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SolveSymbolicPowersOfOnePlusMinusASineThroughTheSine(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)

Method (no overloads)

Summary

Powers of a ± a sin(y) with a symbolic exponent, or beside a fractional power of
g cos(y), and anything else in the sine alone, under u = sin(y):
(1 + sin(y))(1 - sin(y)) is cos(y)^2, so each factor is a power of
1 + u or of 1 - u up to a factor constant on each interval where it is
defined, and dy = du/cos(y) is one more pair of half powers. The cosine's the
same way, by u = cos(y) beside a power of g sin(y).

Remarks

Rubi's (a + b sin)^m (c + d sin)^n files with a^2 = b^2 and
c^2 = d^2 hold about a hundred and twenty problems with a symbolic exponent,
(a + a sin(e + f x))^m sqrt(c - c sin(e + f x)) the plainest, and none was
answered: the half angle at which 1 ± sin(y) is a square, the rule before this
one, wants numeric powers, and the substitution search spent the budget on them. What
is asked in u is (1 + u)^A (1 - u)^B R(u), which is elementary where one
of A, B and A + B is a whole number and the rest is a
polynomial, and is declined where it is not.
The answer is K G(sin(y))/F, for G the antiderivative in u,
F the slope of y and K the powers as written over the form they
were rewritten to and over cos(y): a quotient whose logarithmic derivative is
zero, so the answer holds whatever a, g and the exponents are --
(a (1 + u))^m is not a^m (1 + u)^m for every a, and the quotient
of the two is constant on each interval either way. At the question asked only, and
only where an exponent is symbolic: numeric half powers are the half angle's.
But for a fractional power of the cosine, which the half angle does not read: Rubi's
(e cos(c + d x))^p (a + a sin(c + d x))^m with half-odd p and m were past the budget, the substitution search spending it on the two radicals. In
u they are (1 + u)^A (1 - u)^B with A and B quarter-odd, which
by Chebyshev's theorem is elementary only where A + B = m + p - 1 is whole, and
with every power a number the rule asks only where A, B or A + B is
whole and the rest is a polynomial in u: (a + a sin(y)) sqrt(e cos(y)), which
is elliptic, is declined before anything is asked. Numeric powers are read one level
below the question too, where the polynomial term hands 1/(A B) on as
(A B)^(-1), and a whole power of a product is read as the powers of its factors.
https://github.com/asc-community/AngouriMath/issues/718

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AngouriMath

SolveSymbolicPowersOfOnePlusMinusASineThroughTheSine​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable,​System.​Boolean)

SolveSymbolicPowersOfOnePlusMinusASineThroughTheSine(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)