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dfresenius

  • 3 years ago

Trig substitutions with integrals. Really basic question.

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

  2. dfresenius
    • 3 years ago
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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.

  3. dfresenius
    • 3 years ago
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    |dw:1360033290591:dw|

  4. zepdrix
    • 3 years ago
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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`?

  5. dfresenius
    • 3 years ago
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    oh ya cos^2x

  6. zepdrix
    • 3 years ago
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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.

  7. dfresenius
    • 3 years ago
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    lets find t now, thats where Im lost

  8. dfresenius
    • 3 years ago
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    the other problem i had was real easy, root 2 and 2

  9. dfresenius
    • 3 years ago
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    now im not sure how to use these weird ones

  10. zepdrix
    • 3 years ago
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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 :)\[\large \frac{2}{\sqrt2}=\frac{1}{\cos \theta} \theta \qquad \rightarrow \qquad \cos \theta = \frac{\sqrt2}{2}\]

  11. dfresenius
    • 3 years ago
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    |dw:1360034188203:dw|

  12. dfresenius
    • 3 years ago
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    ahhhh! algebra bites me in the retrice

  13. zepdrix
    • 3 years ago
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    Solution looks good so far :)

  14. dfresenius
    • 3 years ago
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    |dw:1360034363382:dw|

  15. zepdrix
    • 3 years ago
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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.

  16. zepdrix
    • 3 years ago
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    Doesn't sound familiar? :D

  17. dfresenius
    • 3 years ago
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    ya

  18. dfresenius
    • 3 years ago
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    hmm now im lost on the 4

  19. zepdrix
    • 3 years ago
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    \[\large 4=\sec \theta \qquad \rightarrow \qquad \cos \theta=1/4\]Hmmmmm..

  20. zepdrix
    • 3 years ago
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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...

  21. zepdrix
    • 3 years ago
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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.

  22. dfresenius
    • 3 years ago
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    |dw:1360034799626:dw|

  23. zepdrix
    • 3 years ago
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    Hmm I'm not quite sure what you did there :o

  24. dfresenius
    • 3 years ago
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    not sure if that is legal, i multiplied both sides by sec^2/sec^2

  25. zepdrix
    • 3 years ago
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    But it looks like you canceled how a `secant` and a `secant squared` somehow :o

  26. dfresenius
    • 3 years ago
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    ya thats dumb my bad

  27. dfresenius
    • 3 years ago
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    |dw:1360035089559:dw|

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

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