# How do I rewrite this integrals properly?

For a physics project I have to calculate the curvature of a certain particle which deflection dependent on it's current position. I've got three functions:

• $$\theta{(t)}$$ which is the deflection at moment $$t$$;
• $$\phi{(t)}$$ would be $$\phi{(t)} = \alpha - \int_{dt}^{t} \theta{(p)}dp$$ which is the total angle of deflection at moment $$t$$ in which case $$\alpha$$ is the starting angle of deflection;
• $$$$P{(t)}\begin{cases} x{(t)}=v \int_{0}^{t-dt} \cos{(\phi{(p)})} dp\\ y{(t)}=v \int_{0}^{t-dt} \sin{(\phi{(p)})} dp \end{cases}$$$$ which is the formula for the position $$P$$ of the particle at moment $$t$$ and in which $$v$$ is constant. In addition, $$P_{t=0}(0,0)$$.

You probably already see what's wrong with this last formula. For example, a better way to write $$x{(t)}$$ would be $$x{(t)}=v \int_{0}^{t} \cos{(\phi{(p)})} dp$$, which we can do, as a surface with a width of $$0$$ would be $$0$$. However, this is not an option, as $$\phi{(t)}$$ depends on $$\theta{(t)}$$ and $$\theta{(t)}$$ depends on $$x{(t)}$$.

My question basically is, is there any way to rewrite this integral without having to make it an approximation?

EDT: I could clarify this using Riemann-sums: if we denote $$\phi{(t)}$$ as a Riemann-sum, for example, we would get $$\phi{(t)}=\lim_{d \to 0} \sum_{k=0}^{\frac{t-d}{d}} d \cdot \phi{(d + k \cdot d)}$$

• What is $dp$? an infinitesimal length on the particle path or an infinitesimal time interval? – James Arathoon Feb 5 at 1:43
• @JamesArathoon An infinitesmal time interval. – SuperSjoerdie Feb 5 at 8:14

I managed to squeeze some mathematical beauty out of this with a fairly simple trick: if we first calculate the derivative of $$x{(t)}$$ as $$\frac{d}{dt} x{(t)} = \frac{d}{dt} v (\cos{( \Phi{( t-dt)})} - \cos{(\Phi{(0)})}) = v \cos{( \phi{( t-dt)})}$$ as $$\phi{(0)}$$ is not dependent on $$t$$. If we now calculate the antiderivative we get $$x{(t)}= v \int \cos{( \phi{( t-dt)})} dt$$ which is better than what we had as original. Of course it’s clearly still recursive, it’s just “more correct” now.