AngouriMath
SolveAsTheDerivativeOfASineRaisedFromNothingBesidePowersOfTheCosine(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
The derivative of sin(u) prod B_i^(e_i) , for B_i the powers of cosines,
secants and sums of them in the integrand, or ofcos(u) prod B_i^(e_i) beside sines
and cosecants: the sine raised from nothing, and eache_i from two below the
integrand's exponent to one above it. Rubi's(a + a cos(u))^(3/2)/cos(u)^(5/4) is
(4 a^2 sin(u)/(cos(u)^(1/4) sqrt(a + a cos(u))))' . Beside x, by parts once:
x (b + a sin(x))/(a + b sin(x))^2 is x G' for G = -cos(x)/(a + b sin(x)) ,
sox G - integral G , and the integral of a cosine beside a function of the sine is
closed.
secants and sums of them in the integrand, or of
and cosecants: the sine raised from nothing, and each
integrand's exponent to one above it. Rubi's
so
closed.
Remarks
over
rational function of the sine and the cosine, constant where the integrand is the
derivative. Decided at sampled points for each candidate, every symbol pinned, and then
the constant read at the angle whose cosine is 3/5 and sine 4/5, where a rational
function of the two that is constant has the value it has everywhere; the answer is
checked by differentiating it back.
the second integrated
whose integral ran past the budget. Only
where the integrand has no power of the sine itself (or of the cosine, beside sines),
which is what the product-of-powers rule and parts already read, and without x only
for two powers or more, one of them not whole:
angle's, and a power of the cosine alone the reduction's. Of the sine and the cosine
only, not the tangent: a product of the two and their reciprocals is the substitution's.
https://github.com/asc-community/AngouriMath/issues/233
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