Alternative method to solve $x''+x=0$ 
The question is how to solve the equation $x''+x=0$, using the hint to consider $v=dx/dt$ and then to use the chain rule to get $x'' = v\cdot dv/dx$.

My thoughts: This is an equation where the independent variable t does not appear explicitly.
Following the hint and substituting in the main equation I got,
$$v^2 = - (x^2) +C$$
Now how do I solve this equation to get $x= c_1 \sin t + c_2 \cos t$, which I already know, from the more widely used method, is the answer?
 A: You found $v^2=-(x^2)+C$, which says
$$\frac{dx}{dt}=\pm\sqrt{C-x^2}\,.$$
This equation is separable. I'll separate it below, but first I wanted to note that in mechanics (physics) we often encounter second-order equations with no explicit time dependence, and in this case the substitution $v=dx/dt$ is often helpful. It typically yields a first-order equation that can be interpreted as conservation of energy (in this case, $v^2+x^2=C$, as if we had a mass on a spring). To find the time dependence, we typically separate the equation and integrate, as I'll do next. For the particular problem at hand, the method isn't as fast as the approach in the comments; but it's routine.
$$\frac{dx}{\sqrt{C-x^2}}=\pm dt$$
$$\int{\frac{dx}{\sqrt{C-x^2}}}=\pm \int{dt}$$
$$\tan^{-1}\frac{x}{\sqrt{C-x^2}}=\pm t + K$$
$$\frac{x}{\sqrt{C-x^2}}=\tan{(\pm t + K)}\,.$$
Square both sides, rearrange, use the identity $1+\tan^2 = \frac{1}{\cos^2}$, and you get
$$x=\pm C\sin(\pm t + K)\,.$$
All of the $\pm$ possibilities can be captured by some choice of the constants, so we can just write the general solution as
$$x = c\sin(t + k)\,.$$
(For example, to recover $-C\sin(-t+K)$, just take $c = C$ and $k = -K$.)
The answer $x = c\sin(t + k)$ is fine---it has an amplitude and a phase shift---but if you want it as a sum of sine and cosine, use the addition formula for sine to write
$$x=c\sin t\cos k + c\cos t\sin k$$
or 
$$x = c_1\sin t + c_2\cos t$$
where $c_1 = c\cos k$ and $c_2 = c\sin k$.
