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escolas

By completing the square and using a trig substitution, evaluate

  • one year ago
  • one year ago

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  1. escolas
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    \[\int\limits_{}^{}\frac{ x dx }{-x^{2}-2x+3 }\]

    • one year ago
  2. escolas
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    After completing the square I got \[\int\limits_{}^{}\frac{ x dx }{ (x+1)^2 +4 }\]

    • one year ago
  3. escolas
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    But I'm not sure what the mean by using a trig substitution...

    • one year ago
  4. zepdrix
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    Hmm did you complete the square correctly? I feel like there should be a negative on the outside..

    • one year ago
  5. zepdrix
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    \[\large -x^2-2x+3 \qquad = \qquad -(x^2+2x-3)\]Then to complete the square, we'll add and subtract 1.\[\large -(\color{orangered}{x^2+2x+1}-1-3) \qquad = \qquad -(\color{orangered}{(x+1)^2}-4)\]

    • one year ago
  6. escolas
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    See the negative confused me too, I forgot about that. that seems right

    • one year ago
  7. zepdrix
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    If you don't want to factor out the negative, then I guess we would have,\[\large 4-(x+1)^2\]

    • one year ago
  8. zepdrix
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    Which is probably the way we want to write it :)

    • one year ago
  9. escolas
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    What I had originally done was moved the 3 over to the other side and then divided by -1 but I don't think thats right

    • one year ago
  10. escolas
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    But what does it mean by a trig substitution?

    • one year ago
  11. zepdrix
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    So are you familiar with the idea of `U Substitution`? It's a similar idea, just a little bit more complex c:

    • one year ago
  12. escolas
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    I am, is it something like the proof of the area of a circle where you randomly (or seemingly randomly) substitute in a sinx or cosx to magically make things work?

    • one year ago
  13. zepdrix
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    Here is our integral so far, \[\large \int\limits \frac{x\;dx}{4-(x+1)^2}\] Just to make sure we're on the same page.

    • one year ago
  14. zepdrix
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    The purpose of a `Trig sub` is that by applying one, we can get rid of the `addition/subtraction` in the bottom. We can turn it into a single term in the denominator, and it makes it a ton easier to deal with.

    • one year ago
  15. escolas
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    Yep, perfect!

    • one year ago
  16. escolas
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    Ok yea that makes sense I think..

    • one year ago
  17. zepdrix
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    It makes more sense when you have sqrt's in the bottom. But this problem still applies. When we see this form, \[\large a^2-x^2\]We'll make the substitution \(\large x=a\sin\theta\). Here is what will happen when we do that.\[\large a^2-x^2 \qquad = \qquad a^2-(a \sin \theta)^2 \qquad = \qquad a^2-a^2\sin^2 \theta\]\[\large a^2(1-\sin^2\theta) \qquad = \qquad a^2(\color{royalblue}{\cos^2\theta})\]

    • one year ago
  18. zepdrix
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    Lemme know if you're confused about any of that simplification. The blue step was where we applied this identity,\[\large \color{royalblue}{\cos^2\theta+\sin^2\theta=1}\qquad \rightarrow \qquad \color{royalblue}{\cos^2\theta=1-\sin^2\theta}\]

    • one year ago
  19. escolas
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    wow thats awesome...

    • one year ago
  20. escolas
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    so then you have x over that though, where can you go from there?

    • one year ago
  21. zepdrix
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    So we'll have to replace several pieces :) We'll also have to substitute in for the `dx`, so we'll have to take the derivative of our \(\large x=a\sin\theta\).

    • one year ago
  22. escolas
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    oh nvm...you substitute the x in the numerator too...

    • one year ago
  23. escolas
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    I'm slow haha sorry

    • one year ago
  24. zepdrix
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    Yah good call :D

    • one year ago
  25. zepdrix
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    In the problem we're working on, it's just a tiny bit more complicated, \(x\) is not the thing being squared. \((x+1)\) is. So THAT is the thing we'll setting up to substitute.

