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I need help!

  1. Sep 19, 2008 #1
    I need help for this integral


    cosx cosx[tex]^{2}[/tex]
     
  2. jcsd
  3. Sep 19, 2008 #2

    gabbagabbahey

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    Do you mean
    [tex]\int cos(x) cos^2(x) dx[/tex]
    or
    [tex] \int cos(x) cos(x^2) dx[/tex]
    ?
     
  4. Sep 19, 2008 #3

    HallsofIvy

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    Whatever you mean, what have you tried?
     
  5. Sep 21, 2008 #4
    cos(x)cos(x^2)
     
  6. Sep 21, 2008 #5

    Defennder

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    I don't think the anti-derivative can be expressed in elementary functions.
     
  7. Sep 21, 2008 #6

    gabbagabbahey

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  8. Sep 21, 2008 #7
    thanks... sorry for my english...I hope that can understand me...

    This integral [tex]
    \int_{1}[/tex] [tex]^{\infty}[/tex] cos(x)cos(x2) dx

    I have to say that is convergent or divergent. The proffessor gave us a hint (Fresnel Integral)....but I don't know how to use it, He don't discuss it in class :grumpy:. I understand everything but the cos(x2) jeje I don't know how to integrate.:yuck:

    Thanks gabbagabbahey for the link...
     
  9. Sep 21, 2008 #8

    gabbagabbahey

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    Okay, this is going to be ugly, but here it goes:

    [tex]cos(x)cos(x^2)=\frac{cos(x^2-x)+cos(x^2+x)}{2}= \frac{cos \left( x^2-x+\frac{1}{4} - \frac{1}{4} \right)+cos \left( x^2+x+\frac{1}{4} - \frac{1}{4} \right) }{2} = \frac{cos \left( \left( x-\frac{1}{2} \right) ^2 - \frac{1}{4} \right)+cos \left( \left( x+\frac{1}{2} \right) ^2 - \frac{1}{4} \right) }{2} [/tex]


    [tex] = \frac{ cos \left( \left( x - \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x - \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) +cos \left( \left( x + \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x + \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) }{2} [/tex]


    [tex] \Rightarrow \int_1^{\infty} cos(x)cos(x^2)dx [/tex]

    [tex] = \frac{1}{2} \int_1^{\infty} cos \left( \left( x - \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x - \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) +cos \left( \left( x + \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x + \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) dx [/tex]

    [tex] = \frac{1}{2} \left[ cos \left( \frac{1}{4} \right) \int_1^{\infty} cos \left( \left( x - \frac{1}{2} \right) ^2 \right)dx + sin \left( \frac{1}{4} \right) \int_1^{\infty} sin \left( \left( x - \frac{1}{2} \right) ^2 \right)dx + cos \left( \frac{1}{4} \right) \int_1^{\infty} cos \left( \left( x + \frac{1}{2} \right) ^2 \right) dx [/tex]

    [tex] \left{+ sin \left( \frac{1}{4} \right) \int_1^{\infty} sin \left( \left( x + \frac{1}{2} \right) ^2 \right) dx \right] [/tex]


    ...
     
    Last edited: Sep 21, 2008
  10. Sep 21, 2008 #9

    gabbagabbahey

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    ...

    Make the following substitutions:

    [tex]u \equiv x-\frac{1}{2} , \quad v \equiv x+ \frac{1}{2} \quad \Rightarrow du=dv=dx , \quad u: \frac{3}{2} \rightarrow \infty , \quad v: \frac{1}{2} \rightarrow \infty [/tex]

    [tex] \Rightarrow \int_1^{\infty} cos(x)cos(x^2)dx [/tex]
    [tex]= \frac{1}{2} cos \left( \frac{1}{4} \right) \int_{\frac{3}{2}}^{\infty} cos(u^2)du + \frac{1}{2} sin \left( \frac{1}{4} \right) \int_{\frac{3}{2}}^{\infty} sin(u^2)du + \frac{1}{2} cos \left( \frac{1}{4} \right) \int_{\frac{1}{2}}^{\infty} cos(v^2)dv + \frac{1}{2} sin \left( \frac{1}{4} \right) \int_{\frac{1}{2}}^{\infty} sin(v^2)dv [/tex]

    ...
     
    Last edited: Sep 21, 2008
  11. Sep 21, 2008 #10

    gabbagabbahey

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    ...
    Now using the definition of the Fresnel sine and cosine integrals, [tex] S(t),\quad C(t)[/tex] :

    [tex] S(t) \equiv \int_0^t sin(q^2)dq , \quad C(t) \equiv \int_0^t cos(q^2)dq [/tex]

    we get:

    [tex]\int_1^{\infty} cos(x)cos(x^2)dx = \frac{1}{2} cos \left( \frac{1}{4} \right) \left( C(\infty)-C \left( \frac{3}{2} \right) \right) + \frac{1}{2} sin \left( \frac{1}{4} \right) \left( S(\infty)-S \left( \frac{3}{2} \right) \right) + \frac{1}{2} cos \left( \frac{1}{4} \right) \left( C(\infty)-C \left( \frac{1}{2} \right) \right)[/tex]
    [tex] + \frac{1}{2} sin \left( \frac{1}{4} \right) \left( S(\infty)-S \left( \frac{1}{2} \right) \right) [/tex]
    ....
     
    Last edited: Sep 21, 2008
  12. Sep 21, 2008 #11

    gabbagabbahey

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    ...And using the fact that
    [tex] S(\infty) = C(\infty) =\frac{\sqrt{\pi}}{8} [/tex]
    this becomes:

    [tex] \int_1^{\infty} cos(x)cos(x^2)dx = -\frac{1}{2} \left[ cos \left( \frac{1}{4} \right) \left( C \left( \frac{3}{2} \right) + C \left( \frac{1}{2} \right) -\frac{\sqrt{\pi}}{4} \right) + sin \left( \frac{1}{4} \right) \left( S \left( \frac{3}{2} \right) +S \left( \frac{1}{2} \right) -\frac{\sqrt{\pi}}{4} \right) \right]
    [/tex]
     
    Last edited: Sep 21, 2008
  13. Sep 21, 2008 #12
    iah....very ugly...wow!

    thanks thanks... jeje now I will try to understand it!!!!!!!!
     
  14. Sep 21, 2008 #13
    im nanies roommate and we take the class together and the problem is that i dont understand what u did in the first part of this problem and dont know where the cosX^2 - X come from. also i dont understand why u divided everything by 2...please help me
     
  15. Sep 21, 2008 #14

    gabbagabbahey

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    I used the following Trig Identity:

    [tex]cos(A)cos(B)=\frac{cos(A-B)+cos(A+B)}{2}[/tex]
     
  16. Sep 21, 2008 #15
    ok thank you soo much! the thing is that i have being trying it since tuesday! but thanks
     
  17. Sep 21, 2008 #16
    THANK YOU!!!!!!!!!!o:):cool:
     
  18. Sep 22, 2008 #17
    a doubt come into my mind when i was doing the calculus exercise. .....i need to know if in the fresnel integral i can used any limit of integration or only from 0 to infinite....:confused:
     
  19. Sep 22, 2008 #18

    gabbagabbahey

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    If the limits are zero to infinty, then you have S(infinty), but if the limits are 0 to a, you have S(a) as long is a is positive, the fresnel integrals converge.
     
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