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

 Method (no overloads)

Summary

(a + b cos(y) + c sin(y))^n for a^2 = b^2 + c^2, y linear in
x and n half-odd or a negative whole number, in closed form.

Remarks

b cos(y) + c sin(y) is R cos(y - phi) for R = sqrt(b^2 + c^2), so
the base is a (1 ± cos(y - phi)), a square of the half angle as
a + a sin(y) is, and SolveAHalfPowerOfOnePlusASine(AngouriMath.Entity,AngouriMath.Entity.Variable)'s identity
holds for it. Rubi's sqrt(5 + 4 cos(x) + 3 sin(x)) and the rest of its 4.7.7
with a^2 = b^2 + c^2 were declined or past the budget. The answer needs no
phi: with S the base and T = b sin(y) - c cos(y),
T' = S - a, S' = -T and T^2 = S (2a - S), the last of which is
a^2 = b^2 + c^2, and from them
d/dy (T S^(n-1)) = n S^n - a (2n - 1) S^(n-1)
which steps n down to 1/2, where the integral is 2T/sqrt(S), or
up to -1/2 or -1, where it is
(2/sqrt(2a)) atanh(T/(sqrt(2a) sqrt(S))) and T/(a S). Each base is
checked by differentiating it with those three identities alone.
 For a real a of either sign: the derivations use only
S^(3/2) = S sqrt(S) and (sqrt(2a) sqrt(S))^2 = 2a S, which hold for
the principal powers, and the argument of the hyperbolic arctangent stays between
-1 and 1, one less its square being S/(2a), which is not negative
since S has the sign of a. An antiderivative on every interval between
the zeros of S, where T changes sign. a^2 = b^2 + c^2 is decided,
not assumed, and the cosine and the sine are both there: one alone is
SolveAHalfPowerOfOnePlusASine(AngouriMath.Entity,AngouriMath.Entity.Variable)'s.
https://github.com/asc-community/AngouriMath/issues/718

























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