The motion of a pendulum is described by the differential equation
$$ \ddot\theta +\frac gl \sin \theta = 0$$
if we integrate this equation with respect to $\theta$ we obtain
$$ \frac 12 \dot \theta ^2 – \frac gl \cos \theta = C $$
Would anyone please shed some light on how to integrate the first term? It seems that:
$$\int \ddot \theta\,d\theta = \frac 12 \dot \theta ^2$$
Or in other words
$$\int{\frac{d^2\theta}{dt^2}}\,d\theta =\frac{1}{2}\left( \frac{d\theta}{dt} \right) ^2$$
I don't really buy it
Best Answer
There is a tidy trick for that using chain-rule. Remember this once and for all. We have
$$\ddot \theta (t) + {g \over l}\sin \left( {\theta \left( t \right)} \right) = 0$$
where it is a nonlinear second order differential equation. Wow, it seems scary a little as we don't have linearty. This is how we tackle this down
$$\ddot \theta (t) = {{{d^2}\theta } \over {d{t^2}}} = {d \over {dt}}\left( {{{d\theta } \over {dt}}} \right) = {d \over {d\theta }}\left( {{{d\theta } \over {dt}}} \right){{d\theta } \over {dt}} = \dot \theta {d \over {d\theta }}\left( {\dot \theta } \right) = {d \over {d\theta }}\left( {{1 \over 2}{{\dot \theta }^2}} \right)$$
then put this into the equation and integrate with respect to $\theta $.
I just want to say one more thing. When you use the work-energy theorem, you directly obtain the integrated form you wanted. Do you know why this happens? It's because the work-energy theorem is nothing more than integrating the second newton law. If you dig into the proof of work-energy theorem for a particle, you may understand what I mean.