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

 Method (no overloads)

Summary

The substitution u = sin(a) or u = cos(a) where an odd power of the
complement is left standing after the division by du/dx: that power is the
sign of the complement times a power of sqrt(1 - u^2), and the sign is a
constant on every interval between the zeros of the complement, so the integrand
is that constant times an algebraic function of u, which is integrated, and
the sign goes back in as sgn(cos(a)).

Remarks

Charlwood's ln(sin(x)) sqrt(1 + sin(x)) by parts leaves
-2 cos(x)^2/(sin(x) sqrt(1 + sin(x))), which under the sine is
-2 cos(x)/(u sqrt(1 + u)) du, a cosine over: it is -2 sgn(cos(x)) sqrt(1 - u)/u,
whose integral is -4 sqrt(1 - u) + 4 atanh(sqrt(1 - u)) times the sign. And
Charlwood's cos(x)^2/sqrt(1 + cos(x)^2 + cos(x)^4) under the cosine is
-u^2/(sin(x) sqrt(1 + u^2 + u^4)), which is -sgn(sin(x)) u^2/sqrt(1 - u^6) once the roots combine, and -sgn(sin(x)) arcsin(u^3)/3. Exact wherever the
complement is not zero, the generic case.
 A rule of its own and late, after the half-angle substitution: inside the general
substitution it answered 1/(1 + sin(x)) as a sign times a root where the
half-angle substitution gives the tangent of the half angle, and the remainder by
parts left beside that root was nine seconds of search for
ln(sin(x))/(1 + sin(x)), which the tangent answers in a moment. Only where
the complement is a factor of the product and what is left is algebraic in
u: inside a sum, 1/(cos(x) + sin(x)), or beside a logarithm of u,
the search with the sign in it ended nowhere.
 https://github.com/asc-community/AngouriMath/issues/718

























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