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Math Help - trignometric integrals

  1. #1
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    trignometric integrals

    integrate integral of dx/(cosx-1).
    this is how i tried solving it:
    integral dx/(cosx-1)x(cosx+1)/(cosx+1)
    =-integral (cosx + 1)/((sinx)^2)
    =-integral (cosx)/((sinx)^2) + integral (cscx)^2dx
    u=sinx du=cosxdx
    =-[integral du/u^2 dx+ integral (1)dx + integral(cotx)^2dx
    I don't know how to integrate (cotx)^2.
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  2. #2
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    Good start ...

    ∫1/(1-cos(x)) dx

    Multiply the numerator and the denomintor by (1+cos(x))

    = ∫(1+cos(x)) / [(1+cos(x))(1-cos(x))] dx

    = ∫(1+cos(x)) / (1-cosē(x)) dx

    = ∫(1+cos(x)) / sinē(x)) dx

    = ∫(cscē(x) + csc(x)cot(x)) dx

    = -cot(x) - csc(x) + c

    = -[(cos(x) + 1) / sin(x)] + c


    Your start was excellent. It pays to know how to integrate cscē(x).
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  3. #3
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    Hello, twilightstr!

    You had a good start on it . . .


    \int \frac{dx}{\cos x-1}

    We have: . -\int\frac{dx}{1-\cos x}


    Multiply by \frac{1+\cos x}{1+\cos x}\!:\;\;-\int\frac{dx}{1-\cos x}\cdot\frac{1+\cos x}{1+\cos x} \;\;=\;\;-\int\frac{(1+\cos x)\,dx}{1 - \cos^2\!x}


    . . =\;\;-\int\frac{(1+\cos x)\,dx}{\sin^2\!x} \;\;= \;\;-\int\left(\frac{1}{\sin^2\!x} + \frac{\cos x}{\sin^2\!x}\right)\,dx . =\;\;-\int\left(\csc^2\!x  + \csc x\cot x\right)\,dx


    . . =\; -\left(-\cot x - \csc x\right) + C \;\;=\;\;\cot x + \csc x + C

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  4. #4
    MHF Contributor red_dog's Avatar
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    \int\frac{1}{\cos x-1}dx=\int\left(-\frac{1}{2\sin^2\frac{x}{2}}\right)dx=\cot\frac{x}  {2}+C
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