    • one year ago
  26. zepdrix
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    We want to let,\[\large x+1=2\sin \theta\]Take a look at that a minute, let it sink in :D Our \(a\) is 2 right? Because we have \(2^2\) in front of the subtraction.

    • one year ago
  27. escolas
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    Ohh ok yea I was just about to do it and got a little stuck there. that makes more sense. So how do we deal with the numerator then?

    • one year ago
  28. zepdrix
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    Let's write our integral like this, maybe it's easier to read this way,\[\large \int\limits\limits \frac{x}{4-(x+1)^2}\left(dx\right)\]

    • one year ago
  29. zepdrix
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    Oh i see what you're saying. :) Hmm we'll have to be a little sneaky I guess. \[\large x+1=2\sin \theta \qquad \rightarrow \qquad x=2\sin \theta -1\]

    • one year ago
  30. escolas
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    Again...a little slow. that makes perfect sense

    • one year ago
  31. zepdrix
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    Hold on the microwave beeped!! :O brb lol

    • one year ago
  32. escolas
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    haha no worries!!

    • one year ago
  33. escolas
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    So I think I correctly reduced it down to \[\int\limits_{}^{}\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }\]

    • one year ago
  34. escolas
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    Does that seem right so far?

    • one year ago
  35. zepdrix
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    \[\large \int\limits\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }(dx)\]Yah looks good so far :)

    • one year ago
  36. escolas
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    But I need a \[d \theta\]

    • one year ago
  37. zepdrix
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    true story :O

    • one year ago
  38. escolas
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    haha is that just the derivative of x+1?

    • one year ago
  39. escolas
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    no 2sinx

    • one year ago
  40. zepdrix
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    \[\large x+1=2\sin \theta\]We want to take the derivative of both sides, with respect to x. Yah good, both sides.

    • one year ago
  41. escolas
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    \[2\cos \theta\]?

    • one year ago
  42. zepdrix
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    Err maybe we want to take it with respect to theta, then we don't have the chain rule. It won't matter either way, I'm just trying to think about what will be easier to understand.

    • one year ago
  43. zepdrix
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    \[\large dx=2\cos \theta\; d \theta\]That's what you came up with? yah looks good.

    • one year ago
  44. escolas
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    Ok cool this still isn't loooking too great haha

    • one year ago
  45. zepdrix
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    XD

    • one year ago
  46. escolas
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    Oh do the cosines cancel out?

    • one year ago
  47. zepdrix
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    \[\large \int\limits\limits\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }(2\cos \theta \;d \theta)\]It looks like we can make a few cancellations.\[\large \int\limits\limits\limits\frac{ \cancel2\sin \theta-1}{ \cancel4\cos^{\cancel2\;1} \theta }(\cancel2\cancel{\cos \theta} \;d \theta)\]

    • one year ago
  48. escolas
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    the -1 is still messing with it though, isn't it?

    • one year ago
  49. zepdrix
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    \[\large \int\limits \frac{\sin \theta-1}{\cos \theta}d \theta\] Leaving us with this I believe.

    • one year ago
  50. zepdrix
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    Yes it is! But notice that the -1 is now on TOP!

    • one year ago
  51. escolas
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    haha is that better?

    • one year ago
  52. zepdrix
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    When it's in the bottom, that causes a real problem. When it's in the TOP, we can just split it up into a couple of fractions! :)

    • one year ago
  53. escolas
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    that makes sense...so you'd end up with the integral of tanx and the integral of secx? Well not exactly but pretty much, right?

    • one year ago
  54. zepdrix
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    Yes good, tan - sec i think.

    • one year ago
  55. escolas
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    But we're now in terms of theta rather than x, will it be hard to change back?

    • one year ago
  56. zepdrix
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    Which are actually... both... really terrible terms to integrate, you want to memorize both of these :) lol

    • one year ago
  57. zepdrix
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    Yes it will be a little bit tricky to change back, it will require us to setup a triangle and do some quick side work. It's not too bad though.

    • one year ago
  58. escolas
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    secx is secxtanx i think? but i can't remember tanx...

