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Math Help - Integral with trig function in denominator

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    Integral with trig function in denominator

    Another seemingly simple integral I am going blank on:

    \int \frac{1}{cosx-1}\,dx

    I really appreciate any help! I have tried different things but none of them seem to work.
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    Re: Integral with trig function in denominator

    \int\frac1{\cos x - 1}\,dx

    =\int\left(\frac1{\cos x - 1}\cdot\frac{\cos x + 1}{\cos x + 1}\right)\,dx

    =\int\frac{\cos x + 1}{\cos^2x - 1}\,dx

    Check the spoiler if you're still stuck:

    Spoiler:
    =\int\frac{\cos x + 1}{-\sin^2x}\,dx

    =-\int\frac{\cos x}{\sin^2x}\,dx - \int\frac1{\sin^2x}\,dx

    =-\int\left(\sin x\right)^{-2}\cos x\,dx - \int\csc^2x\,dx
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    Re: Integral with trig function in denominator

    Thank you! I actually got that far but stopped because I thought I wasn't getting anywhere. How does the rest of the solution go?
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    Re: Integral with trig function in denominator

    Quote Originally Posted by Ragnarok View Post
    Thank you! I actually got that far but stopped because I thought I wasn't getting anywhere. How does the rest of the solution go?
    For the first integral, let u = \sin x. The second integral should be obvious from the fact that \textstyle\frac d{dx}\left[\cot x\right] = -\csc^2x.
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    Re: Integral with trig function in denominator

    Thank you! I see it now.
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