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Trigonometry Substitution (Integral)

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data
    ∫xtan^-1(x^2)dx


    2. Relevant equations



    3. The attempt at a solution

    I did u = x, du = 1,
    v = ? ,dv = tan^-1(x^2)dx

    I do not know how to get the integral of tan^-1(x^2)
     
  2. jcsd
  3. Feb 23, 2013 #2

    LCKurtz

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    u = x can never simplify an integral. It just changes the name of the variable. Try ##u=x^2##.
     
  4. Feb 23, 2013 #3
    u = x^2
    du = 2xdx
    du/2 = xdx

    ∫xtan(x^2)dx

    (1/2)∫tan(u)du

    (1/2) ln |sec(x^2)| + C

    Correct?
     
  5. Feb 23, 2013 #4

    LCKurtz

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    You don't have to ask. Differentiate it and check if it works. Did you forget an arctangent?
     
  6. Feb 23, 2013 #5
    u = x^2
    du = 2xdx
    du/2 = xdx

    ∫xtan^1(x^2)dx

    (1/2)∫tan^-1(u)du

    I cannot figure out the integral of tan^-1(u)
     
  7. Feb 23, 2013 #6

    SammyS

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    For the integral, [itex]\displaystyle \ \ \int \tan^{-1}(t)\,dt\,,\ [/itex] use integration by parts, with u = tan-1(t) and dv = dt , else look it up. http://en.wikipedia.org/wiki/Inverse_trig_functions#Indefinite_integrals_of_inverse_trigonometric_functions



    Added in Edit:

    An alternate way to approach this problem (It is titled, Trigonometry Substitution (Integral) .) is to use the substitution, x2 = tan(t) .

    It will also require the use of integration by parts.
     
    Last edited: Feb 23, 2013
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