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Antiderivative of cotangent

  1. Mar 15, 2007 #1
    1. The problem statement, all variables and given/known data
    integral (t * csc^2 (t) ) dt

    2. Relevant equations

    3. The attempt at a solution

    t * -cot (t) - integral( 1 * -cot (t)) u= t dv= csc^2(t)
    du= 1 v= - cot (t)
    -t * cot(t) - ??? I don't understand how to find the antiderivative of -cot(t)
  2. jcsd
  3. Mar 15, 2007 #2


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    Integration by parts is the way I'd also go about this.

    Hint: In terms of other trig functions, what is cotangent equal to? You should end up with something that is solvable by substitution.
  4. Mar 15, 2007 #3
    The integral contains singularities whenever sin(x)=0, or x=n PI

    If it's an integral over -T, to T then the integral is zero (by symmetry)
  5. Mar 15, 2007 #4
  6. Mar 16, 2007 #5

    Gib Z

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    [tex]\cot x = \frac{\cos x}{\sin x}[/tex]

    [tex]\int \cot x dx = \int \frac{\cos x}{\sin x} dx[/tex].

    let u= sin x, then du = cos x dx

    [tex]\int \frac{1}{u} du = \ln u + C = \ln (\sin x) + C[/tex]
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