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Math Help - Integration problem

  1. #1
    Newbie matheater's Avatar
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    Integration problem

    hi all
    I am a new member here.Now I want to go to my problem directly,the problem is about an integration,i didn't understand it clearly.Here i have attached the document.It is done thereAttachment 18364 directly.Can anyone help me by showing the procedure step by step?
    thanks
    Attached Thumbnails Attached Thumbnails Integration problem-math.jpg  
    Last edited by matheater; July 28th 2010 at 01:19 PM.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Use substitution z=p*tan(t).
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  3. #3
    Newbie matheater's Avatar
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    thanks a lot, can u help more elaborately? I am still unable to find how they got H=I/(2pi.roh)

    if we use tan(t) then the limit will be changed to -90 to +90 degrees isn't it? then how can we proceed?
    Last edited by mr fantastic; July 28th 2010 at 08:52 PM.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Use substitution z=p*tan(t).

    Step by step solution:

    First of all I will change the your variables(for my own comfort)

    So, z'=x and \ro = a

    Now, we use following substitution x=atan(t), we "put" it in:

    \frac{1}{{(a^2+x^2)}^{\frac{3}{2}}}

    so we get:


    \frac{1}{{{(a^2+a^2tan^2t)}}^{\frac{3}{2}}}=\frac{  1}{{{(a^2(1+tan^2(t))}}^{\frac{3}{2}}}=\frac{1}{{{  (a^2(sec^2(t))}}^{\frac{3}{2}}}=\frac{1}{a^3sec^3(  t)}


    and
    dx=asec^2(t)dt

    so, \int\frac{1}{{{(a^2+x^2)}}^{\frac{3}{2}}}dx=\frac{  1}{a}\int cos(t)dt=\frac{sin(t)}{a}+C

    Now, to return back to x, substitute t=arctan(\frac{x}{a})

    I will leave the last part for you, only trigonometric calculation.
    Last edited by Also sprach Zarathustra; July 28th 2010 at 01:32 PM.
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