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Another indefinite integral question?

  1. Jan 13, 2008 #1
    How to evaluate [tex]\int{\frac{\arccos{x}}{x^2}\,dx}[/tex]?
     
  2. jcsd
  3. Jan 13, 2008 #2
    Try the substition
    [tex]x=\cos u,\,d\,x=-\sin u\,d\,u[/tex]
    and then integrate by parts.
     
  4. Jan 15, 2008 #3
    Perfect..

    [tex]
    x = cos(u)~;~du = -sin(u)
    [/tex]
    also,
    [tex]
    x^2 = cos^2(u)
    [/tex]

    Therefore we have,

    [tex]
    \int\frac{cos^{-1}(x)}{x^2}dx = I = \int\frac{u sin(u)}{cos^{2}(u)}du
    [/tex]

    [tex]
    I = -\int u tan(u)sec(u)du
    [/tex]

    [tex]
    I = -\left[u sec(u) - \int \frac{du}{du} \int sec(u)tan(u)du du\right]
    [/tex]

    [tex]
    I = -\left[u sec(u) - \int sec(u)du\right]
    [/tex]

    [tex]
    I = log_e|sec(u) + tan(u)| - u sec(u) + C
    [/tex]

    Then, you can replace the value of u with [itex]cos(x)[/itex] to get the required integral.
     
  5. Jan 15, 2008 #4

    arildno

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    Blunder there, you are to substitute [itex]u=\cos^{-1}(x)[/itex]
     
    Last edited: Jan 15, 2008
  6. Jan 15, 2008 #5
    [itex]u = \cos^{-1}(x)~;~x = \cos(u)[/itex].. it's the same thing within a restricted domain.
     
  7. Jan 15, 2008 #6

    arildno

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    You wrote substitute u with cos(x), which is wrong.
     
  8. Jan 15, 2008 #7
    ohh.. u mean in the last part.. yeah.. i get it now... ur right :D
     
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