Let $x_0 =0$ and $y_0 \in (0, \pi)$
$$y' = \sin(y) \tag 1 $$
and: $$ y' = x \sin(y) \tag 2 $$
The question is to find the solution of $(1)$ and $(2)$.
I have already shown that $\int \frac 1 {\sin(x)} \, dx = \ln|\tan(\frac x 2 )|$
I have to use the theorem that says , let $\varphi(x)$ the solution of (1) then $ H(\varphi(x)) = F(x) $ with $F(x) = \int_{x_0}^x f(t) \, dt $
$$ H(y)= \int_{y_0}^y \frac 1 {g(s)} \, ds $$
In my case, it is clear that $g(y) = \sin(y)$ for both $(1)$ and $(2)$
$\textbf{My Try}~(1)$
$$ H(y)= \int_{y_0}^y \frac 1 {g(s)} \, ds = \int_{y_0}^y \frac 1 {\sin(s)} \, ds = \ln\left|\tan\left(\frac y 2\right)\right| - \ln \left|\tan\left(\frac{y_0}2\right)\right|$$
$$F(x) = \int_{x_0}^x f(t) \, dt = \int_{x_0}^x 1 \, dt = x-x_0= x$$
Now I must search for the inverse function of $H$:
$$ H^{-1} : H(D) \to D $$
$$H(y) =w \Rightarrow \ln\left|\tan \left(\frac y 2 \right)\right| - \ln\left|\tan\left( \frac{y_0}2 \right)\right| = w$$
$$\ln\left|\tan \left( \frac y 2 \right)\right| = w + \ln\left|\tan\left( \frac {y_0} 2 \right)\right|$$
$$\left|\tan \left(\frac y 2 \right)\right| = \exp \left(w+\ln\left|\tan \left( \frac{y_0} 2 \right) \right|\right)$$
$$\tan\left(\frac y 2\right) = \exp\left(w+\ln\left(\tan\left(\frac{y_0} 2 \right) \right)\right) \text{ because } y \in (0,\pi)$$
$$ y = 2\tan^{-1} \left(\exp\left(w+\ln\left(\tan\left( \frac{y_0} 2 \right) \right) \right) \right)$$
$$H^{-1}(w) =2\tan^{-1}\left(\exp\left(w+\ln\left(\tan\left(\frac{y_0} 2\right)\right)\right)\right)$$
$$ H(\varphi(x)) = F(x) \Rightarrow \varphi(x)= H^{-1}(F(x))$$
$$\varphi(x)= 2\tan^{-1}\left(\exp\left(x+\ln\left(\tan\left( \frac{y_0} 2 \right)\right)\right)\right)$$
Is this correct ?