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Curious Question; find ⌠cotx using integration by parts

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data
    find ⌠cotx using integration by parts with using u= 1/sinx and dv= cosx


    2. Relevant equations
    cotx=cosx/sinx


    3. The attempt at a solution
    u= 1/sinx and dv= cosx dx
    du = -cotxcscx dx v= sinx

    ⌠udv = 1 + ⌠sinx cotx cscx
    sinx and cscx cancel out.

    ⌠cotx = 1 + ⌠cotx
    I=1+I
    0I=1?
     
    Last edited: May 26, 2010
  2. jcsd
  3. May 26, 2010 #2

    Mark44

    Staff: Mentor

    Are you required to use integration by parts? If not, an ordinary substitution will work.

    You have a sign error in this equation ⌠cotx = 1 - ⌠cotx . If IBP is required, try splitting u and dv differently.
     
  4. May 26, 2010 #3
    Hello,
    I know that it could be done with simple substitution. I was just curious of why it won't work by IBP.

    Also, because I've seen ⌠tanx = ln|secx| and as -ln|cosx|

    I thought that ⌠cotx = ln|sinx| would also be -ln|cscx|

    was trying to prove that maybe it is also equal to -ln|cscx|.
     
  5. May 26, 2010 #4

    Mark44

    Staff: Mentor

    Don't omit the differentials and the constants of integration.

    Yes, ⌠cotx dx = ln|sin x| + C = - ln(1/|csc x|) + C. And you need only the properties of logs to show that - ln(1/|csc x|) = ln|sin x|.

    As I already mentioned, this is very much easier when you use substitution, but I believe that you could also use IBP if you're masochistic. I would try u = cos x and dv = csc x dx instead of what you tried.
     
  6. May 27, 2010 #5
    Thank you, Mark44, I realize that it would be too hard to go through that.
     
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