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How to take the integral of this?

  1. Nov 28, 2012 #1
    1. The problem statement, all variables and given/known data

    How do you solve for the integral of 1/(16-20cos4x)?

    2. Relevant equations

    I know the answer but I don't know how to solve it by hand.

    3. The attempt at a solution

    I tried u-substitution and breaking it into two parts with 1/16 - 1/20cos4x but couldn't get anywhere.
     
  2. jcsd
  3. Nov 28, 2012 #2

    Mark44

    Staff: Mentor

    First, you seem to be thinking that a/(b + c) = a/b + a/c. Is 2/(1 + 1) the same as 2/1 + 2/1?

    Second, is what you show the actual problem? It looks to be very difficult to me.
     
  4. Nov 28, 2012 #3
    Oh, my mistake, I see what you mean. But I still can't seem to go anywhere with it.
    Yes, that's the problem exactly as it was stated.
     
  5. Nov 28, 2012 #4

    Mark44

    Staff: Mentor

    And there's a dx in there somewhere?

    Does your integral look like this?

    $$ \int \frac{dx}{16 - 20cos(4x)}$$
     
  6. Nov 28, 2012 #5
    Yup, that's what it looks like. Sorry about the dx omission.
     
  7. Nov 28, 2012 #6
    That seems like a really ugly integral...
     
  8. Nov 28, 2012 #7

    Dick

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    There is a general way to do integrals like this if you can't think of any special tricks that will make it easy. Start by the substitution u=4x, then look at http://en.wikipedia.org/wiki/Weierstrass_substitution
     
  9. Nov 28, 2012 #8

    Mark44

    Staff: Mentor

    I think that the direction to go is to break down the cos(4x) term using the double angle identity for cos(2A) and the basic trig identities. IOW, cos(4x) = cos2(2x) - sin2(2x) = 2cos2(2x) - 1

    Possibly you can get the denominator to look like the difference of squares or something to factor, that you can use partial fractions on.

    However, this might be a daunting task given the mistake you made in post #1.
     
  10. Nov 28, 2012 #9

    lurflurf

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    recall from elementary trigonometry that

    [tex]\frac{1}{16-20\cos(4x)}=\frac{1}{48} \left( \frac{6 \sec^2(2x)}{3tan(2x)-1} - \frac{6 \sec^2(2x)}{3tan(2x)+1} \right)[/tex]
     
    Last edited: Nov 28, 2012
  11. Nov 29, 2012 #10

    Ray Vickson

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    I'm glad you go nowhere this way, because it is WRONG: you cannot write 1/(a-b) as 1/a - 1/b. That violates all known rules of algebra, and you should never, never, never write it.

    RGV
     
  12. Nov 29, 2012 #11
    Is this
    Wolfram Alpha

    the answer that you are given? You can register and see a step-by-step solution.
     
  13. Nov 29, 2012 #12
    [tex]∫\frac{1 dx}{16-20\cos(4x) }[/tex], can be easily solved with substituted method :
    Let u = 16 – 20cos (4x), so we got:

    [tex]∫\frac{1}{u} dx[/tex]*

    By differentiate u we got:
    [tex]\frac{du}{dx} = 80 sin (4x)[/tex]

    [tex]dx = \frac{du}{80 sin (4x)}[/tex]

    use this dx for *
    [tex]∫\frac{du}{u80sin(4x)}=\frac{1}{80sin(4x)}∫\frac{du}{u}=\frac{lnu}{80sin(4x)}=\frac{ln(16-20cos(4x))}{80sin(4x)}[/tex]
     
    Last edited: Nov 29, 2012
  14. Nov 29, 2012 #13

    Curious3141

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    Firstly, we don't show complete solutions here.

    Secondly, your solution is wrong anyway. This step:

    ∫du/(u 80sin⁡(4x)) = 1/(80 sin⁡(4x)) ∫du/u

    is wrong because you cannot bring the variable expression involving sin 4x outside the integral like that. You have to express it in terms of u and then integrate.
     
  15. Nov 29, 2012 #14

    lurflurf

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    ^That is not right.
     
  16. Nov 29, 2012 #15
    [STRIKE]try to see the previous step, we substituted dx with du. therefor our integral not with dx anymore, instead with du. so we can bring any non-u variable expression outside.[/STRIKE]
    I am sorry, my bad.
     
    Last edited: Nov 29, 2012
  17. Nov 29, 2012 #16
    my bad, sorry
     
    Last edited: Nov 29, 2012
  18. Nov 29, 2012 #17

    lurflurf

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    as Curious3141 said

     
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