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

Method (no overloads)

Summary

Fractional powers of a ± a sin(y), or of a ± a cos(y), beside anything
rational in the sine and cosine of y, by the half angle at which they are
squares: 1 + sin(y) is 2 sin(u)^2 and 1 - sin(y) is
2 cos(u)^2 for u = y/2 + pi/4, so (a + a sin(y))^(3/2) is
a^(3/2) 2^(3/2) sgn(sin(u)) sin(u)^3, the sign a constant between the zeros of
the sine that comes out in front of the integral, and what is left is rational in
sin(u) and cos(u) -- sin(y) being 2 sin(u)^2 - 1 and
cos(y) being 2 sin(u) cos(u). For the cosine, u = y/2, with
1 + cos(y) = 2 cos(u)^2 and 1 - cos(y) = 2 sin(u)^2.

Remarks

Rubi's (a + b sin)^m (c + d sin)^n (A + B sin + C sin^2) files with
a^2 = b^2 and c^2 = d^2 -- (A + C sin^2)/((c - c sin)^(3/2) sqrt(a + a sin)) and its kin -- were timeouts: the substitution search has no radical of a
trigonometric function it can rationalise, and SolveAHalfPowerOfOnePlusASine(AngouriMath.Entity,AngouriMath.Entity.Variable) reads one such power alone. This is that rule's identity applied to the whole
integrand at once.
Exact, for any a: 1 + sin(y) is not negative, so (a q)^p = a^p q^p for the principal powers whatever a is, and (2 sin(u)^2)^p is
2^p |sin(u)|^(2p). a^2 = b^2 is decided, not assumed. Whole products
2p only, so that the sign is a power of a sign and what is handed on is a whole
power of the sine; at the top only, as every rule that writes a sign for a function.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

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AngouriMath

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

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