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Homework Help: Integral of secant

  1. Feb 1, 2007 #1
    1. The problem statement, all variables and given/known data

    Basically, I have to find

    [tex]
    \int \frac{1}{cosx} dx
    [/tex]

    by multiplying the integrand by [tex] \frac{cosx}{cosx}


    [/tex]

    I go through and arrive at a solution, but when I differentiate it,
    I get -tan(x)

    something's clearly wrong, but I can't see what it is that I'm doing wrong here...


    2. Relevant equations

    [tex]
    let u = sin(x)[/tex]


    [tex] du = cos(x)dx
    [/tex]

    3. The attempt at a solution

    [tex]
    \int \frac{1}{cosx} dx = \int \frac{cosx}{cos^2x} dx\\
    = \int \frac{cosx}{1-sin^2x} dx\\
    =\int \frac{du}{1-u^2} \\
    =\int \frac{du}{(1-u)*(1+u)} \\
    =\frac{1}{2} * \int \frac {1}{1+u} + \frac {1}{1-u} du\\
    = \frac{1}{2} * (ln(1-u^2}})
    =\frac{1}{2} * (ln(cos^2))


    [/tex]
     
    Last edited: Feb 1, 2007
  2. jcsd
  3. Feb 1, 2007 #2

    StatusX

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    Homework Helper

    Your second to last step (where you actually perform the integration) is wrong. In the second term, remember that u has a minus sign.
     
  4. Feb 1, 2007 #3
    hmm..

    I've never run into anything like this before

    so why does u having a minus sign in front of it pose a problem with what I did in my original solution?

    (thanks for the help)
     
  5. Feb 1, 2007 #4

    cristo

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    Staff Emeritus
    Science Advisor

    Well, the value of your integral will be ln(1+u)-ln(1-u)
     
  6. Feb 1, 2007 #5
    if I kept the minus sign where it was, which is what I did in my original solution, my integral would've been
    [tex] ln(1+u)+ln(1-u)=ln(1-u^2)[/tex]

    StatusX told me to watch out for the negative sign in front of the u, so I factored it out, made sure that u was positive, and then integrated it, which gave me the correct solution.

    the thing is i'm not sure why I couldn't proceed as usual with the minus sign in front of the u
     
  7. Feb 1, 2007 #6

    cristo

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    StatusX said to note the minus sign in front of the u. You can proceed as normal, but noting that [tex]\int\frac{1}{1-u}du=-ln(1-u)[/tex]. In general [tex]\int\frac{1}{f(u)}du=\frac{ln[f(u)]}{df/du}[/tex]. In this case, f(u)=1-u, and so df/du=-1
     
  8. Feb 1, 2007 #7
    ah, ok

    thank you!
     
  9. Feb 2, 2007 #8
    o_O i didn't know we would utilizie such a method to do this integral.

    i've always thought the integral of secant was just sec[x]tan[x]!
     
  10. Feb 2, 2007 #9

    cristo

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    Staff Emeritus
    Science Advisor

    That's the derivative of sec(x)
     
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