Formulation of Bioche's rules in familiar notation

In fact yes, the $dt$ is important in the Bioche's rules.

• 1. means $w(-t)=f(-t)d(-t)=w(t)=f(t)dt\\\implies f(-t)=-f(t)$

Then with the change $u=\cos x$

$f(x)dx=g(\cos x)(−\sin x dx)=g(u)du$

• 2. means $w(\pi-t)=f(\pi-t)d(\pi-t)=w(t)=f(t)dt\\\implies f(\pi-t)=-f(t)$

Then with the change $u=\sin x$

$f(x)dx=g(\sin x)(\cos x dx)=g(u)du$

• 3. means $w(\pi+t)=f(\pi+t)d(\pi+t)=w(t)=f(t)dt\\\implies f(\pi+t)=f(t)$

Then with the change $u=\tan x$

$f(x)dx=g(\tan x)(dx/\cos^2 x)=g(u)du$

You can see that performing the change $t\mapsto -t,\ t\mapsto \pi-t,\ t\mapsto \pi+t$ in the expression in yellow sees both $g$ and the $dx$ group in parenthesis unchanged, justifying this particular substitution.

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So 1. when the integrand behaves as $\sin$ do a substitution in $\cos$

So 2. when the integrand behaves as $\cos$ do a substitution in $\sin$

So 3. when the integrand behaves as $\tan$ do a substitution in $\tan$

In your case, none is working because you have either $\dfrac{-1}{1+\beta}$ or $\dfrac{1}{1-\beta}$ thus you need to fall back to a change in $\tan(\theta/2)$.

Covering the same ground as @zwim while adding a bit more information, for integrals of the form $$\int f(\sin x, \cos x) \, dx$$ where $f$ is a rational function of sine and cosine, I like to refer to $\omega (x) = f(\sin x, \cos x) \, dx$ as a differential form and it is this differential form that must remain invariant under one of the three substitutions: $x \mapsto -x, x \mapsto \pi - x, x \mapsto \pi + x$ if the rules of Bioche are to apply.

For the substitution where the differrntial form is invariant one sets $t = \phi (x)$, where $\phi (x)$ is the function $\cos x$, $\sin x$, or $\tan x$ that also remains invariant under the same substitution.

That is:

1. Set $t = \cos x$ when $x \mapsto -x$ leaves the differential form invariant since $\cos (-x) = \cos x$.

2. Set $t = \sin x$ when $x \mapsto \pi -x$ leaves the differential form invariant since $\sin (\pi -x) = \sin x$.

3. Set $t = \tan x$ when $x \mapsto \pi + x$ leaves the differential form invariant since $\tan (\pi + x) = \tan x$.

In the event that more than one of the initial substitutions leaves the differential form unchanged, the differential form will be unchanged under all three substitutions. In this case any one of the substitutions $t = \cos x$, $t = \sin x$, or $t = \tan x$ may be used but it is usually more efficient to:

4. Set $t = \cos 2x$ since in all three cases $x \mapsto -x, \pi - x$, and $\pi + x$ leave $\cos 2x$ unchanged.

Finally, if none of the initial three substitutions leave the differential form unchanged then as a last resort:

5. Set $t = \tan \frac{x}{2}$.

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Comment

At least in the English speaking world, the Bioche rules do not seem to be widely know. I guess one possible reason for this is an integral consisting of a rational function of sine and cosine can always be found using the rationalising substitution of $t = \tan \frac{x}{2}$. However, when used, one often ends up with an integral requiring cumbersome partial fraction decompositions which can often be avoided by applying the rules of Bioche in cases where it works.

Two written sources (in English) that refer to the Bioche's rules can be found in:

1. Handbook of Integration by Zwillinger on page 108 (though the name "Bioche's rules" is not used here).

2. How to Integrate It: A practical guide to finding elementary integrals by Stewart on pages 190$-$197.