Differential equation of inclined plane I'm having some trouble with the equation $$\frac{d}{dt}\dot{x}=g\sin\Theta \implies \dot{x}(t)=\dot{x}(t=0)+\int_0^t dt'\:g\sin\Theta=\dot{x_0}+g\:t\sin\Theta $$
which appears in page 4 of http://www.astro.caltech.edu/~golwala/ph106ab/ph106ab_notes.pdf. Because of the appearence I know it is not that difficult, but my basis on calculus is weak. Anyway, what exactly he did? And what is the general principle which supports it?
I tried to do the following: $$d\dot{x}=g\sin\Theta\:dt\: \implies \int d\dot{x}=\int g\sin\Theta\:dt \: \implies \dot{x}=g\sin\Theta\:t$$ but I don't know how he got $\dot{x_0}$.
Have a nice day. 
P.S.: He uses $\dot{x_0}$ for the initial velocity

This is just the application of the fundamental theorem of calculus.
That is, for a continuous function $f(x) = \frac{d}{dx}F(x)$ on an interval $(a,b)$, $$\int_a^b f(x)\,dx = F(b)-F(a).$$
For this example, start by integrating both sides from $0$ to $t$
$$\int_0^t \frac{d}{dt} \dot{x}\,dt =\int_0^t g\sin \Theta\, dt.$$
You have already worked out the integral on the right, so applying the fundamental theorem of calculus on both sides gives
$$\dot{x}(t) - \dot{x}(0) = gt\sin\Theta,$$
which leads to the result.
