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

 Method (no overloads)

Summary

A sum of powers of one sine, cosine, secant or cosecant of a linear u, the powers
numbers that are not whole and that differ by whole multiples of two, by the reduction
formula: with u' = 1, (cos sin^(n - 1))' = (n - 1) sin^(n - 2) - n sin^n,
(sin cos^(n - 1))' = n cos^n - (n - 1) cos^(n - 2),
(sin sec^(n - 1))' = (n - 1) sec^n - (n - 2) sec^(n - 2) and
(cos csc^(n - 1))' = (n - 2) csc^(n - 2) - (n - 1) csc^n. Each power is written as
such a derivative and the power two below it, from the highest down, and the sum is
answered where nothing is left of the lowest. Beside x, by parts once: the sum is
Q' for Q a sum of cos sin^(n - 1), whose integral is
sin^n/n, so x Q' is (x Q)' - Q, closed.

Remarks

A power that is not whole is elliptic on its own, and so is each term of Rubi's 4.5.4.1
(3 + 3 sec(c + d x)^2)/sqrt(sec(c + d x)) and of 4.1.10
x/csc(e + f x)^(3/2) - x sqrt(csc(e + f x))/3: split, both were declined.
SolveASymbolicPowerOfATrigonometricFunctionAsTheDerivativeOfAProduct(AngouriMath.Entity,AngouriMath.Entity.Variable) reads a
symbolic exponent and two terms. The identities are the chain rule for the principal
power, which holds wherever the function is not zero, on either side of it, and the
written function is kept: a cosecant's answer is in the cosecant. Not through the
integrator for Q: it wrote sin(x)/sqrt(sec(x))'s integral as a power of
the cosine, which differs from the secant's where the cosine is negative. At most twelve
steps, and x to the first power only, since beside x^2 the remainder x Q is
elliptic again.
https://github.com/asc-community/AngouriMath/issues/233

























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