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Topic:
integration cube root sin x
Replies:
5
Last Post:
Feb 28, 2013 5:09 AM




Re: integration cube root sin x
Posted:
Jul 22, 2001 2:42 AM


We can narrow it down to nontrivial elliptic integrals (known to be nonelementary):
In Int [(sin x)^(1/3) dx] , substitute
x = arcsin(v^(3/2))
and obtain an ugly elliptic integral
(3/2) * Int [ v * (1  v^3)^(1/2) dv]
or, one step further, substitute
x = arcsin((1  w^2)^(3/2))
and get a neater elliptic integral
3 * Int [ (w^2  1) * (3  3 * w^2 + w^4)^(1/2) dw]
The rest can be done by symbolic software.
But the message is: unlike the use of hypergeometric functions, which are likely to end up being elementary for some values of the parameters, this approach proves that the given integral is indeed nonelementary.
Cheers, ZVK(Slavek).
In article <3B5A65D1.ADFEE2DE@home.com>, Doug Magnoli <dmagnoli@home.com> wrote: :And BTW, Mathematica gives the indefinite integral as: : :Int [(sin x)^(1/3) dx] :=  cos(x)*HG2F1[1/2,1/3,3/2,(cos x)^2]*(sin x)^(4/3) / (sin x)^(4/3), : :where HG2F1 is Hypergeometric2F1, defined as: : :HG2F1(a,b;c;z) = sum(k=0,inf) (a_k*b_k / c_k) * z^k / k! : :hth, : :Doug Magnoli : : :Tim 923 wrote: : :> Can this be expressed in elementary functions? I doubt it. :> :> My email address is antispammed. If you want to email me, :> remove the 2 B's after hitting email reply. :> :> Tim 923 :



