SolveAnEvenPowerOfACosineOverAPowerOfABinomialInTheSineByReduction​(AngouriMath.​Entity,​AngouriMath.​Entity.​Variable)

An even power of a cosine over a whole power of a binomial in the sine of the same
linear, cos(y)^p/(a + b sin(y))^n, or of a sine over one in the cosine, with
p of either sign -- a power of the secant is one below the bar -- by the
reduction formulas in the power of the binomial. With u = a + b sin(y),
b^2 cos(y)^2 = (b^2 - a^2) + 2 a u - u^2 lowers a power of the cosine above the
bar, and read for a^2 - b^2, (a^2 - b^2) = 2 a u - u^2 - b^2 cos(y)^2 lowers one below it; what is left is I_n, the integral of 1/u^n, and
(n - 1)(a^2 - b^2) I_n = b cos(y) u^(1 - n) + (2 n - 3) a I_(n - 1) - (n - 2) I_(n - 2) takes it down to I_1, the one integral asked of the rest of the integrator
besides the powers of the sine and the secant the reductions end on. For the cosine,
b sin(y) is -b sin(y) in the last.