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

 Method (no overloads)

Summary

An integrand holding an inverse trigonometric function of the variable itself,
integrated by the substitution that undoes it: x = sin(u) for arcsin(x),
x = tan(u) for arctan(x), x = cos(u) and x = sec(u) for the
other two.

Remarks

e^(arcsin(x)) x^3/sqrt(1 - x^2) had no antiderivative. Under x = sin(u) it
is e^u sin(u)^3, which the product-to-sum rewrite and the closed
exponential-times-trigonometric rule answer between them; x arcsec(x)/sqrt(x^2 - 1) is u sec(u)^2 under x = sec(u), one step of parts. The general
substitution does not find these because it substitutes for a subtree and asks what
is left, and what is left here is x itself, which only the inverse substitution removes -- the same reason SolveByLogarithmSubstitution(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean) exists beside it.
The radical goes by construction.sqrt(1 - x^2) under x = sin(u) is cos(u) on the principal branch, where u lies in [-pi/2, pi/2] and the cosine is not negative; substituting and simplifying instead would leave
sqrt(1 - sin(u)^2), which the simplifier is right not to call cos(u).
So (1 - x^2)^(k/2) becomes cos(u)^k directly, and likewise
(1 + x^2)^(k/2) into sec(u)^k for the tangent, each on its principal
branch. That branch is the one the inverse function is defined on, so the answer holds
wherever the integrand is read through it. For the secant (x^2 - 1)^(k/2) is
|tan(u)|^k, and on the arcsecant's range the tangent has the sign of the
argument, so it becomes tan(u)^k times that sign to the k, as the
cosecant's and the cotangent's roots carry theirs.
 Exactly one inverse function, of the variable or of a linear c x + d in it,
and nothing left in x afterwards; otherwise declined. Handed to the
integrator in u, where the trigonometric rules are. With the linear,
x = (sin(u) - d)/c and dx = cos(u) du/c, and the radical that goes
is of a constant multiple of 1 - (c x + d)^2 -- Rubi's
(d - c^2 d x^2)^p (a + b arcsin(c x))^n -- taken as that multiple's root
times the cosine, the generic case. A power of a + b arcsin(c x) counts
as a power of the inverse: x arcsin(a x)^2 under the sine is
u^2 sin(u) cos(u)/a^2, parts twice against sin(2u), where parts in
x stalled on x^2 arcsin(a x)/sqrt(1 - a^2 x^2).
 https://github.com/asc-community/AngouriMath/issues/718

























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