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Math Help - Finding the integrand

  1. #1
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    Finding the integrand

    <br />
\int_ \frac{\pi}{6}^ \frac{\pi}{2} cot(x)dx<br />

    I ended up switching the cot to the inverse of tangent. Then I put in sines/cosine for tangent. I then did a trig substitution.

    u=sin(x)
    du=cos(x)dx

    However, I end up with this:

    ln1 - ln(1/2)

    ln1 is zero so I'm left with -ln(1/2).

    The answer choices (this is mult.choice)

    A. ln(1/2)
    B. ln(2)
    C. -ln(2-(sq.root of 3))
    D. ln((sq.root of 3) -1)
    E. None of the above

    My answer key says A is the right answer, but I have NEGATIVE ln(1/2). I'm not sure where I went wrong. ??
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Just switch cot(x) = cos(x)/sin(x) Let u =sin(x)

    you end up with ln|sin(x)| an eventually ln1 - ln(1/2) = ln1 - ln1 +ln(2)


    as ln(1/2) = ln1 = ln(2) So B is the right answer
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  3. #3
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    Can you clarify this part?

    - ln(1/2) = - ln1 +ln(2)

    EDIT: btw you're right, my teacher said B is the answer
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  4. #4
    Senior Member
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    Your solution is correct! (Answer choice "A" is not correct.)

    Using some log rules:

    \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)

    You can find that

    -\ln(1/2) = \ln(2)
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