1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    [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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Because you make an equality out of the following non-equality:
  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


    User Avatar

    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?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Integrating cosec
  1. An integral (Replies: 1)

  2. Integral of (Replies: 3)

  3. On Integration (Replies: 4)

  4. An integral (Replies: 2)