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Math Help - Eek Urgent Integral Help!?!

  1. #1
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    Eek Urgent Integral Help!?!

    Need help integrating:

    integral(2cot(x/3)dx)

    for my Calc I take home

    so far, i've gotten as far as

    2integral(cos[x/3]*(1/sin[x/3]))dx

    or am i going in the wrong direction here?
    Last edited by calcidiots101; January 29th 2007 at 07:51 PM.
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  2. #2
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    Quote Originally Posted by calcidiots101 View Post
    Need help integrating:

    integral(2cot(x/3)dx)

    for my Calc I take home

    so far, i've gotten as far as

    2integral(cox[x/3]*(1/sin[x/3]))dx

    or am i going in the wrong direction here?
    In general,
    \int \cot t dt=\int \frac{\cos t}{\sin t}dt
    Let u=\sin t then u'=\cos t
    Thus, by the substitution rule,
    \ln |\sin t|+C

    Thus, when you have,
    \int 2 \cot (x/3) dx = 2\cdot 3 \ln |\sin (x/3)|+C
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  3. #3
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    see edit above, dno if it makes a difference...

    also, could you explain that natural log substitution a bit further? i'm sort of fuzzy on it...
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  4. #4
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    Quote Originally Posted by calcidiots101 View Post
    see edit above, dno if it makes a difference...

    also, could you explain that natural log substitution a bit further? i'm sort of fuzzy on it...
    Just like u=\sin t and then just use the subsitution rule.
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  5. #5
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    thanks for your great help so far...
    but i don't really understand the substitution...

    basically, i have:

    u=sinx
    du=cosxdx

    so then that becomes:

    integral([1/u]du)??
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  6. #6
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    Quote Originally Posted by calcidiots101 View Post
    ...anyone?
    integral{(1/u) du} = ln(|u|) + c

    now substitute back for u.



    or
    2integral(cos[x/3]*(1/sin[x/3]))dx

    put v = sin[x/3]

    dv = cos[x/3] * (1/3) dx

    so that cos[x/3] * dx = 3 dv

    2integral(cos[x/3]*(1/sin[x/3]))dx = 2 integral {3dv / v}

    = 6 integral {dv / v}

    = 6* ln(|v|)+ c

    = 6* ln (| sin[x/3] |)+ c


    some integration formulae are available at

    calculus(integration formulae)
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  7. #7
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    Quote Originally Posted by calcidiots101 View Post
    ...anyone?
    INT.[2cot(x/3)]dx -------------------(i)
    = 2*INT.[cos(x/3) /sin(x/3)]dx -----------(ii)

    You supposed u = sin(x/3)
    So du = cos(x/3) *(1/3) dx
    du = (1/3)cos(x/3)dx

    Back to (ii),
    2*INT.[cos(x/3) /sin(x/3)]dx
    = 2*INT.[1 /sin(x/3)]*cos(x/3)dx
    = 2*INT.[1/u]cos(x/3)dx -----------------(iii)

    Now, cos(x/3)dx is not du.
    The du is (1/3)cos(x/3)dx.
    Meaning, the cos(x/3)dx is 3 times du.

    a) Hence, so that you can use the cos(x/3)dx as "du", you have to divide the cos(x/3)dx by 3.
    In doing that, you also have to multiply the [cos(x/3)dx]/3 by 3---so that you will not alter the (iii).
    Meaning, you divide by 3, and also multiply by 3.
    = 2*INT.[1/u]cos(x/3)dx *(3/3)
    = (2*3)INT.[1/u]*(1/3)cos(x/3)dx
    = 6*INT.[1/u]du
    = 6*ln|u| +C
    Since u = sin(x/3),
    = 6*ln|sin(x/3)| +C ------------answer.

    b) Or, if you want to wrack your brain more, , multiply the du by 3, so that it will match the cos(x/3)dx.
    Now, multiplying du by 3 will not affect, will not alter, the (iii). You are only substituting 3du for cos(x/3)dx into (iii), and 3du = cos(x/3)dx anyway.
    So,
    2*INT.[1/u]cos(x/3)dx
    = 2*INT.[1/u](3du)
    = (2*3)INT.[1/u]du
    etc.....
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  8. #8
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    thanks so much for the help!
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