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Integrating cosec

  1. May 24, 2004 #1
    Int(cosec x)dx Let: u=cos x then: x=arcos u and: dx=-du/sqrt(1-u^2)

    Int(cosec x)dx = -Int{1/sqrt(1-u^2)}^2}du = -Int {du/(1-u^2)}

    = -(1/2)ln(1-u^2) + C

    = -(1/2)ln(sin^[2]x) +C

    = ln(cosec x) +C

    Yet differentiating back gives -cot x.
    Why does this substitution not work?
     
    Last edited: May 25, 2004
  2. jcsd
  3. May 24, 2004 #2

    First:

    [tex] csc x = \frac{1}{sin x} [/tex]

    NOT cos x. So your substitution is wrong to begin with. If you correct that the integration should go like this:

    [tex] u = sin x dx [/tex]

    [tex] du = cos x dx [/tex]

    [tex] sec x du = dx [/tex]

    [tex] \frac{1}{\sqrt{1-u^2}} du = dx [/tex]

    So this makes the integral:

    [tex] \int \frac{1}{u\sqrt{1-u^2}} du [/tex]

    See if you can take it from there.
     
  4. May 25, 2004 #3
    My query was concerning the substitution u = cos x , rather than u = sin x .

    I did make a mistake, arcos x = -1/sqrt(1-x^2). I've edited my original post.
     
    Last edited: May 25, 2004
  5. May 25, 2004 #4

    That is the wrong substitution for integrating cosec(X).[tex] cosec(x) = \frac{1}{sin(x)}[/tex].
     
  6. May 30, 2004 #5
    well your subsitution is didn't work because it is difficult to solve integration with assumption not being in the question
     
  7. May 30, 2004 #6

    arildno

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    Because you make an equality out of the following non-equality:
    [tex]-\int\frac{du}{1-u^{2}}\neq\frac{-1}{2}ln(1-u^{2})+C[/tex]
     
  8. May 30, 2004 #7
    Fine, yes, sorry about that, me being stupid again...but that integral was getting annoying.
     
  9. May 24, 2008 #8
    Help please

    I can't :(. I've spent most of this afternoon trying this question and some other one. I know I could just use the set result, but I want to understand it, and I don't see how to integrate that last result at all :(.
     
  10. May 24, 2008 #9

    Air

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    There is a special method to this.

    [itex]\int \mathrm{cosec} x \ \mathrm{d}x[/itex]

    If you multiply this by [itex]\frac {\mathrm{cosec} x - \cot x}{\mathrm{cosec} x - \cot x}[/itex] and simplify the numerator you will get an integral of...

    [itex]\int \frac{\mathrm{cosec} ^2 x - \mathrm{cosec} x \cot x}{\mathrm{cosec} x - \cot x} \ \mathrm{d}x[/itex]

    Substitute [itex]u = \mathrm{cosec} x - \cot x[/itex] and it should work out beautifully. Carry on from here and post back if you still need help. :smile:
     
  11. May 24, 2008 #10
    Thanks :)!

    Thanks, that worked out really well :). I was wondering though, how would one work it out from the form

    [tex] \int \frac{1}{u\sqrt{1-u^2}} du [/tex]

    ? Could anyone help me see how to integrate it from this?
     
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