Proof that SHM is sinusodial? If we have an object attached to a spring, and the net force on that object is $-kx,$ how do we prove that its motion (if you move the object to  $x\ne 0$) is sinusoidal? I know that you must integrate from $ma=-kx,$ but I don't see how (because you must integrate with respect to time).
 A: $ma=-kx$ is a differential equation that is best solved by inspection. That means, knowing that $a=\ddot x$, you think through your calc 2 and try to remember a function whose second derivative is the original function with a negative sign. "That's it!", you exclaim. "Sin and cos fit that description exactly!"
A: Here is a rigorous derivation of the sinusoidal solution to your mass-spring system. I'm writing this from memory so if someone could take a second look at it that would be great.
The simple harmonic oscillator model for an ideal spring-mass system is one of the most straightforward examples of a differential equation. The net force on a mass attached to a spring, in the simplest picture, is equal to the restoring force giving by Hooke's Law:
$F_{net} = F_s$
$ma = -kx$
$m \frac{d^2 x}{dt^2} + kx = 0$
We have used Newton's second law of motion to replace the net force with an expression containing our dependent variable $x$, the displacement from equilibrium. Because this third equation contains a function and one of its derivatives, it is classified as a differential equation. The solution to a differential equation is not some number as in algebraic equations, but rather a function (in this case, $x(t)$). To be specific, we call this differential equation ordinary because there is only one independent variable (t), second-order because highest derivative is second, linear because the function and its derivative(s) are not variables inside some nonlinear function such as squared or trig, homogeneous because the constant term is zero, and autonomous because the coefficients of the function and its derivatives are constants (rather than functions of t). I hope you can appreciate that this is among the simplest of all differential equations (first order linear homogeneous autonomous ordinary would be the only simpler DE).
For any second-order linear autonomous homogeneous ordinary differential equation (SOLHA ODE) like yours,
$ax'' + bx' + cx = 0$,
We might guess the solution (the function $x(t)$ actually is) might be of the form 
$x(t) = Ce^{rt}$
where $r$ and $C$ are some constants to be determined. We choose the form $e^t$ because it has convenient differentiation patterns, i.e. its derivatives are itself. So if you take the first derivative and second derivative and plug all three functions into our differential equation, you will find 
$a(r^2Ce^{rt}) + b(rCe^{rt}) + c(Ce^{rt}) = 0$
which simplifies to
$ar^2 + br + c = 0$
This is known as the characteristic equation of your differential equation. It allows us to solve for the unknown constant $r$ in our solution for $x(t)$.
Returning to your SHO, the characteristic equation is
$mr^2 + k = 0$
which we can solve by the quadratic formula:
$r = \pm \sqrt{k/m} \times i = \pm \omega i$
where we have defined a new constant
$\omega \equiv \sqrt{k/m}$
which turns out to be the radial/ angular/ natural frequency of our oscillator. And yes, that is the "imaginary" unit $i \equiv \sqrt{-1}$.
For SOLHA ODEs in general, the pair of roots $r$ of the characteristic equation can be real and different, real but identical, or complex (containing an "imaginary" term) conjugates of the form $r = \alpha \pm \beta i$. (The type of roots depends on the discriminant $b^2 - 4ac$ in the quadratic equation.) As I hope you've guessed, the roots of our oscillator are the latter case, being complex due to the imaginary unit $i$.
The solutions for $x$ take visibly different forms according to which of the three types of roots we have from the characteristic equation, but all solutions are based on the exponential solution we guessed earlier. The general solution to any SOLHA ODE with a pair of complex roots like ours, is
$x(t) = C_1e^{r_1 t} + C_2e^{r_2 t} $
$= C_1e^{(\alpha + \beta i)t} + C_2e^{(\alpha - \beta i)t}$
Here $C_1$ and $C_2$ are constants that depend on the initial conditions of the spring-mass system (specifically, two conditions: the initial position and the initial velocity). If the initial conditions for a specific system are known, we can solve for these constants (the task of solving for the constants is common straightforward algebraic task known as an initial value problem). We can rework our above solution using Euler's Formula (one of the most fascinating relationships in mathematics, I encourage you to look into it): 
$e^{ix} = cos(x) + i\times sin(x)$
So thanks to this formula, after a few moves our general solution becomes 
$x(t) = C_1e^{\alpha t}cos(\beta t) + C_2e^{\alpha t}sin(\beta t)$
This is the general solution to a SOLHA ODE when the pair of roots of the characteristic equation is complex. Recall:
In general: $r = \alpha \pm \beta i$
For our SHO: $r = 0 \pm \sqrt{k/m}\times i = \pm \omega i$
So the general solution of $x$ becomes for us
$x(t) = C_1e^{(0)t}cos(\omega t) + C_2e^{(0)t}sin(\omega t) 
=C_1 cos(\omega t) + C_2 sin(\omega t)$
And there you have it! A derivation of the sinusoidal solution to your simple harmonic oscillator model. As you can see, the periodic trig functions came from Euler's Formula, which extracted them from the exponential function, which we used to solve the differential equation, which was derived from simple equating the Newton's second law force to the Hooke's Law restoring force. 
I hope now you have a deeper understanding of this model, and I hope you are relieved to see that our forefathers haven't simply plucked these expressions out of thin air.
If it is an anharmonic, driven, or damped solution you're after, you might consider learning a bit of differential equations course material. Although I suppose if you give enough details, perhaps someone here would be willing to derive the solution for you.
