ago
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    Break it into two integrals.

  67. anonymous
    • 5 years ago
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    and pull the 9/2 all the way out front.

  68. anonymous
    • 5 years ago
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    Oh wow. You're right. Hah! I need to practice these more, for sure.

  69. anonymous
    • 5 years ago
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    Then the only thing you have to do is another sub for z = 2u on the second integral if you don't feel confident doing it in your head.

  70. anonymous
    • 5 years ago
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    Nah it's just division by 2

  71. anonymous
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    indeed.

  72. anonymous
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    \[9/2 ( u - \sin(2u)/u)\] Does this look alright?

  73. anonymous
    • 5 years ago
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    why is that u under the sin(2u) ?

  74. anonymous
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    Oh, sorry that's supposed to be a 2. I have dyslexia, it's a pain in the rear end sometimes.

  75. anonymous
    • 5 years ago
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    yes. Me too.

  76. anonymous
    • 5 years ago
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    With a 2 in the denominator that is correct.

  77. anonymous
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    oh, wait

  78. anonymous
    • 5 years ago
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    shouldn't be a -sin(2u)

  79. anonymous
    • 5 years ago
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    \[\int cos(u)du = sin(u) +C \]

  80. anonymous
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    should be positive.

  81. anonymous
    • 5 years ago
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    the derivative of cos is -sin, but the integral of cos is just sin

  82. anonymous
    • 5 years ago
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    \[u = \sin^{-1}(x/3)\] You're right, I got that mixed up with the rules for sin integration.

  83. anonymous
    • 5 years ago
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    yeah, so that works nicely given what u is.

  84. anonymous
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    So if I were to substitute u, then I would be finished?

  85. anonymous
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    you can simplify probably, but it's up to you

  86. anonymous
    • 5 years ago
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    One quick note, in my packet, the answer is 1/2 (x * (9-x^2)^(1/2) + 9 sin -1 (x/3)) How can it be so different?

  87. anonymous
    • 5 years ago
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    because they plug back in for sin(2u) and use the formula for \(sin(2\theta)\)

  88. anonymous
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    to just get a sin or cos squared which they then swap out for 1-cos or sin squared

  89. anonymous
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    all kinds of fun trig subs going on.

  90. anonymous
    • 5 years ago
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    We can work through it if you want

  91. anonymous
    • 5 years ago
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    well, I'm happy leaving it in the form inverse sine. I'm just curious what would happen when you substitute sin^-1(x/3) into sin(2u)/2

  92. anonymous
    • 5 years ago
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    because I believe she would want it back in the form of x

  93. anonymous
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    I would imagine black holes are created, or something of that variety (when sin and inverse sin collide)

  94. anonymous
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    No, you just get back what you started with. \[sin(sin^{-1}(x/3)) = x/3\]

  95. anonymous
    • 5 years ago
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    What happened to the two?

  96. anonymous
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    it's still there, I'm showing a different example

  97. anonymous
    • 5 years ago
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    Ok, so here it is. \[\frac{9}{2}sin(2u) =\frac{ 9(2)}{2}cos(u)sin(u)\]\[ =9(\sqrt{1-sin^2(u)})sin(u) \]

  98. anonymous
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    and when you plug your u back in your sin will get nuked by the inverse

  99. anonymous
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    Very sweet, I just need a second to absorb.

  100. anonymous
    • 5 years ago
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    Thanks again for your help though, I wish there was some way I could repay you. I hope a challenging problem can suffice since that's all I have to offer.

  101. anonymous
    • 5 years ago
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    the first step comes from: \[sin(2\theta) = cos(\theta)sin(\theta)\] the second step comes from: \[cos\theta = \sqrt{1-sin^2\theta}\]

  102. anonymous
    • 5 years ago
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    oh yes, it's fun! =)

  103. anonymous
    • 5 years ago
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    Maybe somebody will post a linear algebra problem :p I know I enjoy helping people with Calc 1 and Algebra problems, so I imagine it's the same for you, except more advanced.

  104. anonymous
    • 5 years ago
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    indeed =) I'm taking Linear algebra now ;)

  105. anonymous
    • 5 years ago
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    Have with with that! :D I enjoy the idea of precision in math, however the process of learning about it kills me. Especially when there's a concept I can't comprehend.

  106. anonymous
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    just takes time to digest

  107. anonymous
    • 5 years ago
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    Well, I'm off to bed, thanks again for your help. She said that a problem to the one you helped me solve just now will the be the hardest on the exam. Besides that, it's mostly partial fractions, integrals by parts. 1 step integrals. Have a nice night!

  108. anonymous
    • 5 years ago
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    Good luck on the test!

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