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Math Help - calculus proof

  1. #1
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    calculus proof

    Hi, Ive been stuck on this proof problem for awile and was wondering if anyone has any ideas?

    Integral (csc(x)) dx = ? expain how you get the answer.

    **For example in Integral (sec(x)) dx:

    Integral (Sec(x)) (tan(x) + sec(x)) / (tan(x) + sec(x))) dx =

    (sec(x)tan(x) + sec^2(x)) / (tan(x) + sec (x))

    Set u= tan(x) + sec(x) du= (sec^2(x) + sec(x)tan(x)) dx

    If, Integral sec(x) = Integral (1 / u) = ln(abs(u))

    Then, Integral sec(x) = ln(abs(tan(x) + sec(x))) + c

    With the csc(x) proof I am stuck trying to find the function equaling one to times it by in the beginning that will lead to the correct answer.
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  2. #2
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    \int \csc{x} \, dx

    multiply by \frac{\csc{x} + \cot{x}}{\csc{x} + \cot{x}}

    \int \frac{\csc{x}(\csc{x} + \cot{x})}{\csc{x} + \cot{x}} \, dx

    \int \frac{\csc^2{x} + \csc{x}\cot{x}}{\csc{x} + \cot{x}} \,<br />
dx

    -\int \frac{-\csc^2{x} - \csc{x}\cot{x})}{\csc{x} + \cot{x}} \,<br />
dx

    -\ln|\csc{x} + \cot{x}| + C

    \ln \left| \frac{\sin{x}}{1 + \cos{x}} \right| + C
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  3. #3
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    Hello, ulion!

    The routine is similar to that of: \int\sec x\,dx


    \int \csc x\,dx

    Multiply by \frac{\csc x - \cot x}{\csc x - \cot x}

    \int\left( \csc x\cdot\frac{\csc x - \cot x}{\csc x - \cot x}\right)\,dx \;\;=\;\;\int \frac{\csc^2\!x - \csc x\cot x}{\csc x-\cot x}\,dx


    . . = \;\;\int\frac{\overbrace{-\csc x\cot x + \csc^2x}^{du}}{\underbrace{\csc x - \cot x}_u}\,dx \;\;=\;\;\ln|\csc x - \cot x| + C

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  4. #4
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    thankyou for your help
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