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Integral invovling sec

  1. Feb 24, 2008 #1
    evaluate

    [tex]\int2(sec x)^3[/tex] with the limits as -pi/3 to 0

    I tried all sorts of things from breaking it apart to substitution, but known of what I tried work.

    The book shows setting u=sec x & v=tan x

    Then it shows the first step as 2 (sec x tan x) - 2 [tex]\int(sec x) * (tan x)^2 dx[/tex] then evaluate both parts to -pi/3 to 0.

    Which is really what I'm not understanding. How did they integrate the first part & then still have the next part? I'm also not seeing how u & v come into play.

    Guess I'm just plain lost on this one.
     
    Last edited: Feb 24, 2008
  2. jcsd
  3. Feb 24, 2008 #2
    Substitution

    As Griffith's puts it, paraphrased, you can move the derivative from one variable to the other under an integral, and you'll just pick up a minus sign and a boundary term.

    Thus the equation:
    [tex]\int_a^buv'dx=\left.uv\right|_a^b-\int_a^bu'vdx[/tex]
     
  4. Feb 24, 2008 #3
    [tex]\int\sec x(\tan^{2}x+1)dx[/tex]
    [tex]\int\sec x\tan^{2}xdx+\int\sec xdx[/tex]

    [tex]u=\sec x[/tex]
    [tex]du=\sec x \tan x dx[/tex]

    [tex]dV=\tan^{2}xdx[/tex]
    [tex]V=\sec x[/tex]
     
    Last edited: Feb 24, 2008
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