AngouriMath
SolveTwoHalfOddPowersOfSecantSumsByTheHalfAngleSine(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A half-odd power of a ± a sec(y) beside one of c + d sec(y) , integrated in
the half angle's sine, or its cosine for the minus:a + a sec(y) is
2a cos(y/2)^2/cos(y) and c + d sec(y) is (c cos(y) + d)/cos(y) , the two
powers ofcos(y) make a whole one, and in w = sin(y/2) , where
cos(y) = 1 - 2w^2 and dy = 2 dw/cos(y/2) , what is left is rational beside
the one root ofc + d - 2c w^2 . Either sum may be in the cosine instead, and the
cosecant's, with the sine, are the same by the complement.
the half angle's sine, or its cosine for the minus:
powers of
the one root of
cosecant's, with the sine, are the same by the complement.
Remarks
Rubi's sec(e + f x) sqrt(a + a sec(e + f x))/sqrt(c + d sec(e + f x)) and eight more
of 4.5.2.1 and 4.5.2.3 were declined: the half-angle tangent reads a root of the one sum
and not of the other. The constants are not written: the answer is the integrand times the
antiderivative inw over what that differentiates back to, a quotient whose square
is one; and it is kept only where that square is one at the sampled points.
https://github.com/asc-community/AngouriMath/issues/718
of 4.5.2.1 and 4.5.2.3 were declined: the half-angle tangent reads a root of the one sum
and not of the other. The constants are not written: the answer is the integrand times the
antiderivative in
is one; and it is kept only where that square is one at the sampled points.
https://github.com/asc-community/AngouriMath/issues/718
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