Out of curiosity, I was recently dabbling in exponentials of operators, and found out that they often show up in connection with time evolution. So I made up a simple ODE to test it: $$\frac{\mathrm d x}{\mathrm d t} = \sin x.$$
On one hand, we can separate the variables and integrate: $$ \mathrm d t = \frac{\mathrm d x}{\sin x} \implies t = C + \ln \tan x/2. $$ Let's say that $x(t = 0) = \pi/2$ to make it easy. Then $C = 0$ and $$ x(t) = 2 \arctan e^t. $$
On the other hand, I wanted to try out the exponential approach on it. We know that $x(0) = \pi/2$. Suppose that we advance the time by an infinitesimal amount $\delta t$. Then $x(\delta t) \approx x(0) + \delta t \frac{\mathrm d x}{\mathrm d t} = x(0) + \delta t \sin x.$ So the operator $\sin$ looks like a generator of the time evolution. If we exponentiate it, we get an operator that performs a finite time shift ($e^{t \sin}$ should advance the time by $t$), and using that, we get $$ x(t) = e^{t \sin} x(0) = e^{t \sin} \frac{\pi}{2}. $$
Writing out the exponential using the infinite series, I get this series of infinitely nested sines: $$ x(t) = \frac{\pi}{2} + t + \frac{1}{2!} t \sin t + \frac{1}{3!} t \sin(t \sin t) + \frac{1}{4!} t \sin(t \sin(t \sin t)) + \ldots $$
My problem is that after calculating both solutions numerically for $t = 1$, the first approach gives $2 \arctan e \approx 2.43657$, while the infinite series of sines (converges very fast but) gives totally different value of $\approx 3.15024$.
So, somehow the exponential approach didn't work. I'm guessing that I probably violated some condition that needs to be satisfied and that I don't know about (I'm a physics student and I learnt about the exponentials in the QM, where they were covered only very briefly.) I would be grateful if you could point out the mistake and tell me if this approach is just doomed to fail or if it can be mended in some way to do what I think it should do.