# Show equivalent parameterizations of line integral

Suppose that a curve $$\gamma$$ is parametrically defined by two continuously differentiable functions $$f(t),$$ $$a \le t \le b$$ and $$g(u),$$ $$\alpha \le u \le \beta$$.
These functions are called equivalent parameterizations of $$\gamma$$ if there is a continuously differential function $$\phi$$ such that:

1. $$a = \phi(\alpha$$) and $$b = \phi(\beta)$$.
2. $$f(\phi(u)) = g(u),$$ $$\alpha \le u \le \beta$$
3. $$\phi'(u) > 0,$$ $$\alpha < u <\beta$$.

Show that

$$f(t) = (cos(t), sin(t)), 0 \le t \le \frac{\pi}{2}$$ and $$g(u) = \left(\frac{1-u^2}{1+u^2}, \frac{2u}{1+u^2} \right), 0 \le u \le 1$$

are equivalent parameterizations of a quarter-circle

How do I approach this? I'm unfamiliar with equivalent paramterizations and my book touches on it once with an example which isn't helpful.

I thought of placing $$\cos(t)$$ in replacement for $$u$$ in $$\frac{1-u^2}{1+u^2}$$ and $$\sin(t)$$ for $$\frac{2u}{1+u^2}$$, then taking the derivative of $$g(u)$$, though I'm unsure how to proceed with setting up the integrals as I'm given two bounds?

I read the problem and quickly the solution $$\phi(u)=2\arctan(u)$$ came to mind (I'll justify this intuition a little at the end$$^*$$). You can prove this works by means of the double angle formula.
To find a solution on a more systematic manner, assume that there is a function such that $$f\circ\phi=g$$. That is $$\begin{bmatrix}\cos(\phi(u))\\ \sin(\phi(u))\end{bmatrix}=\begin{bmatrix}\frac{1-u^2}{1+u^2}\\ \frac{2u}{1+u^2} \end{bmatrix}$$ Which leads to the system of equations $$\begin{cases}\phi(u)=\arccos\left(\frac{1-u^2}{1+u^2}\right)\\ \phi(u)=\arcsin\left(\frac{2u}{1+u^2}\right)\end{cases}$$
So now you have two candidates for a solution. Just show that either of these satisfies. (Btw, we have that $$\forall u\in [0,1],\;\arccos\left(\frac{1-u^2}{1+u^2}\right)=\arcsin\left(\frac{2u}{1+u^2}\right)$$).
$$^*$$ $$f$$ has trigonometric components and $$g$$ has rational components, a good guess is that $$\phi$$ will be similar to the inverse of a trig function. $$\phi=\arctan$$ seemed like a good candidate since both $$\sin(\arctan(u))$$ and $$\cos(\arctan(u))$$ have $$\sqrt{1+u^2}$$ on the denominator. I just felt like adding a $$2$$ before $$\arctan$$ so that $$\phi(1)=2\arctan(1)=\frac{\pi}{2}$$ (which was required) and all the other conditions sort of just worked.