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dfresenius Group Title

Trig substitutions with integrals. Really basic question.

  • one year ago
  • one year ago

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  1. dfresenius Group Title
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    \[\int\limits_{2\sqrt2}^{4}\frac{ 1 }{ (t^3)\sqrt(t^2-1)dt }\]

    • one year ago
  2. dfresenius Group Title
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    so, I know how to do the entire problem, but, when it comes to plugging in the cos(4) and cos(2root2) i get lost.

    • one year ago
  3. dfresenius Group Title
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    |dw:1360033290591:dw|

    • one year ago
  4. zepdrix Group Title
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    You plugged everything in correctly, I think maybe you just made a tiny boo boo at the end. See how you're squared secant that you're left with is in the `denominator`?

    • one year ago
  5. dfresenius Group Title
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    oh ya cos^2x

    • one year ago
  6. zepdrix Group Title
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    Oh the limits? yah let's not forget that the original limits were in terms of `t`. We can't plug those into our result. We have 2 options. Plug the limits into our original substitution that we made, to find new limits of integration. `Or` we can back-substitute and plug the t values in.

    • one year ago
  7. dfresenius Group Title
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    lets find t now, thats where Im lost

    • one year ago
  8. dfresenius Group Title
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    the other problem i had was real easy, root 2 and 2

    • one year ago
  9. dfresenius Group Title
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    now im not sure how to use these weird ones

    • one year ago
  10. zepdrix Group Title
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    \[\large t=\sec \theta\]The lower limit is \(2\sqrt2\). giving us,\[\large \frac{2}{\sqrt2}=\sec \theta\] Which is a special angle, maybe it just doesn't look like it :)\[\large \frac{2}{\sqrt2}=\frac{1}{\cos \theta} \theta \qquad \rightarrow \qquad \cos \theta = \frac{\sqrt2}{2}\]

    • one year ago
  11. dfresenius Group Title
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    |dw:1360034188203:dw|

    • one year ago
  12. dfresenius Group Title
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    ahhhh! algebra bites me in the retrice

    • one year ago
  13. zepdrix Group Title
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    Solution looks good so far :)

    • one year ago
  14. dfresenius Group Title
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    |dw:1360034363382:dw|

    • one year ago
  15. zepdrix Group Title
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    Umm if I'm remembering my angles correctly, I think that's at \(\dfrac{\pi}{4}\). Sine and Cosine both measure sqrt2/2 when theta is pi/4.

    • one year ago
  16. zepdrix Group Title
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    Doesn't sound familiar? :D

    • one year ago
  17. dfresenius Group Title
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    ya

    • one year ago
  18. dfresenius Group Title
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    hmm now im lost on the 4

    • one year ago
  19. zepdrix Group Title
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    \[\large 4=\sec \theta \qquad \rightarrow \qquad \cos \theta=1/4\]Hmmmmm..

    • one year ago
  20. zepdrix Group Title
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    \[\large \cos \theta=\left(\frac{1}{2}\right)^2\]Hmm no I don't think that's going to get us in the right direction...

    • one year ago
  21. zepdrix Group Title
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    \[\large \cos \theta=1/4 \qquad \rightarrow \qquad \theta=\arccos\left(\frac{1}{4}\right)\]I think we would have to leave it like this.

    • one year ago
  22. dfresenius Group Title
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    |dw:1360034799626:dw|

    • one year ago
  23. zepdrix Group Title
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    Hmm I'm not quite sure what you did there :o

    • one year ago
  24. dfresenius Group Title
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    not sure if that is legal, i multiplied both sides by sec^2/sec^2

    • one year ago
  25. zepdrix Group Title
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    But it looks like you canceled how a `secant` and a `secant squared` somehow :o

    • one year ago
  26. dfresenius Group Title
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    ya thats dumb my bad

    • one year ago
  27. dfresenius Group Title
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    |dw:1360035089559:dw|

    • one year ago
  28. zepdrix Group Title
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    no :c i dont think that works either. I'm pretty sure we have to leave it as the arccosine.

    • one year ago
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