Does Weierstrass substitution work for different arguments in the trig functions?

Question

Suppose we have a two dimensional rational function $R(x,y)$. Then we can use Weierstrass substitution for following integral: $$\int R(\sin(x), \cos(x)) \, dx .$$ But can we use Weierstrass substitution or something similar for following integral? $$\int R(\sin(\alpha x), \cos(x)) \, dx , \qquad \alpha \in \mathbb{R}$$ If $\alpha \in \mathbb{Z}$ we can use the double angle formulas to get to another rational function. But other than that for values like $\alpha = e$, for example, I don't see it happening.

Answer 1

If $\alpha$ is rational, say, $\alpha = \frac{m}{n}$ with $m, n \in \Bbb Z$, the substitution $x = n u$ yields $$\int R(\sin x, \sin \alpha x, \cos x, \cos \alpha x) dx = n \int R(\sin n u, \sin m u, \cos n u, \cos m u) \,du.$$ Using multiple angle and angle sum identities lets us rewrite the latter integral as $$\int S(\sin u, \cos u) \,du$$ for some rational function $S$, hence we can evaluate it using the tangent half-angle (a.k.a. Weierstrass) substitution.

This method does not apply when $\alpha$ is irrational, though some not-quite-trivial integrals with irrational $\alpha$ are still tractable using other techniques, e.g., a product-to-sum formula gives $$\int \sin e x \cos x \,dx = -\frac{\cos (e - 1) x}{2 (e - 1)} - \frac{\cos (e + 1) x}{2 (e + 1)} + C .$$ On the other hand, something like $\int \frac{\sin e x \,dx}{\cos x}$ probably doesn't have a closed-form expression in terms of elementary functions.