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escolas

  • one year ago

By completing the square and using a trig substitution, evaluate

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

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

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

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

  5. zepdrix
    • one year ago
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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.\[\large -(\color{orangered}{x^2+2x+1}-1-3) \qquad = \qquad -(\color{orangered}{(x+1)^2}-4)\]

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

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

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

  9. escolas
    • one year ago
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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

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

  11. zepdrix
    • one year ago
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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:

  12. escolas
    • one year ago
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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?

  13. zepdrix
    • one year ago
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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.

  14. zepdrix
    • one year ago
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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.

  15. escolas
    • one year ago
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    Yep, perfect!

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

  17. zepdrix
    • one year ago
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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.\[\large a^2-x^2 \qquad = \qquad a^2-(a \sin \theta)^2 \qquad = \qquad a^2-a^2\sin^2 \theta\]\[\large a^2(1-\sin^2\theta) \qquad = \qquad a^2(\color{royalblue}{\cos^2\theta})\]

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

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

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

  21. zepdrix
    • one year ago
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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 \(\large x=a\sin\theta\).

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

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

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

  25. zepdrix
    • one year ago
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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.

  26. zepdrix
    • one year ago
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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.

  27. escolas
    • one year ago
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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?

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

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

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

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

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

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

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

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

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

  37. zepdrix
    • one year ago
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    true story :O

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

  39. escolas
    • one year ago
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    no 2sinx

  40. zepdrix
    • one year ago
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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.

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

  42. zepdrix
    • one year ago
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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.

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

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

  45. zepdrix
    • one year ago
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    XD

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

  47. zepdrix
    • one year ago
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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.\[\large \int\limits\limits\limits\frac{ \cancel2\sin \theta-1}{ \cancel4\cos^{\cancel2\;1} \theta }(\cancel2\cancel{\cos \theta} \;d \theta)\]

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

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

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

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

  52. zepdrix
    • one year ago
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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! :)

  53. escolas
    • one year ago
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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?

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

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

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

  57. zepdrix
    • one year ago
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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.

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

  59. zepdrix
    • one year ago
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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.

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

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

  62. zepdrix
    • one year ago
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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.\[\large \int\limits \sec \theta d \theta \quad = \quad \ln|\sec \theta+\tan \theta|\]\[\large \int\limits \tan \theta d \theta \quad = \quad -\ln|\cos \theta|\]

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

  64. zepdrix
    • one year ago
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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.

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

  66. zepdrix
    • one year ago
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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.

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

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

  69. zepdrix
    • one year ago
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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?

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

  71. zepdrix
    • one year ago
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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.

  72. escolas
    • one year ago
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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?

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

  74. escolas
    • one year ago
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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..

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

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

  77. zepdrix
    • one year ago
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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.

  78. zepdrix
    • one year ago
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    Yes good :)

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

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

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

  82. zepdrix
    • one year ago
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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.

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

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

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

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

  87. escolas
    • one year ago
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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..

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