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Integral help

  1. Jan 28, 2007 #1
    I need help with this one:


    I know there must be a way to rewrite it so that i can use 1 - cos^2 = sin^2. I've tried to multiply by cosx + 1 / cosx + 1 and that gets me further but the answer doesn't seem to be the same maybe because of simplification?

    Any hints?

  2. jcsd
  3. Jan 28, 2007 #2
    The integral is equal to:

    [itex] \int \frac {\cos x + 1}{-\sin ^2 x} dx[/itex] I suggest you use some trigonometric identities to change it into
    [itex] \int - \cot x \csc x - \csc ^2 x dx[/itex]
  4. Jan 29, 2007 #3


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    [tex] \int \frac{dx}{\cos x-1}=-\int \frac{dx}{1-\cos x}=-\frac{1}{2}\int \frac{dx}{\sin^{2}\frac{x}{2}} [/tex]

    use a simple substitution and a tabulated integral to get the answer.
  5. Feb 5, 2007 #4
    Last edited: Feb 5, 2007
  6. Feb 5, 2007 #5
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