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Integrating sec x dx

  1. Oct 28, 2008 #1
    1. The problem statement, all variables and given/known data
    By multiplying the integrand sec x dx by [tex]\frac{tan x + sec x}{tan x + sec x}[/tex] find the integral of sec x dx

    2. Relevant equations

    d/dx sec x = tan x.sec x
    d/dx tan x = sec^2 x

    3. The attempt at a solution

    sec x dx([tex]\frac{tan x + sec x}{tan x + sec x}[/tex]) =>

    [tex]\frac{tan x.sec x + sec^2 x}{tan x + sec x}[/tex]dx

    Just noticed the numerator is the derivative of the denominator, so =>

    [tex]\frac{d(sec x + tan x)}{sec x + tan x}[/tex]dx

    Not sure what to do from here...
    Last edited: Oct 28, 2008
  2. jcsd
  3. Oct 28, 2008 #2


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    That's integral of du/u where u=sec(x)+tan(x). What's integral of du/u?
  4. Oct 29, 2008 #3
    Well...integrating the derivative would just return the original function wouldn't it? But in this case it's the reciprocal so it would be 1/sec x + tan x ???
  5. Oct 29, 2008 #4


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    Nooo. Integral of du/u is log(u), isn't it?
  6. Oct 29, 2008 #5
    Oh yeaaaaa...I got confused. I always think of it as 1/x, not dx/x.
    Thanks Richard!
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