Change of variable in $\int_{r_{0}}^{r_{1}}\frac{dr}{r(1-r^{2})}=\int_{0}^{2\pi}dt $

Question

In Strogatz's Book Nonlinear Dynamics and Chaos the example 8.7.1 we have the vector field $\dot{r}=r(1-r^2)$ , $\dot{\theta}=1$ given in polar coordinates. Let $r_0$ and $r_1$ points in the positive real axis. We know that after a time of flight $t=2\pi$ the system completes a return to the x-axis going from $r_0$ to $r_1$. Then $r_1$ satisfies $$\int_{r_{0}}^{r_{1}}\frac{dr}{r(1-r^{2})}=\int_{0}^{2\pi}dt=2\pi.$$ I suppose that $\int_{r_{0}}^{r_{1}}\frac{dr}{r(1-r^{2})}=\int_{r_{0}}^{r_{1}}\frac{dt}{dr}dr$ then we can use integration by sustitution: $$\int_{\varphi(a)}^{\varphi(b)}f(x)dx=\int_{a}^{b}f(\varphi(t))\varphi'(t)dt.$$But i'm not quite sure neither how he choose $\varphi$ for this case nor how he change $r_0$ to $0$ and $r_1$ to $2\pi$. Thanks in advance.

Answer 1

Hint: If you have $r(t)$ such that $r(0) = r_0$, $r(2\pi) = r_1$, you can for sure integrate this thing as: $$ \int_{0}^{2\pi} \frac{\dot{r}(t)}{r(t)(1-r(t)^2)} \, dt = \int_{r_0}^{r_1} \frac{dr}{r(1-r^2)}. $$ So, in terms of substitution formula it means that $\varphi(t) = r(t)$.