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Hard integral

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data
    How to integrate:
    ww682rqwias6riyy5m97.png


    2. Relevant equations



    3. The attempt at a solution
    I used formula: sin^2(t) = ( 1-cos^2(t))
    and now it's:
    9lkj89s3d5cesi48a3ax.png
    Then:
    u=cos(t)
    du=-sin(t)
    8y6dyp0r0o162zbcibba.png

    What to do next?
     
    Last edited: Sep 15, 2013
  2. jcsd
  3. Sep 15, 2013 #2

    arildno

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    We may now rewrite the integrand as:
    [tex]-6\sqrt{s^{2}+1}, s=\frac{\sqrt{5}u}{2}[/tex]
    Now, utilize the trigonometric identity:
    [tex]\tan^{2}(y)+1=\frac{1}{\cos^{2}(y)}[/tex]
    in a creative way.
     
  4. Sep 15, 2013 #3

    CAF123

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    Where did the 3 come from?
    Try using the identity 1 + tan2x = sec2x after some initial manipulation of the integrand.
     
  5. Oct 17, 2013 #4
    Actually the task sounds like this:
    v4jlxgd41iy2lui365kt.png

    Maybe I have to use polar coordinates? Any sugestions? Before I tried with diferent way, but I guess that integration without polar coordinates is too hard.
     
  6. Oct 17, 2013 #5

    arildno

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    L is a segment of an ellipse. You might go polar, but you can eliminate t as well, and express y as a function of x(Hint: sin^+cos^2=1)
     
  7. Oct 17, 2013 #6
    rz4k8k32eudm9k2wr91e.png
    Seems to me that variant when we use y=y(x) is more complicated than variant with polar coordinates.

    Not clear, how did you get there: [tex]s=\frac{\sqrt{5}u}{2}[/tex]
    We didn't learn about triple substitution, but I want to understand, how to get the final result. Can you explain some how?
     
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