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P(cos(x)^3)/(x^6+1) very hard

  1. Feb 1, 2012 #1
    Hi guys

    I'm wondering how shall I solve this primitive/integral

    [tex]\int\frac{\cos^3 x}{x^6+1}dx[/tex]

    I really don't know how to start... which substitution shall I take or if I shall solve it by parts...

    Any hints or tips will be very appreciated

    Thanks in advance
  2. jcsd
  3. Feb 1, 2012 #2
  4. Feb 1, 2012 #3
    I don't know if I need an explicit expression, I just need to find how shall I start...

    I've tried all trigonometric substitutions and nothing :(

    I'm lost
  5. Feb 1, 2012 #4


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    Hey joao_pimentel.

    Have you looked at different tables of integrals? Chances are you will find a table out there somewhere that gives a form that is exact or 'close' to the form you are working with.

    One strategy I have for you is to reduce the trig terms to something that is at least linear. You might have to do a few integrations by parts by differentiating the trig terms and then using trig conversions to put them in the appropriate form.

    The denominator of the term looks like a (1 + x^2) which is used in many tables, although if its in combination with a trig term then that could still make it complicated.
  6. Feb 1, 2012 #5


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    Also having looked at the solution, maybe another suggestion is to consider a transformation to an expontential form (in the form of using expressions of e^(whatever)) so for example cos(x) = 1/2(e^(ix) - e^(ix)).
  7. Feb 1, 2012 #6
    Split the integral in two, in writting cos(x)^3 in terms of cos(x) and cos(3x)
    Split both integrals in six, in writting 1/((x^6)+1) in terms of a sum of six terms c/(x-r) where each r is on the form r=(-1)^(1/6) , i.e. the six complex roots of unit.
    Then, the antiderivatives of each cos(x)/(x-r) and cos(3x)/(x-r) can be formally expressed in terms of the special functions Ci(z) and Si(z) where z is a complex linear function of x.
  8. Feb 1, 2012 #7

    D H

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    That is the factorization of x6-1, not x6+1. The latter factorization is a bit uglier.

    First factor [itex]x^6+1[/itex] as [itex](x^2+1)(x^4-x^2+1)[/itex]. The latter term in turn has the factorization [itex](x^2-1+\sqrt[3]{-1})(x^2-\sqrt[3]{-1})[/itex], so

    [tex]x^6+1 = (x^2+1)(x^2-1+\sqrt[3]{-1})(x^2-\sqrt[3]{-1})[/tex]

    I'll leave the last ugly bit as an exercise for the OP.
  9. Feb 1, 2012 #8
    Thanks a lot guys... it was a big help...

    I shall write cos(x)^3 in terms of a sum of cos(x) and cos(3x) as JJacquelin said and then fatorize x^6+1 as D H said... Then I shall use Ci(z) e Si(z) functions for each term..

    I'll try it :)

    Thanks a lot again

  10. Feb 2, 2012 #9

    I know that [tex]cos^3(x)=\frac{1}{4}\left(cos(3x)+3cos(x)\right)[/tex]


    \int &\frac{\cos^3 x} {(x^2+1) (x^2-1+\sqrt[3]{-1}) (x^2-\sqrt[3]{-1})} dx \\
    & =\frac{1}{4}
    \int \frac{\cos{3x}} {(x^2+1) (x^2-1+\sqrt[3]{-1}) (x^2-\sqrt[3]{-1})} dx +
    \int \frac {\cos{x}} {(x^2+1)(x^2-1+\sqrt[3]{-1} )(x^2-\sqrt[3]{-1})} dx

    Then I need to factorize [tex]\frac{1}{(x^2+1)(x^2-1+\sqrt[3]{-1})(x^2-\sqrt[3]{-1})}[/tex] in partial fractions

    \frac{1} {(x^2+1) (x^2-1+\sqrt[3]{-1})(x^2-\sqrt[3]{-1})} =
    \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2-1+\sqrt[3]{-1})}

    Then I get each fraction [tex]\cos(a x)\frac{\alpha x + \beta}{x^2+\lambda}[/tex]

    How do I use such Ci(z) functions... Actually I confess I'm not aware of such Ci(z) and Si(z) functions... would it be Hyperbolic function???

    Thank you a lot
    Last edited by a moderator: Feb 2, 2012
  11. Feb 2, 2012 #10


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  12. Feb 2, 2012 #11
    I think that it not a good idea to process with partial fractons which denominators are second degree polynomials. This will lead to complicated integrals, not easy to expess in terms of Ci and Si functions.
    I maintain what I already said in my previous post [ Y, 01:19 PM] :
    1/((x^6)+1) is much easier to write as the sum of six terms on the form c/(x-r) where each c is a different complex number easy to find and where the six different r are the roots of (r^6)+1=0, i.e. six complex number easy to find.
  13. Feb 2, 2012 #12

    D H

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    Here's what you said yesterday:
    That's incorrect. Using those (x-r) as the factors, where each r is one of the six sixth roots of unit, is the factorization of x^6-1=0, not x^6+1=0.
  14. Feb 3, 2012 #13
    Obviously :
    r = exp(i(2n+1)pi/6)
    I think that it isn't the key point of the question.
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