integrating both side with different function on each side

Question

I have this equation in a solution. $\theta$ is a function of $t$ $$\ddot\theta=\sin{\theta}$$ and integrate both side to get $$\dot\theta = t\sin{\theta}+C$$ I am not very sure about what happened on the righthand side or is it even correct. I have a vague feeling that it looks like (reverse)chain rule, but I can't write it down to convince myself. Can someone explain it a bit? If it's not correct, I get this, is it right? $$-\frac{1}{\dot\theta}\cos\theta$$

Answer 1

Your integration is incorrect. This is a differential equation, and its solution does not yield to that exact procedure. You can do this: \begin{align*} \ddot{\theta}&=\sin(\theta) \\ \ddot{\theta}\dot{\theta}&=\sin(\theta)\,\dot{\theta} \qquad\text{multiply by }\dot{\theta}\\ \int\dot{\theta}\,\ddot{\theta}\,dt&=\int\sin(\theta)\,\dot{\theta}\,dt \\ \frac{\dot{\theta}^2}{2}&=-\cos(\theta)+C \qquad u \text{ substitution on both sides} \end{align*} That's one integration. The second integration can be written in terms of elliptic integrals: \begin{align*} \dot{\theta}^2&=C-2\cos(\theta) \qquad\text{absorb }2\text{ into constant.} \\ \frac{d\theta}{dt}&=\sqrt{C-2\cos(\theta)} \\ \frac{d\theta}{\sqrt{C-2\cos(\theta)}}&=dt \\ \int\frac{d\theta}{\sqrt{C-2\cos(\theta)}}&=\int dt=t_{f}-t_{i}. \\ \end{align*} And that's about as far as you can go symbolically.

Answer 2

Starting from Adrian Keister's answer, the solution of $$\frac{d\theta}{dt}=\sqrt{C-2\cos(\theta)}$$ is given by $$\theta=2 \text{am}\left(\frac{1}{2} \sqrt{C-2} \left(t+c_1\right)|-\frac{4}{C-2}\right)$$ where appears the amplitude for Jacobi elliptic functions.