What exactly is meant by 'Integrate the Equation of Motion'? Short Question: if a question says to 'integrate the equation of motion', what does it mean?  
Long Question: 
Question: Take a planet of mass $M$ and place a satellite at rest at a distance $R$ from the planet, where $R$ is much greater than the planet's radius.  How long does the satellite take to hit the surface of the planet?  
Part 1 of the question asks the reader to perform dimensional analysis.  This yields
$$\textrm{Time taken }T=C\sqrt{\frac{R^3}{GM}}$$
Part 2 - integrate the equation of motion of the satellite to show that $C=\pi /2\sqrt{2}$.  
As far as I'm aware, the equation of motion for the satellite is 
$$\ddot{r}=-\frac{GM}{r^2}.$$
I've tried solving this differential equation, but to no avail ($r=\frac{9}{2}GMt^{\frac{2}{3}}$ is a particular solution, but I have no idea how to find the more general case; substituting the dimensionless quantity $\kappa=\frac{1}{GM}r^3t^{-2}$ almost worked but not quite).  I also tried using the potential $V=\frac{-GMm}{r}$ to form the equation
$$T=\int_R^0{\frac{dr}{2\sqrt{\frac{GM}{r}-\frac{GM}{R}}}}.$$
However, this integral doesn't look as if it's going to give the right answer.  The $\pi$ in the given expression for $C$ seems to suggest that we're going to get an integral involving a $\sin$ substitution.  
So - what does 'integrate the equation of motion' mean?  Integrate which equation?  And with respect to what?
 A: Regarding the question in the title:  'to integrate the Equation of Motion', or in general, to 'integrate a differential equation' just means to solve it. The expression alludes to the fact that you want to find $r(t)$, the position of the particle as a function of time ("the motion"), but originally you have an equation that relates the motion and its derivatives; so, to "integrate the equation" means, roughly, to get rid of the derivatives ($\dot{r}$, $\ddot{r}$, ...) and find explicitly $r(t)$
A: Multiply both sides of your equation of motion for $\ddot{r}$ by $2 \dot{r}$ then use you knowledge of the chain rule on the right hand side. You will end up integrating a square root but fear not you will be heading in the right direction.
A: As you said, the equation of motion is
$$
\ddot{r}=-\frac{GM}{r^2},
$$
and the total energy $E=\dot{r}^2/2 - GM/r$ is conserved (and equal to $-GM/R$), letting you write
$$
\dot{r}=-\sqrt{2GM}\sqrt{\frac{1}{r}-\frac{1}{R}}
$$
directly; or, in terms of $u=r/R$,
$$
\dot{u}=-\sqrt{\frac{2GM}{R^3}}\sqrt{\frac{1-u}{u}}.
$$
So
$$
T=\sqrt{\frac{R^3}{GM}}\frac{1}{\sqrt{2}}\int_{0}^{1}du\sqrt{\frac{u}{1-u}}.
$$
The integral can be looked up, or you can use the trig substitution $u=\sin^2\theta$ to yield
$$
C =\frac{1}{\sqrt{2}}\int_{0}^{1}du\sqrt{\frac{u}{1-u}}=\frac{1}{\sqrt{2}}\int_{0}^{\pi/2}2\sin^2\theta d\theta=\frac{\pi}{2\sqrt{2}}.
$$
