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Integration by parts where to start

  1. May 31, 2005 #1
    show that INT x sec^2x dx = pi/4 - ln2/2 (between pi/4 and 0)

    pls help i dunno where to start

    i know it is integration by parts - just dunno how i should rearrange it. thanks
  2. jcsd
  3. May 31, 2005 #2


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    If you know part integration,u'll find the notation quite familiar

    [tex] x=u;\sec x dx =dv [/tex]

    BTW,it should read

    [tex] \int_{0}^{\frac{\pi}{4}} x\sec^{2} x \ dx=\frac{\pi}{4}-\frac{\ln 2}{2} [/tex]

  4. May 31, 2005 #3
    Tabular method, so much easier. Enough said.
  5. May 31, 2005 #4


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    Tabular method really only helps when you have several ibp steps. In this case, you only have one. Once you integrate sec^2 x, you get tan x. If you're going to integrate tan x manually (vs just looking at the table), it's easiest to break it into

    [tex]\int{\sin x \frac{1}{\cos x} dx}[/tex]

    and then use u-substitution.
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