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Homework Help: Trig Integral

  1. Mar 20, 2006 #1

    G01

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    [tex] \int_0^{\pi/4} \frac{x\sin(x)}{cos^3(x)}dx [/tex]

    I thought I could do this integral by parts but I keep getting it wrong and I can't find my mistake.

    [tex] \int_0^{\pi/4} x\tan(x)\sec^2(x) dx[/tex]

    let u = xtanx , dv = sec^2x dx

    du = xsec^2x+tanx , v = tanx

    [tex] x\tan^2(x)|_0^{\pi/4} - \int_0^{\pi/4} x\tan(x)\sec^2(x) dx - \int_0^{\pi/4} \tan^2(x) dx [/tex]

    so:

    [tex] \int_0^{\pi/4} x\tan(x)\sec^2(x) dx = \frac{x\tan^2(x)|_0^{\pi/4} - \int_0^{\pi/4} \tan^2(x) dx}{2}[/tex]

    [tex] x\tan^2(x)|_0^{\pi/4} - x|_0^{\pi/4} + \tan(x)|_0^{\pi/4} [/tex]

    This evaluates out to 1 but thats not the answer. Thank you for your help.
     
  2. jcsd
  3. Mar 20, 2006 #2

    0rthodontist

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    Let u = x, dv = tan x sec^2 x. You can integrate the second by substitution, and the x will drop out by differentiation.

    Your error is in the last step, where you have -x + tan x. It should be tan x - x. Also you forgot to divide by 2.
     
    Last edited: Mar 20, 2006
  4. Mar 20, 2006 #3

    G01

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    thanks alot
     
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