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    lol nm i worked it out...
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    (Original post by Dust)
    how can u convert (1-2X^2)^-3

    all into

    1-3(-2X^2)

    ...it doesnt make sense?
    Using the binomial ytheorem, if x is sufficiently small that terms in x^4 and higher can be ignored, (1-2x^2)^(-3) = 1 + 2x^2 (approx)
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    lol yup... thats wat i got in the end...
    thanx

    lol i always panic wen i see cos sin's etc lol
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    okay how can u convert the integral of x cos^2 X into 1/2 of the intergral of X cost 2X +x?

    i can see how u get X cost 2X but where did the +X come from at the end?
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    (Original post by Dust)
    okay how can u convert the integral of x cos^2 X into 1/2 of the intergral of X cost 2X +x?

    i can see how u get X cost 2X but where did the +X come from at the end?
    cos²x = ½(1 + cos2x)
    x.cos²x = ½x(1 + cos2x)
    x.cos²x = ½(xcos2x + x)
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    lol oh yeah.. lol now i feel stupid
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    (Original post by Dust)
    lol oh yeah.. lol now i feel stupid
    obviously you've just been working too hard, that's all
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    (Original post by Fermat)
    obviously you've just been working too hard, that's all
    lol i wish i could believe that.. *shrugs*

    okay i dont get this either: how does this work

    when -2K=ln 1/2
    how does that make k= ln 2??
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    -2k = ln(1/2) = ln1 - ln2 = -ln2
    2k = ln2
    k = (ln2)/2
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    lol ah that makes sense...
    cheers
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    Could be ln(sqrt2).
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    (Original post by ZJuwelH)
    Could be ln(sqrt2).
    how?
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    (Original post by Dust)
    how?
    because,

    plnx = ln(x^p)

    for example,

    5ln2 = ln2 + ln2 + ln2 + ln2 + ln2

    but lnA + lnB = ln(A*B), (by definition of logs)
    So,

    ln2 + ln2 + ln2 + ln2 + ln2 = ln(2*2*2*2*2)
    ln2 + ln2 + ln2 + ln2 + ln2 = ln(2^5)
    5ln2 = ln(2^5)
    =========

    So,

    ½lnx = ln(x^½) = ln(√x)
 
 
 
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