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baldymcgee6

  • 2 years ago

Use an inverse trig substitution on the integral...

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  1. baldymcgee6
    • 2 years ago
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    |dw:1358909488344:dw|

  2. abb0t
    • 2 years ago
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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.

  3. baldymcgee6
    • 2 years ago
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    complete the square in the denominator... okkkk

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

  5. baldymcgee6
    • 2 years ago
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    |dw:1358910770499:dw|

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

  7. baldymcgee6
    • 2 years ago
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    i'm sorry, i forget what to do...

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

  9. abb0t
    • 2 years ago
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    then add it to C.

  10. baldymcgee6
    • 2 years ago
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    |dw:1358910964480:dw|

  11. abb0t
    • 2 years ago
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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.

  12. abb0t
    • 2 years ago
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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.

  13. baldymcgee6
    • 2 years ago
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    |dw:1358914184400:dw|@abb0t

  14. abb0t
    • 2 years ago
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    Now use u-substitution.

  15. baldymcgee6
    • 2 years ago
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    of what?

  16. baldymcgee6
    • 2 years ago
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    u - x-1 ?

  17. abb0t
    • 2 years ago
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    yeah

  18. baldymcgee6
    • 2 years ago
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    u = x-1 ***

  19. baldymcgee6
    • 2 years ago
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    i don't see where that takes us.

  20. baldymcgee6
    • 2 years ago
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    |dw:1358914405152:dw|

  21. baldymcgee6
    • 2 years ago
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    @zepdrix ?

  22. zepdrix
    • 2 years ago
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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.\[\large \int\limits\frac{dx}{\left[2(x-1)^2-3\right]^{3/2}}\]

  23. zepdrix
    • 2 years ago
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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.\[\large \int\limits\limits\frac{dx}{\left[2\left((x-1)^2-\dfrac{3}{2}\right)\right]^{3/2}}\]

  24. zepdrix
    • 2 years ago
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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 \(\dfrac{\sqrt3}{\sqrt2}\).

  25. zepdrix
    • 2 years ago
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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, \[\large \left[2\left(\color{salmon}{(x-1)^2}-\dfrac{3}{2}\right)\right]^{3/2} \qquad \rightarrow \qquad \left[2\left(\color{salmon}{\frac{3}{2}\sec^2\theta}-\frac{3}{2}\right)\right]^{3/2}\]

  26. zepdrix
    • 2 years ago
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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,\[\large \sqrt{27}\tan^3 \theta\]

  27. zepdrix
    • 2 years ago
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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.

  28. zepdrix
    • 2 years ago
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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,\[\large \frac{1}{\sqrt{27}}\cdot\frac{\sqrt3}{\sqrt2}\int\limits\limits\frac{\sec \theta \tan \theta \; d \theta}{\tan^3\theta}\]

  29. zepdrix
    • 2 years ago
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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.

  30. zepdrix
    • 2 years ago
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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.

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