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

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Zdislav V. Kovarik

Posts: 3,419
Registered: 12/6/04
Re: integration cube root sin x
Posted: Jul 22, 2001 2:42 AM
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We can narrow it down to non-trivial elliptic integrals (known to be

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 non-elementary.

Cheers, ZVK(Slavek).

In article <>,
Doug Magnoli <> 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!
:-Doug Magnoli
:Tim 9-23 wrote:
:> Can this be expressed in elementary functions? I doubt it.
:> My email address is anti-spammed. If you want to email me,
:> remove the 2 B's after hitting email reply.
:> Tim 9-23

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