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Homework Help: Another Indefinite Integral question

  1. Jul 18, 2010 #1
    1. The problem statement, all variables and given/known data
    ∫cos(x)/sin^2(x)*dx


    2. Relevant equations



    3. The attempt at a solution
    Based off my earlier question, where is my error please.
    u=sinx du=cosx*dx
    ∫u^-1*du sin(x)^-1 1/sin(x) + C -or-csc(x) + C

    Thanks
     
  2. jcsd
  3. Jul 18, 2010 #2

    hunt_mat

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    your integral now becomes:
    [tex]
    \int\frac{du}{u^{2}}=-\frac{1}{u}+C=-\frac{1}{\sin x}+C
    [/tex]
     
  4. Jul 18, 2010 #3
    But where does the -1/u com from? Since the dx of sin(x) is cos(x) and the sin(x) was in the denominator, I know to subtract 1 from the original sin^2(x) in the denominator instead of add. But I keep coming up with 1/sin(x)
     
    Last edited: Jul 18, 2010
  5. Jul 19, 2010 #4

    hunt_mat

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    You should know that:
    [tex]
    \int u^{n}du=\frac{1}{n+1}u^{n+1}+C\quad n\neq -1
    [/tex]
    Just take n=-2 for the answer you require.
     
  6. Jul 19, 2010 #5

    hunt_mat

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    I think I see your question (sorry)
    The integral is:
    [tex]
    \int\frac{\cos x}{\sin^{2}x}dx
    [/tex]
    let [tex]u=\sin x[/tex], then du/dx=cos x and the integral reduces to the one I posted earlier.
     
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