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SolveByTrigonometricTowerAnsatz(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

A rational function of x and of sin(x) and cos(x) together, with
an exponential e^(a x) in front or not, closed by the ansatz
F = e^(a x) P/Q with P and Q polynomials in the three.

Remarks

Timofeev's x^2/(x cos(x) - sin(x))^2 is ((x sin(x) + cos(x))/(x cos(x) - sin(x)))' and (2x + sin(2x))/(cos(x) + x sin(x))^2 is (2x sin(x)/(x sin(x) + cos(x)))';
neither had an antiderivative. Nothing reads them: the half-angle substitution
wants no bare x, by parts goes round in a circle, and no subtree is a
substitution. They are the logarithm tower's ansatz with the sine and cosine for
the logarithm. The ring of polynomials in x, sin(x) and cos(x) is Q[x, s, c]/(s^2 + c^2 - 1), an integral domain with the basis
x^i c^k, x^i s c^k, and it is closed under the derivative --
(s c^k)' = c^(k+1) - k (1 - c^2) c^(k-1) -- so F' = N/D is the identity
(a P Q + P' Q - P Q') D = N Q^2 in that ring, one equation per basis element
and linear in the coefficients of P. Exact, so a solution is an answer and
its absence a decline; the derivative of what comes out is checked against the
integrand at sampled points all the same. Q is tried from the integrand's
written denominator as the other ansätze try theirs, the factors with their powers
lowered by one and then as they are. The half-angle tangent was tried for the
tower first, and every substitution of it leaves powers of 1 + t^2 above and
below that nothing cancels; the ring needs no substitution.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

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AngouriMath

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

SolveByTrigonometricTowerAnsatz(AngouriMath.Entity,AngouriMath.Entity.Variable)