AngouriMath
SolveASecantPowerOverAPowerOfACosineAndASineInPowersOfABinomialInTheTangent(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
A power of a secant over a whole power of a cos(y) + b sin(y) , the two powers
together odd,sec(y)^m/(a cos(y) + b sin(y))^n , in powers of the binomial
w = a + b tan(y) : the quotient is sec(y) (1 + tan(y)^2)^k/w^n with
2 k + 1 = m + n , and with tan(y) = (w - a)/b the power
(1 + tan(y)^2)^k = ((w - a)^2 + b^2)^k/b^(2 k) = sum d_j w^j . A term below the
bar,sec(y)/w^i , is cos(y)^(i - 1)/(a cos(y) + b sin(y))^i , and one above
it a polynomial in the tangent beside the secant; each is its own question.
together odd,
bar,
it a polynomial in the tangent beside the secant; each is its own question.
Remarks
Where m + n is even the quotient is a rational function of the tangent beside
sec(y)^2 , which the substitution answers; odd, it took the half-angle, and with
symbols fora and b Rubi's 4.7.2 sec(c + d x)^4/(a cos + b sin)^3 and sec^3/(a cos + b sin)^4 ran out of time, and sec^6/(a cos + b sin) ,
answered, was 5.6 thousand characters. Fromm + n = 7 , where the half-angle
stops answering in time.
https://github.com/asc-community/AngouriMath/issues/233
symbols for
answered, was 5.6 thousand characters. From
stops answering in time.
https://github.com/asc-community/AngouriMath/issues/233
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