Are there other good ways to look at this differential equation or is this it?

Question

\begin{align} \tan \frac \alpha 2 & = u \\[4pt] \alpha & = 2\arctan u \\[4pt] d\alpha & = \frac{2\,du}{1+u^2} \\[4pt] \sin\alpha & = \sin(2\arctan u) = 2(\sin\arctan u) (\cos \arctan u) \\[4pt] & = 2\,\frac u {\sqrt{1+u^2}} \cdot \frac 1 {\sqrt{1+u^2}} = \frac{2u}{1+u^2} \end{align} $$ \frac{\hspace{6cm}}{} \qquad \S \qquad \frac{\hspace{6cm}}{} $$

The well known substitution above is for the moment my preferred approach to solving the differential equation $$ \frac{d\alpha}{\sin\alpha} = \frac{d\beta}{\sin\beta}. \tag{differential equation} $$ One gets $$ \frac{d\alpha}{\sin\alpha} = \frac{\left( \dfrac{2\,du}{1+u^2} \right)}{\left( \dfrac{2u}{1+u^2} \right)} = \frac{du} u $$ and writing $v = \tan \dfrac\beta 2$, we then have $$ \frac{du} u = \frac {dv} v $$ so that $\log u = \log v + \text{constant}$ and so $u = v\times \text{constant}$, and finally $$ \tan \frac\alpha 2 = \text{constant} \times \tan \frac\beta 2. \tag{solution} $$ So my question is whether there are other ways to approach this that are either better or otherwise worth some attention.

Answer 1

From the first time I heard about Weierstrss substitution, it also became my favored way for solving things such as $$\int \frac{dx}{a+b \sin(x)+c\cos(x)}=-\frac{2 }{\sqrt{b^2+c^2-a^2}}\tanh ^{-1}\left(\frac{(a-c) \tan \left(\frac{x}{2}\right)+b}{\sqrt{b^2+c^2-a^2}}\right)$$

For things such as $$I=\int \frac{dx}{\sin(x)}=2i\int \frac{dx}{e^{ix}-e^{-ix}}$$ the change of variable $$e^{ix}=t \implies x=-i \log(t)\implies dx=-i\frac{dt} t$$ is also interesting since it makes $$I=2\int \frac {dt}{t^2-1}=\log\left(\frac{1-t}{1+t} \right)= -2 \tanh ^{-1}(t)$$