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SolveByExponentialHalfAngleAnsatz(AngouriMath.Entity,AngouriMath.Entity.Variable)

Method (no overloads)

Summary

e^(a x) times a rational function of sin(x) and cos(x), closed by
an ansatz in the half-angle tangent: F = e^(a x) P(t)/Q(t) with t = tan(x/2).

Remarks

Timofeev's e^x (1 - sin(x))/(1 - cos(x)) had no antiderivative, and it is
-e^x cot(x/2). Nothing here reads it: the exponential times a trigonometric
rule wants a polynomial in sine and cosine, the half-angle substitution wants no
exponential, and by parts goes round in a circle. Under t = tan(x/2) the
trigonometric part is a rational function r(t), and the integrand is
e^(a x) r(t) with dt/dx = (1 + t^2)/2; differentiating the ansatz,

F' = e^(a x) [a P/Q + (P'Q - P Q') (1 + t^2)/(2 Q^2)]

and asking that it equal e^(a x) n(t)/d(t) is one polynomial identity,
[a P Q + (P'Q - P Q')(1 + t^2)/2] d = n Q^2, linear in the coefficients of
P. Q is tried from d's written factors, as the exponential ansatz
tries its denominators; the identity is exact, so a solution is an answer and the
absence of one a decline. Timofeev's has r = (1 - t)^2/(2 t^2), Q = t,
P = -1.
https://github.com/asc-community/AngouriMath/issues/718

AngouriMath

Table of contents

AngouriMath

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

SolveByExponentialHalfAngleAnsatz(AngouriMath.Entity,AngouriMath.Entity.Variable)