Why is this the result of this integral? This problem has to do with the non-approximated solution of the motion of a simple pendulum. I'm asking here instead of at the physics forum because I have a mathematical question.
Anyways, the differential equation for a simple pendulum is:
$$\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}+\omega_0^2\sin\theta=0$$
Multiplying both sides by $\frac{\mathrm{d}\theta}{\mathrm{d}t}$ and integrating over $t$:
$$\int\left( \frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}\right)\mathrm{d}\theta=\omega_0^2\cos\theta+C$$
The next step simply states that, therefore:
$$\frac{1}{2}\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2-\omega_0^2\cos\theta=K$$
My problem is with the first term in the left hand side. Why is that the result of the integral? It implies that it's integrating something like:
$$\int\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)\mathrm{d}\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)=\frac{1}{2}\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 + C$$
But I fail to see how that makes any sense with the given steps. Am I missing something obvious here?
 A: $$\int\left( \frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}\right)\mathrm{d}\theta=\omega_0^2\cos\theta+C$$
$$\int  \dfrac {d  }{dt}\left( \frac{\mathrm{d}\theta}{\mathrm{d}t}\right)\mathrm{d}\theta=\omega_0^2\cos\theta+C$$
$$\int  d\left( \frac{\mathrm{d}\theta}{\mathrm{d}t}\right)\dfrac {d\theta}{dt}=\omega_0^2\cos\theta+C$$
$$\dfrac 12  \left( \frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2=\omega_0^2\cos\theta+C$$
A: More generally if $\ddot{\theta}=-V'(\theta)$ then $\dot{\theta}\ddot{\theta}=-V'(\theta)\dot{\theta}$ so, applying $\int dt$, $\frac12\dot{\theta}^2=-V(\theta)+C$ with $C$ an integration constant. All the source you read did was write $\int\dot{\theta}\ddot{\theta}dt$ as $\int\ddot{\theta}d\theta$.
Physically, $C=\frac12\dot{\theta}^2+V(\theta)$ is a conserved energy, with $\frac12\dot{\theta}^2$ ($V$) kinetic (potential).
A: Notice that you can make the problem simpler if solving $$\theta''+\omega^2\sin(\theta)=0$$ you start switching the variables. This gives
$$-\frac{t''}{[t']^3}+\omega^2\sin(\theta)=0$$ Using the reduction of order $p=t'$
$$\frac {p'}{p^3}=\omega^2\sin(\theta)\implies t'=p=\pm\frac 1{ \sqrt{c_1+2 \omega ^2 \cos (\theta )}}$$
A: Here is a more rigorous method. Consider $$\ddot{\theta}=-\omega_0^2\sin(\theta),$$ and if you multiply by $2\dot{\theta},$ you get $$2\dot{\theta}\ddot{\theta}=-2\omega_0^2\sin(\theta)\dot{\theta}.$$ By the chain rule, $$2\dot{\theta}\ddot{\theta}=\frac{\mathrm{d}}{\mathrm{d}t}\dot{\theta}^2,$$ and $$\sin(\theta)\dot{\theta}=\frac{\mathrm{d}}{\mathrm{d}t}[-\cos(\theta)].$$ Therefore, $$\frac{\mathrm{d}}{\mathrm{d}t}\dot{\theta}^2=2\omega_0^2\frac{\mathrm{d}}{\mathrm{d}t}\cos(\theta),$$ and thus, $$\dot{\theta}=2\omega_0^2\cos(\theta)+K,$$ which is the result desired.
