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How do i figure out this integral

  1. Aug 24, 2008 #1
    1. The problem statement, all variables and given/known data
    ∫ 1/cos theta
    before it was ∫ sec theta but i know that sec theta is equal to 1/cos theta

    would i write it as 1/2 ∫ cos theta

    and then 1/2 sin theta
     
  2. jcsd
  3. Aug 24, 2008 #2

    rock.freak667

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    try multiplying sec(theta) by

    [tex]\frac{sec\theta +tan\theta}{sec\theta +tan\theta}[/tex]


    then try a substitution.
     
  4. Aug 24, 2008 #3
    where did you get sec theta + tan theta/ sec theta+ tan theta from
     
  5. Aug 24, 2008 #4
    is this what i do (sec theta) (sec theta+ tan theta)/ Sec theta( sec theta + tan theta)
     
  6. Aug 24, 2008 #5

    rock.freak667

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    You should get

    [tex]\frac{sec\theta(sec\theta +tan\theta)}{sec\theta + tan\theta}[/tex]


    The expand out the numerator.


    It's an easier way to find the integral faster rather than another method.
     
  7. Aug 24, 2008 #6
    Ok thank you.
     
  8. Aug 25, 2008 #7

    HallsofIvy

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    By the way, [itex]\int dx/cos(x)[/itex] involves an odd power of cosine so the "standard" method for that situation will work. Multiply both numerator and denominator by cos(x) to get
    [tex]\int \frac{dx}{cos x}= \int \frac{cos x dx}{cos^2 x}= int \frac{cos x dx}{1- sin^2 x}[/tex]

    Now let u= sin x so du= cos x dx and 1- sin2 x= 1- u2

    [tex]\int \frac{dx}{cos x}= \int \frac{du}{1- u^2}= \int \frac{du}{(1- u)(1+ u)}[/itex]

    and you can use "partial fractions" to integrate that.
     
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