1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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

    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
    Last edited: May 26, 2010
  2. jcsd
  3. May 26, 2010 #2


    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
    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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook