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Evaluate the integral (inverse trig functions)

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    [23/4, 2] 4/(x√(x4-4))

    2. Relevant equations
    ∫ du/(u√(u2 - a2)) = 1/a(sec-1(u/a) + c

    3. The attempt at a solution
    I first multiplied the whole thing by x/x. This made the problem:
    4x/(x2√(x4 - 4))

    Then I did a u substitution making u = x2. Therefore, du = 2xdx. I multiplied by 2 to get 2du = 4xdx

    The problem then becomes 2∫du/(u√(u2 - 4))

    Solving the integral I got 2[(1/2)sec-1(x2/2)] from [23/4, 2]

    I plug in the bounds and get 2[(1/2)sec-1(2) - (1/2)sec-1(26/4/2)

    This is where I'm lost. The second sec does not seem like a nice number and I'm assuming my professor would make the problem come out nicely as he always has. I'm pretty sure I made a mistake somewhere because of this but I don't know where.
     
  2. jcsd
  3. Feb 3, 2015 #2

    jambaugh

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    Science Advisor
    Gold Member

    Try simplifying the fractional exponent and rewriting inverse secant in terms of inverse cosine ( arcsec(x) = arccos(1/x)).
     
  4. Feb 3, 2015 #3
    Thank you so much! I should have noticed the reducing.

    Finishing the problem:

    sec-1(2) - sec-1(21/2)

    sec-1(2) - sec-1(√(2))

    This equals:
    arccos(1/2) - arccos(1/√(2))
    arccos(1/2) - arccos(√(2)/2)
    pi/3 - pi/4 = pi/12
     
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