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Math Help - Integral of 1/cot(x) + 2

  1. #1
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    Integral of 1/cot(x) + 2

    pls what is the integral of 1/cotx+2

    thanks.

    let me make it clearer guys;

    the equation is dy/dx = cot(y+x)-1

    solution is Cos(y+x)=ce-x


    my problem is the process leading to the solution and was solved using variable saperable differencial equation


    thanks for ur assistance.
    Last edited by lawochekel; July 28th 2011 at 03:01 AM. Reason: for clearity
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: integration equation! pls help

    You have to be more clear, do you have to calculate the integral:
     \int \frac{1}{\cot(x)+2}dx
    or
    \int \left(\frac{1}{\cot(x)}+2\right)dx
    ?
    To start, write \cot(x)=\frac{\cos(x)}{\sin(x)}
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  3. #3
    Member anonimnystefy's Avatar
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    Re: integration equation! pls help

    it is probably the first one because the second one is the same as inegral of tan(x)+2.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: integration equation! pls help

    To solve the integral, I think it's better to wright: \cot(x)=\frac{1}{\tan(x)}.
     \int \frac{dx}{\cot(x)+2} = \int \frac{dx}{\frac{1}{\tan(x)}+2}

    Let \tan(x)=t then \frac{dx}{\cos^2(x)}=dt \Leftrightarrow dx=dt\cdot \cos^2(x)

    We want now \cos^2(x) in function of t so, we use the fact that: \tan(x)=t \Leftrightarrow x=\arctan(t).

    And \cos^2[\arctan(t)]=\left(\frac{1}{\sqrt{1+t^2}}\right)^2=\frac{1}{1+  t^2}

    Now the integral becomes:
    \int \frac{dt}{\left(\frac{1}{t}+2\right)\cdot \left(1+t^2)}}

    Now you can try to split up in partial fractions.
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  5. #5
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    Re: integration equation! pls help

    Your ODE is

    \dfrac{dy}{dx} = \cot (x + y ) - 1

    If you let u = x + y then

    \dfrac{du}{dx} = \cot u

    which leads to the separation

    \dfrac{du}{\cot u} = dx or \tan u du = dx.
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