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Csc x dx

  1. Nov 12, 2003 #1
    How do you find &int; cscx dx?
  2. jcsd
  3. Nov 12, 2003 #2


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    Look it up in an integral table? {ln(tan(x/2))}
  4. Nov 12, 2003 #3


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    By seeing the trick used to find &int; sec x dx and applying it to this problem.

    (Your textbook almost certainly shows how to compute the integral of sec x, and if it doesn't, look up the answer in an integral table and work backwards to discover the trick)
  5. Nov 13, 2003 #4
    &int; csc x dx = &int; csc x * (csc x + cot x)/(csc x + cot x) dx
    =&int; ((csc x)^2 + csc x cot x)/(csc x + cot x) dx
    =-ln | csc x + cot x | + constant

    hope it answers your question....
    Last edited by a moderator: Nov 13, 2003
  6. Nov 13, 2003 #5


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    Now, that's just scary! :smile:

    Here's how I would do it: csc(x)= 1/sin(x) so this is the same as &int; dx/sin(x)- an odd power of x. Back in calculus, I learned that when you have an odd power of either sin(x) or cos(x), you can use one of them with dx and use sin2+ cos2= 1 for the rest: here sin(x) is in the denominator so I would write it as
    &int;sin(x)dx/sin2= &int; sin(x)dx/(1- cos2(x)).
    Now let u= cos(x) so du= sin(x)dx and the integeral becomes
    &int; du/(1-u3). The denominator factors as (1-u)(1+ u) so- partial fractions. 1/(1-u2)= (1/2)/(1-u)+ (1/2)/(1+u).
    The integral is now (1/2)&int;du/(1-u)+ (1/2)&int;du/(1+u)= -(1/2)ln|1-u|+ (1/2)ln|1+u|+C= (1/2)ln(|1+u|/|1-u|)+ C= (1/2)ln((1+sin(x))/(x-sin(x)))+ C= ln(&radic;((1+sin(x))/(1-sin(x)))+ C
    Hmmm, maybe Reihhard-t's solution isn't so scary after all!
  7. Nov 14, 2003 #6
    I do not see the flow of your intergration.

    but the most important ,when you do math,you must try the most simple,short, and clear....and get the best result.

    what I did actually just multiply with 1 which is &int;(csc x + cot x)/(csc x + cot x) ....and then just carry two more steps,and done...

    that's math all about
  8. Nov 14, 2003 #7
    OK how about treating &int; csc x dx as &int; 1/sinx dx?

    Here's what I had to far, I had 2 approaches

    &int; 1/sinx dx = &int; (1/sinx)*cosx/cosx dx

    = &int; (1/u)(1/cosx) du = &int;1/sqrt(u^2 - u^4) du.

    I don't know how to continue from there, or is it even continuable at all?

    2nd approach:

    &int;(1/sinx) dx = &int;sinx/sin^2x dx = &int;sinx/(1-cos^2x) dx
    assuming u = cos x, du = -sinx

    = &int; 1/(u^2-1) du

    Then I stop there... don't know how to continue. Or is it continuable??
  9. Nov 14, 2003 #8


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    You can use partial fractions to continue that, once you learn them.
  10. Nov 14, 2003 #9
    HallsofIvy lost a negative u= cos(x) so du= -sin(x)dx
  11. Nov 15, 2003 #10

    Can you teach me? Or someone please teach me?
  12. Nov 15, 2003 #11
    actually all equation can be integrated,but for you case optimus,your further integration will be very complicated,and frankly speaking ,I don't know how to integrate more,to get exactly like my previous answer.
    so I recommend you to implement my previos method.that one is actually a math trick.it is fast and incisive
  13. Nov 15, 2003 #12
    OK heres what i got after trying partial fractions.

    Is that correct?
  14. Nov 15, 2003 #13


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    I think those twos should be one-halves.

    Also, don't forget your absolute value signs!

    And, finally, you need to back substitute to change the 'u's back into 'x's.
  15. Nov 16, 2003 #14
    yeah.you've come with great idea optimus,just slight careless like what is mention by hurkvl.It is both(1/2) instead of 2,and ending it up with variable x together with modulus sign.

    it becomes (1/2)ln |(cos x - 1)/(cos x + 1)|

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