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Romero

  • 3 years ago

Help with integration. Can someone show me how to do this.

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  1. Romero
    • 3 years ago
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    \[\int\limits_0^{2 pi} (\sqrt{1/(3 pi)}+e^{i x}\sqrt{1/(6 pi)} )^2 dx = 2/3 \]

  2. Romero
    • 3 years ago
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    The inside is squared.

  3. Romero
    • 3 years ago
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    The e^ix is really bugging me.

  4. UnkleRhaukus
    • 3 years ago
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    \[\int\limits_0^{2\pi}\sqrt{\frac{1}{3\pi}}+e^{ix}\sqrt{\frac{1}{{(6\pi)}^2}}\text d x=\frac 2 3\]

  5. UnkleRhaukus
    • 3 years ago
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    is that right?

  6. Romero
    • 3 years ago
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    no no The whole thing inside the integral is squared.

  7. Romero
    • 3 years ago
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    That's why I have a parentheses in the beginning it just doesn't show it that well when i used equations.

  8. Romero
    • 3 years ago
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    yes yes.

  9. UnkleRhaukus
    • 3 years ago
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    \[\int\limits_0^{2\pi}\left( {\sqrt{\frac{1}{3\pi}}+e^{ix}\sqrt{\frac{1}{6\pi}}}\right)^2\text d x=\frac 2 3\]

  10. Romero
    • 3 years ago
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    So do I just foil it out and find the integral?

  11. inkyvoyd
    • 3 years ago
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    turn e^(ix) into cos x+i sin x.

  12. inkyvoyd
    • 3 years ago
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    If it really bothers you that much. I would just treat i as a constnat. Replace i with g and just integrate it like a constant.

  13. inkyvoyd
    • 3 years ago
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    Then, replace the g's back with i's. Remember that i^2=-1

  14. Romero
    • 3 years ago
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    If I carry out the foil I know that I can split it into three different integral with the first one being integrating 1/3pi If I do that I can simply take out 1/3pi since it's a constant and I get (1/3pi)(2pi-0)=2/3 which is the answer but how does that other part cancel out?

  15. christbot
    • 3 years ago
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    Can you copy and paste your integral it into Wolfram Alpha? It will show you how to do it step-by-step

  16. Romero
    • 3 years ago
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    tried that

  17. inkyvoyd
    • 3 years ago
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    I would input that into mathematica, but unfortunately my laptop with its installation is being fixed atm

  18. christbot
    • 3 years ago
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    My calc 2 teacher avoided imaginary numbers. So I'm really curious now.

  19. christbot
    • 3 years ago
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    Mathematica is on my dead mac... Ugh

  20. Romero
    • 3 years ago
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    Oh I think I know what I did wrong. I was suppose to multiply the inside function with it's complex conjugate so it cancels out all imaginar values.

  21. christbot
    • 3 years ago
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    Anyone have Mac OS X up. Grapher could probably do it.

  22. christbot
    • 3 years ago
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    Change it to polar form maybe?

  23. Romero
    • 3 years ago
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    No the thing is that I was suppose to multiply imaginary function with -imaginary and that just cancels out all the imaginaries.

  24. UnkleRhaukus
    • 3 years ago
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    what do i enter into grapher? @Christbot

  25. christbot
    • 3 years ago
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    Wait, wait never about Grapher... Here http://en.wikipedia.org/wiki/Integration_using_Euler%27s_formula

  26. christbot
    • 3 years ago
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    You guys familiar with Euler's formula?

  27. Romero
    • 3 years ago
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    Yes but I don't need to use it at this point.

  28. christbot
    • 3 years ago
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    Have you tried foiling that giant square and breaking up the integral and placing the e^ix before the integral?

  29. Romero
    • 3 years ago
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    That was the original problem I just wrote it wrong.

  30. christbot
    • 3 years ago
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    Ouch!

  31. Romero
    • 3 years ago
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    I actually got 1 with the addition of some leftover integrals that should cancel out yet I don't really know why.

  32. Romero
    • 3 years ago
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    Ouch indeed I hate my life right now. lol

  33. Romero
    • 3 years ago
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    shouldn't post your phone number on the internet lol and I don't have an iphone. :(

  34. christbot
    • 3 years ago
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    I'm just facebooking myself the input... I'll work to copy and paste, lol

  35. christbot
    • 3 years ago
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    What the hell?! Wolfram won't integrate it?!

  36. christbot
    • 3 years ago
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    Can't we let 1 = trig identity? Wow...

  37. christbot
    • 3 years ago
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    bam I got it as an indefinite integral! Gotta wait for my slow phone, sorry!

  38. christbot
    • 3 years ago
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  39. Romero
    • 3 years ago
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    TY!!

  40. christbot
    • 3 years ago
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    Want the last bit or you got it?

  41. Romero
    • 3 years ago
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    Go it!

  42. christbot
    • 3 years ago
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    High five!

  43. Romero
    • 3 years ago
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    *high five* thanks again!! really helped me out!

  44. christbot
    • 3 years ago
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    No prob! I love a good problem.

  45. christbot
    • 3 years ago
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    Oh the web version works for free. Yeah, sometime you just gotta tinker with the input. I'm deleting the image links, but here is my mathy blog, FWIW http://mathstem.blogspot.com/ I'll delete my cell# too, lol!

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