Results 1 to 7 of 7

Math Help - Difficult integral.

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    6

    Difficult integral.

    Can any one suggest a solution or method for solving this integration with respect to x

    <br />
\int\frac{1}{\left[a-b\cos(k.x)\right]^{4}}dx<br />

    Regards
    James
    Last edited by mr fantastic; June 2nd 2010 at 04:36 AM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by James4321 View Post
    Can any one suggest a solution or method for solving this integration with respect to x

    <br />
\int\frac{1}{\left[a-b\cos(k.x)\right]^{4}}dx<br />

    Regards
    James
    Not elegant, but WolframAlpha gives:

    integrate 1&#x2f;&#x28;a-b cos&#x28;k x&#x29;&#x29;&#x5e;4 - Wolfram|Alpha

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2010
    Posts
    6
    Quote Originally Posted by CaptainBlack View Post
    Thank you Giordano for your prompt reply. I tried this already. Unfortunately, WolframAlpha result is wrong.

    James
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by James4321 View Post
    Can any one suggest a solution or method for solving this integration with respect to x

    <br />
\int\frac{1}{\left[a-b\cos(k.x)\right]^{4}}dx<br />

    Regards
    James
    I'd like to make this more general , let the power to be  n

    To handle such 'troublesome' integral , first thing to do is to reduce it to a simplier form :


     \frac{1}{(-b)^n} \int \frac{dx}{ [ (-a/b) + \cos(kx) ]^n }

    Let  kx = t the integral becomes

     \frac{1}{(-b)^n k }\int \frac{dt}{ [A+ \cos(t) ]^n } ~~~ A = -a/b

    it looks better now and we wish to obtain a reduction formula .

    Let

     I_n = \int \frac{dt}{ [A+ \cos(t) ]^n }

     A I_n = \int \frac{(A + \cos(t))-\cos(t)}{ [A+ \cos(t) ]^n } ~dt

     = I_{n-1} - \int \frac{\cos(t)}{ [A+ \cos(t) ]^n }~dt

    Use integration by parts ,

     = I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n \int \frac{\sin^2(t)}{ [A+ \cos(t) ]^{n+1} } ~dt

     = I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n \int \frac{1-\cos^2(t)}{ [A+ \cos(t) ]^{n+1} } ~dt

     = I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n I_{n+1} - n\int \frac{\cos^2(t)}{ [A+ \cos(t) ]^{n+1} } ~dt

     = I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n I_{n+1} - n\int \frac{ [(A+\cos(t)) -A ]^2}{[A+ \cos(t) ]^{n+1} } ~dt

     =I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n I_{n+1} - n\int \frac{ (A+\cos(t))^2 -2A(A+\cos(t)) + A^2}{[A+ \cos(t) ]^{n+1} } ~dt


     = I_{n-1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n } + n I_{n+1} - nI_{n-1} +2An I_n - A^2 n I_{n+1}

    Thus

     AI_n = (1 - n) I_{n-1} + 2AnI_n + n(1-A^2) I_{n+1} - \frac{\sin(t)}{ [A+ \cos(t) ]^n }

     n(1-A^2) I_{n+1} = A(1-2n)I_n + (n-1)I_{n-1} + \frac{\sin(t)}{ [A+ \cos(t) ]^n }

    If  A^2 = 1 , the reduction only consists of two variables but in general , it consists of three variables :  I_{n-1} , I_n , I_{n+1} , that means we have to find  I_1 , I_2 . By substituting  n=1 we obtain :


     (1-A^2)I_2 = -A I_1 + 0 + \frac{\sin(t)}{ A+ \cos(t)}

    I_2 = \frac{A}{A^2-1} I_1 - \frac{\sin(t)}{ ( A^2 -1) [ A + \cos(t) ] }

    I give you the case for  A^2 = 1 , in the above expression , we asumme  A^2 \neq 1 .

     I_1 = \int \frac{dt}{ A + \cos(t) }


     = \int \frac{dt}{ A( \cos^2(t/2) + \sin^2(t/2) ) + \cos^2(t/2) - \sin^2(t/2) }

     = \int \frac{dt}{ (A+1) \cos^2(t/2) +(A-1) \sin^2(t/2) }

     = \frac{1}{A-1} \int \frac{ \sec^2(t/2) ~dt}{ \tan^2(t/2) + \frac{A+1}{A-1} }

    Sub.  \tan(t/2) = u

     \sec^2(t/2)~dt = 2du

     I_1 = \frac{2}{A-1} \int \frac{du}{ u^2 +\frac{A+1}{A-1} }

    If you know the sign of  A^2 - 1 ~~ ( a^2 -b^2 ) , you should be able to continue ...
    Last edited by simplependulum; June 2nd 2010 at 09:31 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2010
    Posts
    6
    Thank you very much for this brilliant suggestion!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    Quote Originally Posted by James4321 View Post
    Thank you Giordano for your prompt reply. I tried this already. Unfortunately, WolframAlpha result is wrong.

    James
    "Giordano" is not Captain Black's name. Giordano Bruno is the author of the quote he gives in his signature.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2010
    Posts
    6
    Sorry about this! I would also like to thank Simplependulum for his elegant and complete solution.

    James4321
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. difficult integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 2nd 2010, 02:05 AM
  2. Difficult Integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 14th 2010, 12:53 AM
  3. a difficult integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 25th 2009, 06:16 AM
  4. Difficult Integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 8th 2009, 07:03 AM
  5. difficult integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 3rd 2007, 02:59 PM

Search Tags


/mathhelpforum @mathhelpforum