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Math Help - Primitive of 1 / (p + q cos a x)

  1. #1
    Super Member Matt Westwood's Avatar
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    Primitive of 1 / (p + q cos a x)

    No. 390.

    This is what the book says:

    \displaystyle \int \frac {\mathrm d x} {p + q \cos a x} = \begin{cases} \displaystyle \frac 2 {a \sqrt {p^2 - q^2} } \arctan \left({\sqrt {\frac {p - q} {p + q} } \tan \dfrac {a x} 2}\right) + C & : p^2 > q^2 \\ \displaystyle \frac 1 {a \sqrt {q^2 - p^2} } \ln \left\vert{\frac {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q + p} {q - p} } } {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q + p} {q - p} } } }\right\vert + C & : p^2 < q^2 \\\end{cases}


    ... but I'm not getting the same when I work through it, as follows:

    Weierstrass substitution:

    u = \tan \dfrac {a x} 2

    delivers:

    \int \frac {\mathrm d x} {p + q \cos a x} = \frac 2 {a \left({p + q}\right)} \int \frac {\mathrm d u} {u^2 + \dfrac {p - q} {p + q} }


    Let p^2 > q^2.

    Then:

    \dfrac {p - q} {p + q} > 0

    Thus, let \dfrac {p - q} {p + q} = d^2.

    Then:

    \int \frac {\mathrm d x} {p + q \sin a x}r = \frac 2 {a \left({p + q}\right)} \int \frac {\mathrm d u} {u^2 + d^2}

    = \frac 2 {a \left({p + q}\right)} \frac 1 d \arctan \frac u d + C (by standard integrals)

    = \frac 2 {a \left({p + q}\right)} \frac 1 {\sqrt {\dfrac {p - q} {p + q} } } \arctan \left({\frac {\tan \dfrac {a x} 2} {\sqrt {\dfrac {p - q} {p + q} } } }\right) + C (substituting for u and d)

    = \frac 2 {a \sqrt {p^2 - q^2} } \arctan \left({\sqrt {\frac {p + q} {p - q} } \tan \dfrac {a x} 2}\right) + C
    \blacksquare

    I can't reconcile the \sqrt {\dfrac {p + q} {p - q} } given here with the  \sqrt {\dfrac {p - q} {p + q} } as given above.


    Now let p^2 < q^2.

    Then: \dfrac {p - q} {p + q} < 0

    Thus, let:

    -\dfrac {p - q} {p + q} = d^2

    or:

    \dfrac {q - p} {q + p} = d^2


    This forces us to write the integrand in the form:

    \int \frac {\mathrm d x} {p + q \sin a x} = \frac 2 {a \left({p + q}\right)} \int \frac {\mathrm d u} {u^2 - d^2}

    = \frac 2 {a \left({p + q}\right)} \frac 1 {2 d} \ln \left\vert{\frac {u - d} {u + d} }\right\vert + C (standard integral)

    Substituting for u and d:

    = \frac 2 {a \left({p + q}\right)} \frac 1 {2 \sqrt {\dfrac {q - p} {q + p} } } \ln \left\vert{\frac {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q - p} {q + p} } } {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q - p} {q + p} } } }\right\vert + C

    = \frac 1 {a \sqrt {q^2 - p^2} } \ln \left\vert{\frac {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q - p} {q + p} } } {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q - p} {q + p} } } }\right\vert + C after simplifying
    \blacksquare


    Again, this is different from how it appears in the source work.

    The question is: who's got it wrong -- me or Murray R. Spiegel in his Mathematical Handbook of Formulas and Tables? He has been known to be wrong before.
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  2. #2
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    Re: Primitive of 1 / (p + q cos a x)

    The book says

    \int \frac{1}{5-4 \cos  x} \, dx =\frac{2}{3} \arctan (3 \tan  (x/2))

    The book also says

    \int \frac{1}{-5+4 \cos  x} \, dx =\frac{2}{3} \arctan (3 \tan  (x/2))

    This can't be right
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  3. #3
    Super Member Matt Westwood's Avatar
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    Re: Primitive of 1 / (p + q cos a x)

    Quote Originally Posted by Idea View Post
    The book says

    \int \frac{1}{5-4 \cos  x} \, dx =\frac{2}{3} \arctan (3 \tan  (x/2))

    The book also says

    \int \frac{1}{-5+4 \cos  x} \, dx =\frac{2}{3} \arctan (3 \tan  (x/2))

    This can't be right
    Same applies with the version I worked out:

    \int \frac{1}{5-4 \cos  x} \, dx =  \frac{2}{3}\arctan \left({\sqrt {\frac{5 + (-4)}{5 - (-4)}} \tan  (x/2)}\right) = \frac{2}{3}\arctan \left({\sqrt {\frac{1}{9}}  \tan  (x/2)}\right) = \frac{2}{3}\arctan \left({\frac{1}{3} \tan  (x/2)}\right)


    \int \frac{1}{-5 + 4 \cos  x} \, dx = \frac{2}{3}\arctan \left({\sqrt {\frac{-5 + 4}{-5 - 4}}  \tan  (x/2)}\right) = \frac{2}{3}\arctan \left({ \sqrt {\frac{-1}{-9}}  \tan  (x/2)}\right) = \frac{2}{3}\arctan \left({\frac{1}{3}  \tan  (x/2)}\right)

    So whatever the resolution of the fact that \int (-f) \, dx = \int f \, dx, it's unrelated to the wrongness / rightness of the integral.
    Last edited by Matt Westwood; August 31st 2014 at 11:22 AM.
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