Question: I am unable to solve this nonlinear differential equation:
$$\dot{x}(t)^2+x(t)=\sin(x)^2$$
I tried with Maple without success.
Is it possible to solve it without the use of numerical methods? Thanks.

Answer: 
If $x(t)$ is of class $C^2$ you can differentiate the equation to get $$2\dot{x} \ddot{x}+\dot{x}=\sin(2x) \dot{x} $$

$\dot{x}=0$ gives constant solutions $x(t)=x\_0$, which by the original ODE must satisfy $x\_0=\sin(x\_0)^2$, thus reducing to $x\_0=0$ alone.

Other solutions satisfy $$2 \ddot{x}=\sin{2x}-1 $$ 

and this equation seems to be within the capability of Maple (an implicit solution can be obtained).

Another approach could be separating the variables: $$\dot{x}^2=\sin^2 x-x \\\dot{x}= \pm \sqrt{\sin^2{x}-x} \\ \frac{dx}{\sqrt{\sin^2 x-x}}=\pm dt \\ \int \frac{dx}{\sqrt{\sin^2 x-x}}=\pm(t-t\_0)$$
Which also gives an implicit solution.
Note that $\sin^2x -x\geqslant0$ if and only if $x\leqslant0$ hence one can rewrite this as $\dot x=\varepsilon u(x)$ on every interval where $x\lt0$, with $u:\xi\mapsto\sqrt{\sin^2\xi-\xi}$ and $\varepsilon=\pm1$. Thus, assuming that $x(0)=x\_0$ with $x\_0\lt0$,
$$
\int\_{x\_0}^{x(t)}\frac{\mathrm d\xi}{u(\xi)}=\varepsilon t,
$$
for every $t$ such that the LHS yields $x(t)\lt0$. This defines implicitely $x(t)$ for every $t$ such that $x(t)$ exists.
Two cases arise. If $\varepsilon=+1$, then $x$ is increasing and $x(t)\to0$ when $t\to\tau(x(0))$, where, for every $y\lt0$,
$$
\tau(y)=\int\_{y}^{0}\frac{\mathrm d\xi}{u(\xi)},
$$
thus, $x$ is defined on $[0,\tau(x\_0))$. If $\varepsilon=-1$, then $x$ is decreasing and $x(t)\to-\infty$ when $t\to+\infty$ because the integral
$$
\int\frac{\mathrm d\xi}{u(\xi)}
$$
diverges at $-\infty$. Both cases are actually the same, since the maximal interval of definition of $x$ is $(-\infty,\tau(x\_0))$ when $\varepsilon=+1$ and $(-\tau(x\_0),+\infty)$ when $\varepsilon=-1$. Thus, one follows the same and unique solution, starting at some point given by $x\_0$, either in the direction where the solution stops to be defined after a finite time or in the direction where it is defined forever. 

Said differently, there exists a unique increasing function $\gamma:(-\infty,0)\to(-\infty,0)$ such that $\gamma(0-)=0-$ and $\gamma(-\infty)=-\infty$, and, for every $x\_0\lt0$ and every $t$ such that $x(t)$ exists, $$x(t)=\gamma(\varepsilon t+\gamma^{-1}(x\_0)),
$$
either with $\varepsilon=+1$ or $\varepsilon=-1$. And an explicit formula for $\gamma$ is
$$
\int\_{\gamma(t)}^0\frac{\mathrm d\xi}{u(\xi)}=-t.
$$