From: "James Van Buskirk" <torsop@ix.netcom.com>
Subject: Re: Integral of (cosX)^2 / (1+AsinX)^4 for A<1
Date: Mon, 8 May 2000 18:37:46 -0600
Newsgroups: sci.math,sci.math.symbolic
Summary: [missing]

Peter L. Montgomery wrote in message ...
>In article <391294D5.5D2ED036@ic.ac.uk> "Mr. SISAVATH Sourith"
<s.sisavath@ic.ac.uk> writes:


>>I am looking for the integral from 0 to 2pi of
>>(cosX)^2 / (1+AsinX)^4   dX    for A<1
>>
>>I am looking for the analytical solution of this equation.
>>I know this can be obtained using the residue theorem,
>>but its resolution requires a lot of calculations
>>and taking into account that I also need
>>
>>(cosX)^2 / (1+AsinX)^6   dX    for A<1


This is one of those integrals that just cries out for Kepler's
angle.  Let x = y+pi/2 in the first integral.  Then we get
Integral from 0 to 2*pi of sin(x)**2/(1+A*cos(x))**4 dx.
Using Kepler's angle:
sin(z) = sqrt(1-A**2)*sin(y)/(1+A*cos(y))
cos(z) = (A+cos(y))/(1+A*cos(y))
So that (1+A*cos(y))*(1-A*cos(z)) = 1-A**2
dz = sqrt(1-A**2)*sin(y)/(1+A*cos(y)) dy
This is transformed to:
Integral from 0 to 2*pi of sin(z)**2*(1-A*cos(z)) dz/(1-A**2)**(5/2)
and taking into account that the average value of sin(z)**2 is
1/2 and that of the term with an odd power of cos(z) is 0, we
get pi/(1-A**2)**(5/2).
The same approach transforms the second integral to:
Integral from 0 to 2*pi of sin(z)**2*(1-A*cos(z))**3 dz/(1-A**2)**(9/2)
and recalling that sin(z)*cos(z) = sin(2*z)/2, we get
pi*(1-3*A**2/4)/(1-A**2)**(9/2).