AngouriMath
SolveByTrigonometricPowerSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
An integrand that is a power of the sine times a power of the cosine, of one common
argument linear in the variable, turned into a polynomial by whichever of
u = cos , u = sin and u = tan the two exponents admit.
argument linear in the variable, turned into a polynomial by whichever of
Remarks
and reading it that way is what makes one rule out of a family Rubi spreads over
several sections:
are integers of either sign, so the same three cases below cover all of them.
Laurent polynomial — a sum of powers of
because that is what makes the rule closed. It asks the integrator nothing.
p odd and at least one:u = cos ,du = -sin dx , so one sine goes
intodu andsin^(p-1) is(1 - u^2)^((p-1)/2) . Leaves
-(1 - u^2)^((p-1)/2) u^q , andq may be anything.
q odd and at least one:u = sin , the mirror of it.
both even and p + q at most-2 :u = tan , under which
cos^2 = 1/(1 + u^2) anddx = du/(1 + u^2) , leaving
u^p (1 + u^2)^(-(p+q)/2 - 1) — a polynomial exactly whenp + q is at
most-2 , which is the condition.tan^2 sec^4 is this case and neither
of the others.
zero, where the tangent substitution leaves a negative power of
than a polynomial. Those are the ordinary
power-reduction rules already answer, so the boundary costs nothing. A power of the
secant or cosecant alone reaches SolveBySecantPowerReduction(AngouriMath.Entity,AngouriMath.Entity.Variable) first,
which gives a shorter answer for it.
#1265: a rule
that answers a sub-integral which used to come back unanswered lets the search that
asked for it carry on, and that cost lands on integrands the rule never fires on.
Angouri © 2019-2023 · Project's repo · Site's repo · Octicons · Transparency · 4378 pages online