/*	Howell - starting with given expression : 
/$		∫[∂(θ),0 to 2*π: sin(θ)/(1 - β^2*sin(θ)^2)^(3/2)}  
/*	From "Howell - Background math for Lucas Universal Force, Chapter 4.odt"
	Recurring integrals from http://integral-table.com
/$		∫[∂(θ),0 to Of: sin(θ)^2)   
		=		O/2 - sin(2*O)/4 | from 0 to Of
		=  O ^2/2 - sin(2*O)/4 | from 0 to Of
		= (Of^2/2 - sin(2*Of)/4) - (0 - 0) 

/*	Take a look at (1 - b^2*sin^2(O))^(-3/2) 
	As a hunch, take the derivative 
/$		∂(∂(θ): (1 - β^2*sin(θ)^2)^( - 1/2)
		= (-1/2)*(1 - β^2*sin(θ)^2)^(-3/2)*∂(∂(θ): (1 - β^2*sin(θ)^2))  
		= (-1/2)*(1 - β^2*sin(θ)^2)^(-3/2)*(-1)*β^2*2*sin(O)*∂(∂(θ): sin(θ))  
		= (-1/2)*(-2)*β^2*sin(O)*cos(θ)*(1 - β^2*sin(θ)^2)^(-3/2)  
		=             β^2*sin(O)*cos(θ)*(1 - β^2*sin(θ)^2)^(-3/2)  

/*	27Jun2016  Now look at original integral
	As a hunch, take sin(O)*(1 - b^2*sin^2(O))^(-1/2)
/$         ∂[∂(θ): sin(O)*(1 - β^2*sin(θ)^2)^( - 1/2)] 
      =  ∂[∂(θ): sin(O)]*(1 - β^2*sin(θ)^2)^(-1/2)   + sin(O)*∂[∂(θ): (1 - β^2*sin(θ)^2)^( - 1/2)] 
      =          cos(O) *(1 - β^2*sin(θ)^2)^(-1/2)   + sin(O)*[ β^2*sin(O)  *cos(θ)*(1 - β^2*sin(θ)^2)^(-3/2) ] 
      =          cos(O) *(1 - β^2*sin(θ)^2)^(-1/2)   +          β^2*sin(O)^2*cos(θ)*(1 - β^2*sin(θ)^2)^(-3/2)  
      =   [      cos(O) *(1 - β^2*sin(θ)^2)^1   + β^2*sin(O)^2*cos(θ)  ]      *(1 - β^2*sin(θ)^2)^(-3/2)  
      =   [  cos(O)  - cos(O)*β^2*sin(θ)^2)^1   + β^2*sin(O)^2*cos(θ)  ]      *(1 - β^2*sin(θ)^2)^(-3/2)  
      =      cos(O)     *[1 - β^2*sin(θ)^2)^1   + β^2*sin(O)^2       ]      *(1 - β^2*sin(θ)^2)^(-3/2)  
      =      cos(O)     *(1 - β^2*sin(θ)^2)^(-3/2)  
/*  Nuts - cos instead of sin (50% chance of wrong, and I did wrong)
  Now try :  
/$         ∂[∂(θ): cos(O)*(1 - β^2*sin(θ)^2)^( - 1/2)] 
      =  ∂[∂(θ): cos(O)]*(1 - β^2*sin(θ)^2)^(-1/2)   + cos(O)*∂[∂(θ): (1 - β^2*sin(θ)^2)^( - 1/2)] 
      =         -sin(O) *(1 - β^2*sin(θ)^2)^(-1/2)   + cos(O)*[ β^2*sin(O)  *cos(θ)  *(1 - β^2*sin(θ)^2)^(-3/2) ] 
      =         -sin(O) *(1 - β^2*sin(θ)^2)^(-1/2)   +          β^2*sin(O)  *cos(θ)^2*(1 - β^2*sin(θ)^2)^(-3/2) 
      =  [      -sin(O) *(1 - β^2*sin(θ)^2)^1        + β^2*sin(O)*cos(θ)^2 ]  *(1 - β^2*sin(θ)^2)^(-3/2) 
      =          sin(O)*[-1 + β^2*sin(θ)^2)          + β^2       *cos(θ)^2 ]  *(1 - β^2*sin(θ)^2)^(-3/2) 
      =          sin(O)*[-1 + β^2*(sin(θ)^2) + cos(θ)^2) ]  *(1 - β^2*sin(θ)^2)^(-3/2) 
      =          sin(O)*[-1 + β^2                      ]  *(1 - β^2*sin(θ)^2)^(-3/2) 

/* Therefore (NOTE!  Lucas specifies dO, 0 to 2*π, but it must be 0 to π  ??!!??
/$     ∫[∂(θ),0 to π: sin(θ)/(1 - β^2*sin(θ)^2)^(3/2)}  
     =    cos(O) *(1 - β^2*sin(θ)^2)^(-1/2) / (-1 + β^2)  | from 0 to π
     =  { cos(π) *(1 - β^2*sin^2(π))^(-1/2) / (-1 + β^2) }  
      - { cos(0) *(1 - β^2*sin^2(0))^(-1/2) / (-1 + β^2) } 
     =  {   (-1) *(1 - β^2*0       )^(-1/2) / (-1 + β^2) }  
      - {   ( 1) *(1 - β^2*0       )^(-1/2) / (-1 + β^2) } 
     =       -2 /(-1 + β^2)  
     =        2 / (1 - β^2)  


/*++++++++++++++++++++++++++++++++++++++
/*add_eqn  "Lucas_Typo_or_omission   
   04_42
   Special integral with binomial series 
/$		(1 - β^2*sin(θ)^2)^(3/2)  

/$L		∫[∂(φ),0 to 2*π: ∫[∂(θ),0 to π: ET*(1 - λ(v))*r*sin(O)/(1 - β^2*sin(θ)^2)^(3/2)]} =  4*π*Q*(1 - λ(v))/(1 - β^2)  
		∫[∂(θ),0 to 2*π: sin(θ)/(1 - β^2*sin(θ)^2)^(3/2)) =  2/(1 - β^2)  
/*	and  
/$		E0(r)  =  q/r^2   
/*	so
/$L		λ(v) = β^2    

/$H		∫[∂(φ),0 to 2*π: ∫[∂(θ),0 to π: E0*(1 - λ(v))*r*sin(O)/(1 - β^2*sin(θ)^2)^(3/2)]*r} =  4*π*Q*(1 - λ(v))/(1 - β^2)  
		∫[∂(θ),0 to 2*π: sin(θ)/(1 - β^2*sin(θ)^2)^(3/2)) =  2/(1 - β^2)  
/*	and
/$		E0(r)  =  q/r^2   
/*	so
/$		λ(v) = β^2    

/*	OK - easy, but Lucas uses wrong 2*π rather than π upper limit on sinO/(1 - b^2*sin^2(O))^(3/2) integral, and uses ET rather than E0, and is missing an r!   
	HOWEVER : my 4-37 is missing the lambda term - so my earlier resilts dont work with this (maybe some other twisted argument will do for me?).