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Math Help - difficult integral

  1. #1
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    difficult integral

    hey this integral is really bothering me, the main bit being the k in the middle of it. thanks to anyone who can help


    0∫π 1/((k + cosx)^2) dx where k>1

    cheers!
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by miranda View Post
    hey this integral is really bothering me, the main bit being the k in the middle of it. thanks to anyone who can help


    0∫π 1/((k + cosx)^2) dx where k>1

    cheers!
    \displaystyle{\int\limits_0^\pi\dfrac{dx}{(k+\cos{  x})^2}.}

    \dfrac{1}{(k+\cos{x})^2}=\dfrac{1}{\left(k\sin^2\d  frac{x}{2}+k\cos^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}\right)^2}=

    =\dfrac{1}{\left((k-1)\sin^2\dfrac{x}{2}+(k+1)\cos^2\dfrac{x}{2}\right  )^2}=\dfrac{1}{(k+1)^2}\dfrac{1}{\cos^4\dfrac{x}{2  }\left(\dfrac{k-1}{k+1}\tan^2\dfrac{x}{2}+1\right)^2}=

    =\dfrac{1}{(k+1)^2}\dfrac{1+\tan^2\dfrac{x}{2}}{\l  eft(a\tan^2\dfrac{x}{2}+1\right)^2}\dfrac{1}{\cos^  2\dfrac{x}{2}},~a=\dfrac{k-1}{k+1}.

    Substitution

    \tan\dfrac{x}{2}=u~\Rightarrow~\dfrac{dx}{\cos^2\d  frac{x}{2}}=2\,du
    Last edited by DeMath; August 26th 2010 at 02:12 PM. Reason: typo
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  3. #3
    Senior Member DeMath's Avatar
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    So we have

    \displaystyle{\int\limits_0^\pi\dfrac{dx}{(k+\cos{  x})^2}=\dfrac{2}{(k+1)^2}\mathop{\lim}\limits_{b\t  o\infty}\int\limits_0^b\dfrac{1+u^2}{(au^2+1)^2}\,  du=\ldots=}

    =\left.{\dfrac{2}{(k+1)^2}\mathop{\lim}\limits_{b\  to\infty}\left(\dfrac{a-1}{2a}\dfrac{u}{au^2+1}+\dfrac{a+1}{2a^{3/2}}\arctan\!\left(\sqrt{a}\,u\right)\right)}\right  |_0^b=

    =\left.{\dfrac{2}{(k+1)^2}\mathop{\lim}\limits_{b\  to\infty}\left(\dfrac{1}{1-k}\dfrac{u}{au^2+1}+\dfrac{k(k+1)^{1/2}}{(k-1)^{3/2}}\arctan\!\left(\sqrt{a}\,u\right)\right)}\right  |_0^b=

    =\dfrac{k\,\pi}{(k^2-1)^{3/2}},~k>0.
    Last edited by DeMath; August 26th 2010 at 01:12 PM.
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