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baldymcgee6

Use an inverse trig substitution on the integral...

  • one year ago
  • one year ago

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  1. baldymcgee6
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    |dw:1358909488344:dw|

    • one year ago
  2. abb0t
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    For inverse trig function, you want to get it into a form that looks like arctan, arcsin, arcos, etc., The first step is to complete the square.

    • one year ago
  3. baldymcgee6
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    complete the square in the denominator... okkkk

    • one year ago
  4. baldymcgee6
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    you're going to have to refresh my memory on how to do that :/

    • one year ago
  5. baldymcgee6
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    |dw:1358910770499:dw|

    • one year ago
  6. abb0t
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    \[ax^2+bx+c \] \[(\frac{ b }{ 2 })^2 \]

    • one year ago
  7. baldymcgee6
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    i'm sorry, i forget what to do...

    • one year ago
  8. abb0t
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    divide by two and square the answer of the 2nd term (b).

    • one year ago
  9. abb0t
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    then add it to C.

    • one year ago
  10. baldymcgee6
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    |dw:1358910964480:dw|

    • one year ago
  11. abb0t
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    Yeah. Now, you can factor it to get something that looks like an inverse trig, so that you get: \[(x \pm a)^2 + 1\] where you will tell what inverse trig function you have. Try it.

    • one year ago
  12. abb0t
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    Essentially, you might need to do more work, like factor out a number or two. But remember that you have it in a square root! so I think it might be arcos? Just a guess. It wont be arctan since arctan does;t have a square root.

    • one year ago
  13. baldymcgee6
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    |dw:1358914184400:dw|@abb0t

    • one year ago
  14. abb0t
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    Now use u-substitution.

    • one year ago
  15. baldymcgee6
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    of what?

    • one year ago
  16. baldymcgee6
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    u - x-1 ?

    • one year ago
  17. abb0t
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    yeah

    • one year ago
  18. baldymcgee6
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    u = x-1 ***

    • one year ago
  19. baldymcgee6
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    i don't see where that takes us.

    • one year ago
  20. baldymcgee6
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    |dw:1358914405152:dw|

    • one year ago
  21. baldymcgee6
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    @zepdrix ?

    • one year ago
  22. zepdrix
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    Let's not do a U-sub, that seems rather unnecessary. We'll already be doing a trig sub, so it's better if we can limit the number of substitutions that we apply. Your factoring looks good. So we're at this point.\[\large \int\limits\frac{dx}{\left[2(x-1)^2-3\right]^{3/2}}\]

    • one year ago
  23. zepdrix
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    Ok this next part is going to be a little tricky to understand, let me see if I can put it into the right words. There are 3 primary forms that we look for when making a trig sub. We'll look at the one that specifically applies to this problem. Ignore the 2 coefficient on X in the bottom for a moment. Our intention is to get rid of the addition or subtraction in the bottom. If we can do that, simplify it down to ONE term, we can easily apply the exponent on the outside to it. If we see:\[\large x^2-a^2\]We want to let \(\large x=a\sec \theta\) That will do this for us,\[\large x^2-a^2 \qquad \rightarrow \qquad a^2\sec^2 \theta-a^2 \qquad \rightarrow \qquad a^2(\sec^2\theta-1)\] From there, we can recall that \(\large \sec^2\theta-1=tan^2\theta\) and make the switch. \[\large a^2(\tan^2\theta)\] Now you might be saying, "But zep! We don't have an a^2! We just have a 3!" Well baldy this is where it gets a little tricky. We need to get that ugly 2 off of our x term.\[\large \int\limits\limits\frac{dx}{\left[2\left((x-1)^2-\dfrac{3}{2}\right)\right]^{3/2}}\]

    • one year ago
  24. zepdrix
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    "But zep! We still don't have a nice \(a\) value!" Well that's just the way it works out sometimes. We DO have an \(a\), it's just a little ugly :D So in our case, our \(a\) value is going to be \(\dfrac{\sqrt3}{\sqrt2}\).

    • one year ago
  25. zepdrix
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    We'll make the following substitution, \[\large (x-1)=\frac{\sqrt3}{\sqrt2}\sec \theta\] So our denominator will do something like this, \[\large \left[2\left(\color{salmon}{(x-1)^2}-\dfrac{3}{2}\right)\right]^{3/2} \qquad \rightarrow \qquad \left[2\left(\color{salmon}{\frac{3}{2}\sec^2\theta}-\frac{3}{2}\right)\right]^{3/2}\]

    • one year ago
  26. zepdrix
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    \[\large \left[3\left(\sec^2\theta-1\right)\right]^{3/2} \qquad \rightarrow \qquad \left[3\left(\tan^2\theta\right)\right]^{3/2}\]Which simplifies to,\[\large \sqrt{27}\tan^3 \theta\]

    • one year ago
  27. zepdrix
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    Giving us an integral of,\[\large \frac{1}{\sqrt{27}}\int\limits\frac{dx}{\tan^3\theta}\]From here, we'll still need to deal with the dx.

    • one year ago
  28. zepdrix
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    \[\large (x-1)=\frac{\sqrt3}{\sqrt2}\sec \theta \qquad \rightarrow \qquad dx=\frac{\sqrt3}{\sqrt2}\sec\theta\tan\theta \; d \theta\] So our integral becomes,\[\large \frac{1}{\sqrt{27}}\cdot\frac{\sqrt3}{\sqrt2}\int\limits\limits\frac{\sec \theta \tan \theta \; d \theta}{\tan^3\theta}\]

    • one year ago
  29. zepdrix
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    So from here, cancel out a couple of tangents. Then convert everything to sines and cosines. At that point you'll have a couple options. You can do a nice easy U-sub, Or You can write your terms as \(\cot\theta \csc\theta\) which is a fairly easy integral.

    • one year ago
  30. zepdrix
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    There a bit more to be done to get through this. But I'm tired, and you're offline anyway XD lol.

    • one year ago
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