Interval of definition of the solutions of $\dot x=e^x\sin x$ I'm trying to prove that the solutions of this ODE

$\dot x=e^x\sin x$

are defined in $\mathbb R$.
I'm really new on this subject, I'm trying to use the Picard Theorem, but this function is not locally Lipschitz, I need help here.
 A: Draw a phase diagram:


*

*If $x(0)=n\pi$ for some integer $n$, then $x(t)=n\pi$ for every $t\geqslant0$ hence $x$ is defined on $[0,\infty)$.

*If $x(0)=x_0$ with $2n\pi\lt x_0\lt(2n+2)\pi$ for some integer $n$, then $x(t)\to(2n+1)\pi$ when $t\to T$, where $[0,T)$ is the maximal "forward" interval of definition of the function $x$.
Furthermore,
$$
\int_{x_0}^{x(t)}\frac{\mathrm du}{\mathrm e^u\sin u}=t,
$$
hence the fact that $T=+\infty$ is equivalent to the divergence of the integral
$$
\int^{(2n+1)\pi}\frac{\mathrm du}{\mathrm e^u\sin u},
$$
which is obvious since $\mathrm e^u\to c\ne0$ and $\sin u\sim(2n+1)\pi-u$ when $u\to(2n+1)\pi$.


The same argument used "backwards" shows that $x$ is defined on $\mathbb R$.
A: Local existence and uniqueness
As Erick Wong noted in comments, the function $e^x\sin x$ is continuously differentiable and therefore locally Lipschitz. This ensures the local existence and uniqueness of solutions.
Stationary solutions
Some solutions are stationary: $x\equiv \pi n$, $n\in\mathbb Z$. Every non-stationary solution begins with initial value $x(t_0)=x_0$, where $\pi n < x_0<\pi(n+1)$ for some $n\in\mathbb Z$. By uniqueness, solution curves do not cross: therefore, $\pi n < x(t)<\pi(n+1)$ for all times.
Uniformity
Now that you know that the values of $x$ remain within interval $[\pi n,\pi(n+1)]$, you can give a uniform estimate for $M$ and $L$ in the statement of the Picard–Lindelöf theorem. For example, you can take a larger interval $I = [\pi n-1,\pi(n+1)+1]$ (so you have room for $b=1$)*, and then let $M=\sup_I |e^x\sin x|$ and $L=\sup_{I}|(e^x\sin x)'|$. The theorem then gives you $\epsilon>0$ (depending only on $n$) such that the solution of initial value problem $x(t_0)=x_0$ with $\pi n < x_0<\pi(n+1)$ exists in the  interval $(t_0-\epsilon, t_0+\epsilon)$.
Global existence
Now that you have a fixed $\epsilon$ independent of $t_0$, you can apply the Picard–Lindelöf theorem repeatedly to get global existence.
Remark
To take the suprema over an interval larger than $[\pi n,\pi(n+1)]$  is mathematically unnecessary: the behavior of the right hand side outside of $[\pi n,\pi(n+1)]$ is irrelevant. But it would take some effort to argue this point, while enlarging the interval takes no effort at all.
A: You seem to be having trouble with this. I am switching to an easier problem,
$$ y' = \cos y.   $$
Here the independent variable is $x,$ so you can draw the result in $xy$ graph paper. 
Now, there are horizontal lines at $$y = \frac{- \pi}{2}, \; y = \frac{ \pi}{2}, \;y = \frac{3 \pi}{2}, $$ because cosine is zero there and $y$ stays constant. 
The thing to expect is that solutions between constant solutions will start near one line and approach the other. And, indeed, for $ \frac{- \pi}{2} < y < \frac{ \pi}{2},$ each solution can be written
$$   y = \arcsin \tanh (x + C),  $$ for some constant $C.$ Such a curve resembles an $\arctan$  curve, and $y$ is exactly zero when $x = -C.$
For $ \frac{ \pi}{2} < y < \frac{3 \pi}{2},$ each solution can be written
$$   y =\pi - \arcsin \tanh (x + C).  $$ Here   $y$ is exactly $\pi$ when $x = -C.$ Also, these resemble an upside down arctan curve. 
This pattern continues for $y$ on successive intervals of height $\pi.$ 
