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Thread: 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; Jul 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:
    $\displaystyle \int \frac{1}{\cot(x)+2}dx$
    or
    $\displaystyle \int \left(\frac{1}{\cot(x)}+2\right)dx$
    ?
    To start, write $\displaystyle \cot(x)=\frac{\cos(x)}{\sin(x)}$
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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: $\displaystyle \cot(x)=\frac{1}{\tan(x)}$.
    $\displaystyle \int \frac{dx}{\cot(x)+2} = \int \frac{dx}{\frac{1}{\tan(x)}+2}$

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

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

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

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

    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

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

    If you let $\displaystyle u = x + y$ then

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

    which leads to the separation

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