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

 Method (no overloads)

Summary

Bioche's first two rules: a rational function of sin(x) and cos(x) that is odd in the sine is a rational function of u = cos(x) times
sin(x) dx = -du, and one odd in the cosine the mirror of it; with radicals
of polynomials in the two even in the function admitted as coefficients.

Remarks

Bondarenko's 1/(cos(x) + cos(3x))^5 is 1/(4 cos(x)^3 - 2 cos(x))^5, odd in
the cosine, and under u = sin(x) it is 1/((1 - u^2)^3 (2 - 4u^2)^5), a
rational function with a denominator of degree sixteen; the half-angle
substitution, which answers any rational function of the two, makes one of
degree thirty in the tangent of the half angle and did not return within the
budget. The parity is read off the polynomials: with N/D the integrand over
the two and f the function it is odd in, N/(D f) is even in f exactly when, after clearing D against D(-f) where D is neither
even nor odd, every power of f above is odd and every one below is even --
and then each f^2 is 1 - u^2. Exact, and the substitution is a
bijection on each interval between the zeros of f, the standing caveat on
every trigonometric substitution here.
 In front of the half-angle substitution, for the smaller rational function and
the shorter answer -- sin(x)/(1 + cos(x)^2) is -arctan(cos(x)) here and
a page in the half-angle tangent -- and behind everything that answers a power
product or a homogeneous quotient in its own terms.
 https://github.com/asc-community/AngouriMath/issues/718

























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