    • one year ago
  59. zepdrix
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    No, Integrating, secxtanx gives secx. Not the other way around :) We're going to get a couple of logs from both of these terms.

    • one year ago
  60. escolas
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    crap haha how do you integrate them then? are they just formulas?

    • one year ago
  61. escolas
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    Ah I found them in the back of my book

    • one year ago
  62. zepdrix
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    I don't really wanna go through the steps of integrating them right now XD It'll take too much time lol. So yes, for now let's just refer to the formulas.\[\large \int\limits \sec \theta d \theta \quad = \quad \ln|\sec \theta+\tan \theta|\]\[\large \int\limits \tan \theta d \theta \quad = \quad -\ln|\cos \theta|\]

    • one year ago
  63. zepdrix
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    Oh ok cool c:

    • one year ago
  64. zepdrix
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    the tangent can also be written as \[\large \ln|\sec \theta|\]By bringing the negative in as a power. Depending on which way you want to write it.

    • one year ago
  65. escolas
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    Are the absolute values gonna make things even more difficult?

    • one year ago
  66. zepdrix
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    No, I don't think we need to worry about that. We'll include the bars in our final answer but that's the jist of it. If we had a definite integral, and were plugging values in, we might want to be extra careful, but not in this case.

    • one year ago
  67. escolas
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    Ok cool, so now i solve for theta right?

    • one year ago
  68. escolas
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    \[\theta=\sin^{-1} \frac{ x+1 }{2 }\]

    • one year ago
  69. zepdrix
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    \[\int\limits \tan \theta - \sec \theta \; d \theta\] \[= \ln|\sec \theta|-\ln|\sec \theta+ \tan \theta|+C\] So this is what we came up with so far right?

    • one year ago
  70. escolas
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    Right! and does that make sense?

    • one year ago
  71. zepdrix
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    Let's not do that, if we had a lonesome \(\theta\) somewhere in our solution, then we would have to include the arcsine function. But since we have only trig functions in our solution, we'll do some triangle math.

    • one year ago
  72. escolas
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    Ah ok, I just figured we could get the angles theta was with triangles and then take the tan and sec of those! What do we need to do?

    • one year ago
  73. zepdrix
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    |dw:1359837750633:dw|\[x+1=2\sin \theta \qquad \rightarrow \qquad \sin \theta=\frac{x+1}{2} \qquad \rightarrow \qquad \sin \theta=\frac{opposite}{hypotenuse}\]

    • one year ago
  74. escolas
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    then use pythagorean to find that that side is well i'm too dumb to do that in my head actually. haha i have to write it down..

    • one year ago
  75. zepdrix
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    I think we end with something likeeeeee\[\large \sqrt{2^2-(x+1)^2} \qquad = \qquad \sqrt{4-(x+1)^2}\]

    • one year ago
  76. escolas
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    Yep just wrote it out! So then I find the secant and tangent of that triangle?

    • one year ago
  77. zepdrix
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    Notice how that value under the square root actually matches the denominator we started with? That will always happen.. If I'm remembering correctly. You should always see your initial substitution.. thing, in the triangle somewhere.

    • one year ago
  78. zepdrix
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    Yes good :)

    • one year ago
  79. escolas
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    wow thats interesting....and i can just leave the answer like that, right?

    • one year ago
  80. escolas
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    well the logs can simplify to one over the other actually right?

    • one year ago
  81. zepdrix
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    |dw:1359838191896:dw|

    • one year ago
  82. zepdrix
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    Hmm yah you could combine the logs i suppose, but not much simplification is needed after you've put it back in x.

    • one year ago
  83. zepdrix
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    Just don't forget that +C! lol

    • one year ago
  84. escolas
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    Yea it'll probably make it more messy if anything

    • one year ago
  85. escolas
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    ah thank you so much!

    • one year ago
  86. zepdrix
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    No prob \c:/ hopefully this all makes a bit more sense now!

    • one year ago
  87. escolas
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    it makes a ton more sense haha, sadly i never learned that so I was very confused when i saw those directions..

    • one year ago
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