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Math Help - 2 integrals I dont know how to start

  1. #1
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    2 integrals I dont know how to start

    I have these two integrals I need to evaluate but not sure how to start them.

    Can any1 show me?

    1. integral 1/(x^6-1) dx

    2. integral sqrt(1-cos(ax)) dx
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  2. #2
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    I think for Q1 can be started by using partial fractions.
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  3. #3
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    For the first try partial fractions

    \dfrac{1}{x^6-1} = \dfrac{1}{6}\, \dfrac{1}{ x-1} - \dfrac{1}{6}\, \dfrac{1}{ x+1}-\dfrac{1}{6}\,{<br />
\dfrac {x+2}{{x}^{2}+x+1}}+\dfrac{1}{6}\,{\dfrac {x-2}{{x}^{2}-x+1}}

    and the second, try using the identity

    \cos ax = 1 - 2 \sin^2 \frac{ax}{2}.
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  4. #4
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    im stuck on how u got that partial fraction decomposition?
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  5. #5
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    (x^6-1) =

    (x^2-1)(x^4+x^2+1) =

    (x-1)(x+1)(x^4+x^2+1)

    and then

    (x-1)(x+1)(x^2-x+1)(x^2+x+1)
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  6. #6
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    so far with this i have

    1/x^6-1 = A/x-1 + B/x+1 + Cx+D/x^2+x+1 + Ex+F/x^2-x+1

    I have found A = 1/6, B = -1/6, D = -2, and F = 4/3

    I dont know if they are right but I am finding it difficult to work out the other constants.
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  7. #7
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    the answer I get for Q2 is:

    2*sqrt((cos(ax))+1)/a + C

    can any1 verify this?
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