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

 Method (no overloads)

Summary

tan(y) tan(2y) written sec(2y) - 1, and the integrand asked again in the
one argument 2y.

Remarks

tan(y) is (1 - cos(2y))/sin(2y), so tan(y) tan(2y) is
(1 - cos(2y))/cos(2y), which is sec(2y) - 1 wherever both sides are
defined; at the zeros of cos(y) the left is -2 as a limit, which the right
is. Rubi's 4.7.7 has sec(2(a + b x))^k sqrt(c tan(a + b x) tan(2(a + b x))) and
its kin, a half-odd power of c sec(2y) - c beside powers of the secant or cosine
of 2y, which SolveByTheHalfAngleTangentBesideAHalfOddPowerOfOnePlusASecant(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) answers in that one argument; with two arguments nothing read them.
 The same question in another spelling, so asked as the same question. The answer is
what that rule gives, exact where the integrand is real.
https://github.com/asc-community/AngouriMath/issues/718

























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